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1

Welcome!

Welcome to CRUX MATHEMATICORUM with MATHEMATICAL MAYHEM. We hope that those of you new to any part of our journal will enjoy all the material. We see this merger as bene cial to all our readers. We now o er a broader range of problems. High School students and teachers will nd more material at that level. Please note that we will accept solutions to problems in the MAYHEM section only from students. These are the people that we wish to stimulate. When they nd they are successful with the MAYHEM and SKOLIAD problems, we hope they will then nd a further challenge with the problems in the OLYMPIAD corner, and, eventually, with the PROBLEMS section. Good luck, and we look forward to hearing from you. You will note that the majority of the members of the Crux Editorial Board are continuing in their previous jobs. Sadly, Colin Bartholomew has taken early retirement from Memorial University, and I shall miss him. However, we are pleased to welcome Clayton Halfyard as Associate Editor. There will be a short pro le of him elsewhere in this issue. We are also delighted to welcome Naoki Sato as Mayhem Editor and Cyrus Hsia as Assistant Mayhem Editor. These two Canadian IMO medallists bring with them a wealth of experience of problem solving, and many years of experience of writing for MAYHEM. There are some minor administrative changes: the table of contents (which is larger) is now on the back outside cover; and the ordering of the various sections has been changed. There is no change in the quantity of material that has been recently published in CRUX. Whereas previously MAYHEM appeared ve times per year, it now appears eight times per year. The annual quantity of material will be approximately the same. Here are some details that will help you in cross-referencing previous material: 1. Please note that the volume number is consecutive for CRUX. 2. Material occurring in previous volumes of CRUX and material occuring in CRUX with MAYHEM will be referenced as [year: page no]. For example, the last page of the last issue of CRUX is [1996: 384], and the rst page of CRUX with MAYHEM is [1997: 1]. 3. Material occurring in previous volumes of MAYHEM will be referenced as [MAYHEM volume: issue, page no: year]. For example, [MAYHEM 8: 5, 28: 1996] is the last page of the last issue of MAYHEM. Bruce Shawyer Editor-in-Chief

2

Bienvenue!

Bienvenue au CRUX MATHEMATICORUM with MATHEMATICAL MAYHEM. Nous esperons que ceux qui ne connaissent pas l'une des nouvelles parties de notre journal en apprecieront tout le contenu. A notre avis, la fusion de ces deux magazines sera tout a l'avantage de nos lecteurs. La gamme des problemes publiesy est plus vaste qu'auparavant, et les e leves et les professeurs du secondaire y trouveront davantage de problemes a leur niveau. Nous vous signalons au passage que nous accepterons uniquement les solutions aux problemes de la section MAYHEM provenant des e tudiants; apres tout, c'est eux que nous voulons stimuler! Lorsqu'ils pourront reussir les problemes des sections MAYHEM et SKOLIAD, nous esperons qu'ils s'attaqueront a ceux de la partie OLYMPIAD, et m^eme, un peu plus tard, a ceux de la section PROBLEMS. Bonne chance et au plaisir de recevoir de vos nouvelles! Vous aurez remarque que la majorite des membres de l'equipe e ditoriale du Crux sont demeures en poste. Malheureusement, Colin Bartholomew a pris une retraite anticipee de l'Universite Memorial; il me manquera. Nous sommes toutefois heureux d'accueillir Clayton Halfyard au poste de redacteur adjoint. Vous trouverez son pro l ailleurs dans ce numero. Nous sommes e galement enchantes d'avoir parmi nous Naoki Sato et Cyrus Hsia, qui occupent respectivement les postes de redacteur et de redacteur adjoint du Mayhem. Ces deux medailles canadiens de l'OIM apportent avec eux tout un bagage d'expertise en resolution de problemes, et des annees d'experience a la redaction du MAYHEM. Vous constaterez en outre quelques petits changements d'ordre administratif : la table des matieres (elargie) est desormais en quatrieme de couverture et les sections ne sont plus dans le m^eme ordre. Le volume d'information du CRUX sera comparable a la quantite d'information qu'il contenait avant la fusion. Le MAYHEM, qui paraissait cinq fois l'an, sera publie huit fois par annee; ainsi, le volume annuel de contenu demeurera approximativement le m^eme. Voici quelques details sur la facon de noter les renvois : 1. Les numeros de volume du CRUX continuent dans la m^eme sequence. 2. On notera ainsi les renvois a des volumes pre-fusion du CRUX et aux volumes du CRUX with MAYHEM : [anne: no de page]. Par exemple, la derniere page du dernier numero du CRUX sera [1996: 384], et la premiere page du CRUX with MAYHEM sera [1997: 1]. 3. On notera ainsi les renvois a des volumes pre-fusion du MAYHEM : [volume du MAYHEM: no, no de page: annee]. Par exemple, [MAYHEM 8: 5, 28: 1996] est la derniere page du dernier numero du MAYHEM. Le redacteur en chef Bruce Shawyer

3

THE ACADEMY CORNER No. 8 Bruce Shawyer

All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 In this issue, courtesy of Waldemar Pompe, student, University of Warsaw, Poland, we print an international contest paper for university students. Please send me your nice solutions.

INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS July 31 { August 1996, Plovdiv, Bulgaria First day | August 2, 1996 1. Let for j = 0; 1; : : : ; n, aj = a + jd, where a , d are xed real numbers. Put 0

0

0a a a ::: BB a a a ::: A=B a a a B@ : : : : : : : : : :: :: :: an an, an, : : : Calculate det A | the determinant of A. 0

1

2

1

0

1

2

1

0

1

2. Evaluate the integral

2

an an, an, ::: a

1 2

1 CC CC : A

0

Z

sin nx dx ; , (1 + 2x ) sin x

where n is a natural number. 3. A linear operator A on a vector space V is called an involution if A = E, where E is the identity operator on V . Let dim V = n < 1. (i) Prove that for every involution A on V there exists a basis of V consisting of eigenvectors of A. (ii) Find the maximal number of distinct pairwise commuting involutions on V . 2

4 nP , 4. Let a = 1, an = n akan,k for n  2. k Show that (i) limsup jan j =n < 2, = ; n!1 (ii) limsup jan j =n  2=3. n!1 5. (i) Let a, b be real numbers such that b  0 and 1 + ax + bx every x 2 [0; 1]. Prove that  Z n dx = ,1=a if a < 0; lim n (1 + ax + bx ) + 1 if a  0: n!1 1

1

1

=1

1

1 2

1

2

1

 0 for

2

0

(ii) Let f : [0; 1] ! [0; 1) be a function with continuous second derivative and f 00 (x)  0 for every x 2 [0; 1]. Suppose that

Z L = nlim n (f (x))n dx !1 1

0

exists and 0 < L < +1.

Prove that f 0 has a constant sign and L =



,

0 xmin 2[0;1] jf (x)j

1

:

6. Upper content of a subset E of the plane R  R is de ned as

C(E) = inf

(X n

i=1

diam (Ei)

)

where the in mum is takenn over all nite families of sets E ; E ; : : : ; En S in R  R such that E  Ei. 1

2

i=1

Lower content of E is de ned as

K(E) = supflength (L)g

such that L is a closed line segment onto which E can be contracted. Show that (i) C (L) = length (L), if L is a closed line segment. (ii) C (E )  K(E ). (iii) equality in (ii) need not hold even if E is compact. Hint: If E = T [ T 0 where T is the triangle with vertices (,2; 2), (2; 2) and (0; 4), and T 0 is its re ection about the x-axis, then C (E ) = 8 > K(E).

5 Remarks: All distances used in this problem are Euclidean. Diameter of a set E is diam (E ) = supfdist (x; y ) j x; y 2 E g. Contraction of a set E to a set F is a mapping f : E ! F such that dist (f (x);f (y ))  dist (x; y ) for all x; y 2 E . A set E can be contracted onto a set F if there is a contraction f of E to F which is onto, that is such that f (E ) = F . Triangle is de ned as the union of the three line segments joining its vertices, so that it does not contain the interior.

Problems 1 and 2 are worth 10 points, problems 3 and 4 are worth 15 points, problems 5 and 6 are worth 25 points. You have 5 hours. Please write the solutions on separate sheets of paper. Good luck!

6

THE OLYMPIAD CORNER No. 179 R.E. Woodrow

All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. Another year has passed, and the rst with Bruce Shawyer as Editorin-Chief. He has made the transition pleasant and easy. Special thanks go to Joanne Longworth whose TEX skills have made changed fonts and formats easy to incorporate. Thanks also go to the many contributors to the two Corners including: Miguel Amengual Covas Cyrus C. Hsia Dieter Ruo Sefket Arslanagic Murray Klamkin Toshio Seimiya Mansur Boase Derek Kisman Michael Selby Seung-Jin Bang Ted Lewis D.J. Smeenk Christopher Bradley Joseph Ling Daryl Tingley Francisco Bellot Rosado Beatriz Margolis Panos E. Tsaoussoglou Paul Colucci Stewart Metchette Ravi Vakil Hans Engelhaupt Richard Nowakowski Edward T.H. Wang Tony Gardiner Michael Nutt Hoe Teck Wee Solomon Golomb Siu Taur Pang Chris Wildhagen Gareth Grith Bob Prielipp Siming Zhan Georg Gunther Chandan Reddy Let me open with a major apology. In the November 1996 number of the corner we gave twelve of the problems proposed to the jury but not used at the 36th International Mathematical Olympiad held in Canada. Those familiar with the process of selection will know that the problems do not initiate with the host country. They come from proposers in other countries, and the responsibility of the host country selection committee is to re ne and select from these submissions the ocial list of problems proposed to the jury. Over the years I have loosely referred to the problems proposed to the jury at the Xth International Olympiad in Y as the \Y-problems" for short. However, when that became part of a longer more ocial sounding sub-title \Canadian Problems for consideration by the International Jury," I should have seen that it read as if the original proposers are from Canada, thus insulting the creators. In retrospect I do not understand why that interpretation did not jump o the page. To the many creative non-Canadians who submitted problems for possible use at the 36th IMO my sincere apologies.

7 As an Olympiad Contest this issue we give the problems of the 44th Mathematical Olympiad from Latvia. My thanks go to Richard Nowakowski, Canadian Team Leader to the IMO in Hong Kong for collecting the problems for me.

LATVIAN 44 MATHEMATICAL OLYMPIAD Final Grade, 3rd Round Riga, 1994

1. It is given that cos x = cos y and sin x = , sin y. Prove that sin 1994x + sin 1994y = 0.

2. The plane is divided into unit squares in the standard way. Consider a pentagon with all its vertices at grid points. (a) Prove that its area is not less than 3=2. (b) Prove that its area is not less than 5=2, if it is given that the pentagon is convex. 3. It is given that a > 0, b > 0, c > 0, a + b + c = abc. Prove that at least one of the numbers a, b, c exceeds 17=10. 4. Solve the equation 1!+2!+3!+    + n! = m in natural numbers. 5. There are 1994 employees in the oce. Each of them knows 1600 of the others. Prove that we can nd 6 employees, each of them knowing all 5 others. 1st SELECTION ROUND 1. It is given that x and y are positive integers and 3x + x = 4y + y. Prove that x , y , 3x + 3y + 1 and 4x + 4y + 1 are squares of integers. 2. Is it possible to nd 2 di erent pairs of natural numbers (ai; bi) such that the following 2 properties hold simultaneously: 1 + 1 +  + 1 (1) ab ab a 1994 b 1994 = 1, (2) (a + a +    + a 1994 ) + (b + b +    + b 1994 ) = 3 ? 3. A circle with unit radius is given. A system of line segments is called a cover i each line with a common point with the circle also has some common point with some of the segments of the system. (a) Prove that the sum of the lengths of the segments of a cover is more than 3, (b) Does there exist a cover with this sum less than 5? 4. A natural number is written on the blackboard. Two players move alternatively. The rst player's move consists of replacing the number n on the blackboard by n=2, by n=4 or by 3n ( rst two choices are allowed only if they are natural numbers). The second player's move consists of replacing the number n on the blackboard by n + 1 or by n , 1. The rst player wants the number 3 to appear on the blackboard (no matter who writes it down). Can he always achieve his aim? 3

2

2

1994

1

1

1

2

2

2

2

2

2

1

2

2

1995

8

5. Three equal circles intersect at the point O and also two by two at the points A, B , C . Let T be the triangle whose sides are common tangents of the circles; T contains all the circles inside itself. Prove that the area of T is not less than 9 times the area of ABC . 2nd SELECTION ROUND 1. It is given that 0  xi  1, i = 1; 2; : : : ; n. Find the maximum of the expression x x xn + +    + x x : : : xn + 1 x x x : : : xn + 1 x x : : : xn, + 1 : 2

1

2

3

1

3

4

1

2

1

2. There are 2n points on the circle dividing it into 2n equal arcs. We must draw n chords having these points as endpoints so that the lengths of all chords are di erent. Is it possible if: (a) n = 24, (b) n = 1994? 3. A triangle ABC is given. From the vertex B, n rays are constructed intersecting the side AC . For each of the n +1 triangles obtained, an incircle with radius ri and excircle (which touches the side AC ) with radius Ri is constructed. Prove that the expression r r : : : rn R R : : : Rn 1

2

1

2

+1 +1

depends on neither n nor on which rays are constructed. 3rd SELECTION ROUND 1. A square is divided into n cells. Into some cells \1" or \2" is written so that there is exactly one \1" and exactly one \2" in each row and in each column. We are allowed to interchange two rows or two columns; this is called a move. Prove that there is a sequence of moves such that after performing it \1"-s and \2"-s have interchanged their positions. 2. Let aij be integers, jaij j < 100. We know that the equation 2

a x + a y + a z + a xy + a xz + a yz = 0 has a solution (1234; 3456; 5678). Prove that this equation also has a solution with x, y , z pairwise relatively prime which is not proportional to the 11

2

22

2

33

2

12

13

23

given one. 3. Let ABCD be an inscribed quadrilateral. Its diagonals intersect at O. Let the midpoints of AB and CD be U and V . Prove that the lines through O, U and V , perpendicular to AD, BD and AC respectively, are concurrent.

9 Next we give ve Klamkin Quickies. My thanks go to Murray Klamkin, the University of Alberta, for supplying them to us. Next issue we will give the \quick" solutions along with another ve of his special teasers.

FIVE KLAMKIN QUICKIES October 21, 1996

1. For x; y;z> 0, prove thatx 1

1

(i) 1 + (x + 1)  1 + x(x + 2) , (ii) [(x + y )(x + z )]x [(y + z )(y + x)]y [(z + x)(z + y )]z  [4xy ]x [4yz ]y [4zx]z . 2. If ABCD is a quadrilateral inscribed in a circle, prove that the four lines joining each vertex to the nine point centre of the triangle formed by the other three vertices are concurrent. 3. How many six digit perfect squares are there each having the property that if each digit is increased by one, the resulting number is also a perfect square? 4. Let ViWi, i = 1; 2; 3; 4, denote four cevians of a tetrahedron V V V V which are concurrent at an interior point P of the tetrahedron. Prove that PW + PW + PW + PW  max ViWi  longest edge: 5. Determine the radius r of a circle inscribed in a given quadrilateral if the lengths of successive tangents from the vertices of the quadrilateral to the circle are a; a; b; b; c; c; d; d, respectively. 1

2

3

4

1

2

3

4

We now turn to solutions from the readers to problems posed in the May 1995 number of the corner on the Sixth Irish Mathematical Olympiad, May 8, 1993 [1995: 151{152].

SIXTH IRISH MATHEMATICAL OLYMPIAD May 8, 1993 | First Paper (Time: 3 hours)

1. The real numbers , satisfy the equations

, 3 + 5 , 17 = 0; Find + . 3

2

, 3 + 5 + 11 = 0: 3

2

 Solutions by Sefket Arslanagic, Berlin, Germany; by Beatriz Margolis, Paris, France; by Vedula N. Murty, Andhra University, Visakhapatnam, India; by D.J. Smeenk, Zaltbommel, the Netherlands; by Panos E. Tsaoussoglou, Athens, Greece; and comment by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Margolis's solution.

10 De ne f (x) = x , 3x + 5x. We show that if f ( ) + f ( ) = 6, then + = 2. Since f (x) = (x , 1) + 2(x , 1) + 3, we have 3

2

3

f ( ) , 3 = ( , 1) + 2( , 1) f ( ) , 3 = ( , 1) + 2( , 1) 3

3

Adding gives

0 = ( , 1) + ( , 1) + 2( + , 2) = ( + , 2)[( , 1) + ( , 1)( , 1) + ( , 1) + 2] 3

3

2

2

and, since the second factor is positive, we obtain the result. (See Olympiad Corner 142 and its solution.) [Wang's Comment:] The problem is strikingly similar to problem 11.2 of the XXV Soviet Mathematical Olympiad, 11th Form [1993: 37]. The method used by Bradley given in the published solution [1994: 99] works for the present problem as well and, in fact, yields the same answer: + = 2. 3. The line l is tangent to the circle S at the point A; B and C are points on l on opposite sides of A and the other tangents from B , C to S intersect at a point P . If B , C vary along l in such a way that the product jABj  jAC j is constant, nd the locus of P . Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; and by D.J. Smeenk, Zaltbommel, the Netherlands. We use Smeenk's solution.

P

t

t D

r

p B

=2 =2

p

I r

r

E S

q

=2

=2

q C H A Let S be the incircle s(I; 2) of 4BCP . We denote \PBA = , \PCA = AB = pAC = q

11 with pq = k , a constant. Let S touch BP and CP at D and E respectively. For 4PEI we have \EIP = ( + ). Thus 2

1 2

t = r tan 12 ( + ) = (ppq+,qr)r : The semiperimeter of 4BCP is p + q + t = p + q + (ppq+,qr)r = pqpq(p,+rq) : The area, F , of 4BCP is r pqpq(p,+rq) = 12 (p + q)PH; where PH is the altitude to BC . It follows immediately that 2k r : PH = pq2pqr = ,r k ,r So the locus of P is a line parallel to BC . 4. Let a ; a ; : : : ; an, be real numbers, where n  1, and let f (x) = xn + an, xn, +    + a be such that jf (0)j = f (1) and each root of f is real and satis es 0 < < 1. Prove that the product of the roots does not exceed 1=2n. 2

2

2

2

2

2

2

2

0

1

1

1

2

1

0

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Let f (x) = (x , )(x , ) : : : (x , n ) where the i denote Q(1 , the) =n real roots of f , i = 1 ; 2 ; : : : ; n . Then from j f (0) j = f (1) we get i Q . (All products are over i = 1; 2; : : : ; n.) Using the Arithmetic Mean{ i Geometric Mean Inequality we then get 1

2

Y  Y Y  i + (1 , i)  1 i = i(1 , i)  = 2 2n Q from which i  n follows. Equality holds if and only if i = for all i = 1; 2; : : : ; n. 2

2

2

1 2

1 2

May 8, 1993 | Second Paper (Time: 3 hours)

3. For non-negative integers n, r the binomial coecient ,nr denotes

the number of,ncombinations ,  of n objects chosen r at a time, with the convention that = 1 and nr = 0 if n < r. Prove the identity 0

1 n , r + 1r , 1 n X d d,1 = r d =1

12 for all integers n, r with 1  r  n. Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We use a combinatorial argument to establish the obviously equivalent identity      k n,r+1 r,1 X n = () d r,d r d where k = minfr;n , r + 1g. It clearly suces to demonstrate that the left hand side of () counts the number of ways of selecting r objects from n distinct objects (without replacements). Let jS j = r , 1. For each xed d = 1; 2; : : : ; k, any selection of d objects from S (SnS ) together with any selection of r , d objections from S would yield a selection of r objects from S. The total number of such selections is ,n,dr ,rr,,d. Conversely, each selection of r objects from S clearly must arise in this manner. Summing over d = 1; 2; : : : , () follows. 4. Let x be a real number with 0 < x < . Prove that, for all natural numbers n, the sum n , 1)x sin x + sin33x + sin55x +    + sin(2 2n , 1 =1

2

1

2

2

+1

1

is positive.  Solutions by Sefket Arslanagic, Berlin, Germany; and by Vedula N. Murty, Andhra University, Visakhapatnam, India. We give Murty's solution. We use mathematical induction. Let n sin(2k , 1)x X (2k , 1) : k S (x) = sin x > 0 for x 2 (0; ). Thus the proposed inequality is true for n = 1. Let Sr (x) > 0 for r = 1; 2; : : : ; n , 1. We will deduce that Sn(x) > 0 for x 2 (0;). Suppose that Sn(x )  0 ford some x 2 (0; ), and that Sn (x) attains its minimum at x = x . Hence dx [Sn (x)]x x0 = 0. That is n X 0 Sn(x ) = cos((2k , 1)x ) = 0;

Sn(x) =

=1

1

0

0

0

0

so that

=

0

k=1

n X 2 sin x Sn0 (x ) = 2 cos((2k , 1)x ) sin x 0

0

0

k=1 n X

0

= [sin(2kx ) , sin((2k , 2)x )] k = sin 2nx : 0

=1

0

0

13 Thus Sn0 (x ) = 0

nx0 = 0 implying sin2nx0 = 0. Hence x0

sin 2 2 sin

  2 3  (2 n , 1) x 2 2n ; 2n ; 2n ; : : : ; 2n : It is easily veri ed that at each of these values Sn (x ) > 0, a contradiction. Hence Sn (x) > 0 for x 2 (0; ). 0

0

Editor's Note: Both solutions used the calculus. Does anyone have a more elementary solution? We complete this number of the Corner with solutions by our readers to problems of the 1992 Dutch Mathematical Olympiad, Second Round given in the June 1995 number of the Corner [1995; 192{193].

1992 DUTCH MATHEMATICAL OLYMPIAD Second Round

September 18, 1992

1. Four dice are thrown. What is the chance that the product of the numbers equals 36? Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. There are four di erent kinds of outcomes in which the product is 36: each of f1; 1; 6; 6g and f2; 2; 3; 3g can occur in = 6 ways; f1; 4; 3; 3g can occur in = 12 ways; and f1; 2; 3; 6g can occur in 4! = 24 ways. Hence the probability that the product equals 36 is 4 = . 2. In the fraction and its decimal notation (with period of length 4) every letter represents a digit. Di erent letters denote di erent digits. The numerator and denominator are mutually prime. Determine the value of the fraction: 4! 2!

4! 2!2!

48 6

1 27

ADA = :SNELSNELSNELSNEL : : : KOK

[Note. ADA KOK is a famous Dutch swimmer. She won gold in the 1968 Olympic Games in Mexico. SNEL is Dutch for FAST.] Solution by the Editor. ADA = :SNEL. Then 10 x = SNEL:SNELSNEL : : : Let x = KOK and 10 x , x = SNEL. So 4

4

SNEL SNEL SNEL x = SNEL 9999 = 11  909 = 33  303 = 99  101 :

Taking KOK = 909 we obtain SNEL = 11  ADA, and A = L, which is impossible. Taking KOK = 303, we obtain SNEL = 33  ADA, so 3A < 10, as the product has four digits and 3A = L. Because S 6= L 6= 0, 3  3A < 9,

14 giving A = 1 or A = 2. Now A = 1 gives L = 3 = K , which is impossible, so A = 2. This gives L = 6, and D  2, so there is a carry. This gives D  4, as A = 2, K = 3. ADA = :SNEL is For D = 4, KOK = :7986, a solution. For D = 5, ADA = 252 is not coprime to KOK = 303. For D = 6, SNEL = 8646 and N = L. For D = 7, SNEL = 8976 and D = E . For D = 8, SNEL = 9036 and O = N . For D = 9, SNEL = 9636 and N = L. Taking KOK = 101 gives SNEL = 99  ADA forcing A = 1 for SNEL to have four digits, but then A = K . Thus the only solution is = :7986. 3. The vertices of six squares coincide in such a way that they enclose triangles; see the picture. Prove that the sum of the areas of the three outer squares (I, II and III) equals three times the sum of the areas of the three inner squares (IV, V and VI). 242 303

242 303

aa " aaaa """ TTT T II I VI TT """T aaa ,@ PPPT" aaa, , V @@P IV  @@ ,,PPB @,B B BB III BBB BB

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; and by Vedula N. Murty, Andhra University, Visakhapatnam, India. We give Murty's solution. Let the gure on the next page be labelled as shown: Let MN = x , LU = x , UY = x , AC = x , AB = x , BC = x , \MBN = , \LAU = , \V CY = , \BAC = A, \ACB = C and \ABC = B . Then we have + =  , + A =  , + C =  1

3

2

4

x = x + x , 2x x cos 9 > > = x = x + x , 2x x cos > : : : > x = x + x , 2x x cos ; 2 1

2 6

2 5

5

6

2 2

2 4

2 6

4

6

2 3

2 4

2 5

4

5

5

6

(1)

15

Pa

X " aaaa " TT " " a TT aaN Y"" VI TTT II T"T W I " T B,@ PCPPTPT"""" Q aaa aaa ,, @@  IV  V aa, @ PP M@@ V ,,APP U @@,, BB LB B BB III BB BB BB  BB T S

x = x + x , 2x x cos B 9 > > = x = x + x , 2x x cos C > : : : > x = x + x , 2x x cos A ; 2 4

2 5

2 6

5

6

2 4

2 4

2 6

4

6

2 6

2 4

2 5

4

6

(2)

From (2), we have

x + x + x = 2x x cos A + 2x x cos B + 2x x cos C = ,2x x cos , 2x x cos , 2x x cos : : : (3) 2 4

2 5

2 6

4

5

4

5

5

6

5

4

6

6

4

6

From (1), we have

x + x + x = 2(x + x + x ) , 2x x cos ,2x x cos , 2x x cos 2 1

2 2

2 3

2 4

4

2 5

2 6

5

5

4

6

6

using (3)

= 2(x + x + x ) + x + x + x = 3(x + x + x ): That is, Area of (I + II + III ) = 3 Area of (IV + V + V I ). 2 4 2 4

2 5 2 5

2 6 2 6

2 4

2 5

2 6

4. For every positive integer n, n? is de ned as follows: (

n=1 n? = 1 n for for n2 n, ? p p Prove 1992 < 1992? < 1992. (

4 3

1)

16 Solutions by Vedula N. Murty, Andhra University, Visakhapatnam, India; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's solution and his remark. Using the more convenient notation f (n) for n?, we show that in general p p

n + 1 < f (n) < 43 n () p for all even n  6. In particular, for n = 1992, we would get 1993 < f (1992) < p1992. First note that f (n) = f nn, = nn, f (n , 2) for all n  3. If N = 2k where k  2, then multiplying f (2q ) = q,q f (2q , 2) for q = 2; 3; : : : ; k, 4 3

(

1)

we get

1 2 2

1

f (2k) = 34  65    2k2,k 1  f (2)               = 12  43  65    2k2,k 1 > 32 54 67    2k2+k 1 :

Hence

(f (2k)) > 1  32  54  6 (2 k2,k 1)  3  52  74  6 (2 k2+k 1) = 2k + 1; 2

from which it follows that

p f (n) = f (2k) > 2k + 1 = pn + 1: On the other hand, for k  3 we have  2  4   6   2k , 2  2(2k) = 3 5 7    2k , 1  2k  2  5   7   2k , 1  < 3 6 8    2k 2k:

(1)

Hence

 2  4  6    (2k , 2) 5  7    (2k , 1) (f (2k)) < 3  5  7    (2k , 1)  6  8    2k  (2k)   = 32  4  2k; 2

2

2

2

from which it follows that

p f (n) = f (2k) < 43 2k = 43 pn:

The result follows from (1) and (2).

(2)

17 Remark: Using similar arguments, upper and lower bounds for f (n) when n is odd can also be easily derived. In fact, if we set P =    k,k (usually denoted by k,k ) then various upper and lower bounds for P abound in the literature; for example, it is known that 1 3 5 (2 2 4 6 2

(2 1)!! (2 )!!

s

1)

s

1 5 P  1 3 2 4k + 1 2 2k + 1 and

q 1 < P  p1n : (n + ) 1 2

(Compare, for example, x3.1.16 on p. 192 of Analytic Inequalities by D.S. Mitronovic.) Clearly, each pair of these double inequalities would yield corresponding upper and lower bounds for the function f (n) considered in the given problem. 5. We consider regular n-gons with a xed circumference 4. We call the distance from the centre of such a n-gon to a vertex rn and the distance from the centre to an edge an . a) Determine a ; r ; a ; r . b) Give an appropriate interpretation forpa and r . c) Prove: a n = (an + rn ) and r n = a n rn . 4

4

8

8

2

1 2

2

2

an

2

2

rn

Let u ; u ; u ; u ; : : : be de ned as follows: 0

1

2

3

u = 0; u = 1; un = 21 (un, + un, ) for n even and un = pun,  un, for n odd: 0

1

2

2

dia.

1

1

d) Determine: limn!1 un . Solution by Vedula N. Murty, Andhra University, Visakhapatnam, In-

Let O be the centre of the regular n-gon. Let A A denote one side of the regular n-gon 1

2

18

O

rn

an

A A Then we have \A OA = n , \OA A = \OA A =  , n . Thus r , ,, ! jA A j = rn + rn , 2rn cos 2n r = 2rn(1 , cos 2n ) r = 4rn sin n = 2rn sin n : 1

2

1

1

2

2

1

2

2

2

2

2

1

2

2

2

2

2

The circumference of the regular n-gon is 2nrn sin n = 4 whence

In particular

2 ; rn = n sin  n    an = rn sin 2 , n = rn cos n = n2 cot n :

p r = 12 sin1  = 22 ;

a = 24 cot 4 = 12 ;

4

4

4

2 = 1 : r = 8 sin  4 sin  8

8

Now, cos  = p = 1 , 2 sin  gives 1

4

so

8

2

8

2

q p sin 8 = 12 2 , 2;

r = 14 p 2 p = 12  p 1 p ; 2, 2 2, 2 s p 1 2 = 1 1p p2;  a = r cos 8 = 4 22 + p , 2 42, 2 8

and

8

8

since cos  = 2 cos  , 1. For (b), r = 1, a = 0 as the 2-gon is a straight line with O lying at the middle of A and A . 2

4

2

1

8

2

2

19 For (c), we have

  an + rn = rn 1 + cos n = 2rn cos 2n 4 cos  = n sin  2n n 4 = 2n sin  cos  cos 2n = n2 cot 2n : n n Thus (an + rn ) = n cot( n ) = a n , and 2

2

2

2

1 2

1

2

2



2



cos n = 1 ; n 2 = 1 a nrn = n1 cos     sin n n sin n n sin n cos n n sin n so pa n rn = n 2n = r n . For (d), note u = 0, u = 1, and u = . For n  2 we have that un is either the arithmetic or geometric mean of un, and un, and in either case lies between them. It is also easy to show by induction that u ; u ; u ; : : : form an increasing sequence, and u ; u ; u ; : : : form a decreasing sequence with u l  u s for all l; s  0. Let limk!1 u k = P and limk!1 u k = I . Then P  I . We also have from u n = (u n, + u n, ) that P = (I + P ) so that I = P and limn!1 un exists. Let limn!1 un = L. With a = 0 and r = 1, let u k = a k+1 and u k = r k+1 , for k = 0; 1; 2; : : : . From (c), u = a 1 = a = 0 and u = r 1 = r = 1. Also for n = 2k + 2, u k = a k+1+1 = a  k+1 = (a k+1 + b k+1 ) = (u k + u k ); that is un = (un, + un, ) and for n = 2k + 3 u k = up k = r k+1+1 = r k+1 = a k+1  r k+1 = pa k+1+1  r k+1 = pu k  u k p so un = un,  un, . Thus un and un satisfy the same recurrence and it follows that L = limk!1 a k+1 = limk!1 r k+1 . Now, from the solution 2

2

1 sin

2

2

2

2

2

2

2

2

2

2

2

0

1

1 2

2

1

0

2

4

1

2

3

5

2 +1

2 +1

2

1 2

2

2

2

0

1 2

2 +1

2 +3

2

2

)

2( +1)

1

1 1 2

2 2

2

2

2

2

2

2

1

2

1

2( +1)+1 2(2

2 +1

2

2

2

2 +2

2

1 2

2

2

1 2

2

2

2(2

2

2

)

2

2 +1

2

2

to (c),

2



2 =2 n ; rn = n sin   sin  n n  so limn!1 rn =  since n ! 0. Therefore limn!1 un =  . 2

2

That completes the column for this issue. Olympiad season is approaching. Send me your contests and nice solutions.

20

BOOK REVIEWS Edited by ANDY LIU Shaking Hands in Corner Brook and other Math Problems, edited by Peter Booth, Bruce Shawyer and John Grant McLoughlin, published by the Waterloo Mathematics Foundation, Waterloo, 1995, 153 pages, paperback, ISBN 0-921418-31-0. Reviewed by Robert Geretschlager and Gottfried Perz.  One representative from each of six regions met in Corner Brook, Newfoundland, to discuss math problems. Each of these delegates shook hands with each other delegate. How many handshakes were there? This is the rst of many problems that can be found in this collection of problems from the Newfoundland and Labrador Teachers' Association (NLTA) Senior Mathematics League, which conveniently o ers an explanation as to the whereabouts of Corner Brook. The NLTA Math League was started in 1987, and has since developed into a very interesting competition at the regional level. A number of aspects make this competition di erent from most math competitions. First of all, it is purely a team competition, with four students from each participating school comprising a team. There is no individual ranking, and so students are motivated to work together at nding solutions. For each of ten questions posed, a team can receive ve points for a correct team answer. If the members of a team cannot agree on the correct answer, they can submit individual answers, for which their team can get one point each, if correct. Finally, there is a relay question, made up of four parts. In the relay, each part yields an answer, which is necessary to be able to solve the next part (much as in the American Regions Mathematics League (ARML), which may be better known to many readers). The relay section can yield a maximum of 15 points (made up of ve points for the solution and extra points for solving the problems in a short time), for a possible total of 65 points. If teams end up with the same point sum, a tie breaker question is posed. The concept behind this competition is geared to fostering cooperative problem solving, something that is generally ignored in olympiad-style competitions. The level of diculty of the problems posed is adequate to the time allowed (usually from 3 to 10 minutes per question) and the intentions of the competition, and ranges from fairly easy to pre-olympiad level. The book is divided into sections covering regular questions, relay questions, tie breakers and solutions. The problems are in a random order, and no indication is given of which questions were posed at which competition. Perhaps at least one example of ten speci c questions posed at one competition, and the order they were posed in, might have been of interest.

21 Here are a few problems to whet your appetite:  How many three digit numbers include at least one seven but have no zeros?  Al, Betty, Charles, Darlene and Elaine play a game in which each is either a frog or a moose. A frog's statement is always false while a moose's statement is always true. Al says that Betty is a moose. Charles says that Darlene is a frog. Elaine says that Al is not a frog. Betty says that Charles is not a moose. Darlene says that Elaine and Al are di erent kinds of animals. How many frogs are there?  Triangle ABC is isosceles, with \ABC = \ACB. There are points D; E and F on BC; CA and AB, respectively, that form an equilateral triangle. Given that \AFE = x and \CED = y , calculate \BDF in terms of x and y . The book has a very pleasing layout, with the cover showing the densest packing of seven circles in an equilateral triangle. The solutions are nicely presented, and in several cases, alternate solutions are given, occasionally labeled the \routine way" and the \subtle" or \smart way". Shaking Hands in Corner Brook should be of interest to anyone involved with high school mathematics, either in competitions, or simply seeking enrichment material for the interested student. Copies of the above reviewed book may be obtained from: Canadian Mathematics Competition Faculty of Mathematics, University of Waterloo Waterloo, Ontario, Canada. N2L 3G1 The cost is $12 in Canadian funds (plus 7% GST for shipping to Canadian addresses). Cheques or money orders in Canadian funds should be made payable to: Canadian Mathematics Competition. All pro ts from the sale of this book are for the Newfoundland Mathematics Prizes Fund.

22 Teachers interested in providing a lively and stimulating high school mathematics competition for their students may be interested in participating in a NLTA Senior Mathematics League in their own area. Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario, a regular contributor to CRUX, is already a participant. Meetings can be held within individual schools, or between teams from more than one school. Sample games and other information on how the NLTA Senior Mathematics League is organised may be obtained free from: Dr. Peter Booth Department of Mathematics and Statistics Memorial University of Newfoundland St. John's, Newfoundland, Canada. A1C 5S7 Tel: int+ 709{737{8786 Fax: int+ 709{737{3010 email: [email protected] Schools that participate on a regular basis will be sent questions and detailed solutions ve times per year (October, November, February, March and May). There is an annual fee of $50 (Canadian funds) for each group of schools participating. Cheques or money orders should be made payable to Newfoundland Mathematics Prizes Fund.

23

A Probabilistic Approach to Determinants with Integer Entries Theodore Chronis It is well known that the probability for an integer number to be odd is equal to the probability for the number to be even. What about determinants? What is the probability for a rectangular matrix with integer entries to have odd determinant? More generally, if m is a natural number, what is the probability for which det A  mi mod m, where mi is chosen from the set f0; 1; 2; : : : ; m , 1g? I have the following problem to propose; I hope you will nd it interesting. Let A be an n  n matrix whose elements are integers. What is the probability the determinant of A is an odd number? Solution: Let A = [aij ] ; i; j = 1; : : : ; n. It is obvious that det A  det ([aij mod 2]) mod 2: So the problem is to nd the probability that the determinant of an n  n matrix with elements from the set f0; 1g is an odd number. Let An be an n  n matrix with elements from the set f0; 1g. Let also N (det An) be the number of odd n  n determinants, and P (det An ) be the corresponding probability. Let

0 k k :: :: k 1 BB a a :: :: a nn CC f (K = fk ; k ; : : : ; kng) = N B : C B@ :: :: CA ; : a a :: :: a n n nn where k ; k ; : : : ; kn 2 f0; 1g are xed. P Then N (det An ) = (f (K )), where the sum is calculated for all the 2n , 1 possible permutations K = fk ; k : : : ; kn g with ki 2 f0; 1g, i = 1; : : : ; n and k ; k ; : : : ; kn not all zero. (Note that f (0; 0; : : : ; 0) = 0:) 1

1

1

2

21

22

1

2

2

2

1

1

2

2

2

24 Lemma: f (K ) = 2n, N (det An, ), where k + k + : : : + kn 6= 0. Proof. It is well known that a determinant remains unchanged if from the elements of one of its columns we substract the corresponding elements of another column. It is also obvious that the same is true for N (det An ). Additionally, a determinant just changes sign if we interchange two columns, while N (det An ) remains unchanged. So 1

1

1

2

0 k k :: :: k 1 0 1 0 :: :: 0 n BB a a :: :: a n CC BB a a :: :: a n : : NB : C =NB : B@ :: :: C B A @ : : : a a :: :: a a a :: :: a : n n nn n n nn () f (K ) = 2n, N (det An, ): Hence N (det An ) = 2n, N (det An, )(2n , 1). Of course N (det A ) = 1 and so n Y N (det An) = 2n n, = (2i , 1): 1

2

21

22

1

2

2

21

22

1

2

1

1

2

1 C CC CA

1

1

1

(

1) 2

i=1

Q

Finally, P (det An ) = N n2An () P (det An ) = ni (1 , 2,i ). Q Note: The in nite sequence ni (1 , 2,i ); n = 1; 2; : : : is decreasing and bounded below by 0, so limn!1 P (det An ) exists. Using Mathematica, we found that (det

)

2

=1

=1



nlim !1 P (det An ) = 0:288788095086602421278899721929 : : : :

Ayras 15, Ki sia Athens 14562, GREECE e-mail: [email protected]

25

THE SKOLIAD CORNER No. 19 R.E. Woodrow

The problem set we give in this issue comes to us with our thanks from Tony Gardiner of the UK Mathematics Foundation, School of Mathematics, University of Birmingham. The Nat West Junior Mathematical Challenge was written Tuesday, April 26, 1994 by about 105,000 students. Students from England and Wales must be in school year 8 or below. The use of calculators, calendars, rulers, and measuring instruments was forbidden.

1994 NAT WEST UK JUNIOR MATHEMATICAL CHALLENGE

Tuesday, April 26, 1994 | Time: 1 hour 1. 38 + 47 + 56 + 65 + 74 + 83 + 92 equals A. 425 B. 435 C. 445 D. 456

E. 465.

2. What is the largest possible number of people at a party if no two of them have birthdays in the same month? A. 11 B. 12 C. 13

D. 23

3. I have $500 in 5p coins. How many 5p coins is that? A. 100

B. 500

C. 1000

D. 2500

E. 334. E. 10000.

4. What was the precise date exactly sixty days ago today? (No calendars!) A. Friday 25th February B. Saturday 26th February C. Friday 26th February D. Saturday 27th February E. Tuesday 26th February.

5.

You have to nd a route from A to B moving horizontally and vertically only, from one square to an adjacent square. Each time you enter a square you add the number in that square to your total. What is the lowest possible total score for a route from A to B ? A. 28

B. 29

C. 30

D. 31

3 9 B 8 5 6 9 11 7 A 8 10 E. 34.

26

6. On a clock face, how big is the angle between the lines joining the centre to the 2 O'clock and the 7 O'clock marks? A. 160 B. 150 C. 140

D. 130

E. 120.

7. Gill is just six and boasts that she can count up to 100. However, she

often mixes up nineteen and ninety, and so jumps straight from nineteen to ninety one. How many numbers does she miss out when she does this? A. 70 B. 71 C. 72 D. 78 E. 89.

8.

In how many ways can you join the two shapes shown here to make a gure with a line of symmetry? A. 0

B. 1

C. 2

D. 3

E. 4.

9. If you divide 98765432 by 8, which non-zero digit does not appear in your answer? A. 2 B. 4

C. 6

D. 8

E. 9.

10. How many numbers between 20 and 30 (inclusive) cannot be written

as a multiple of 5, or as a multiple of 7, or as the sum of a multiple of 5 and a multiple of 7? A. 1 B. 2 C. 3 D. 5 E. 6.

11. Four children are arguing over a broken toy. Alex says Barbara broke

it. Barbara says Claire broke it. Claire and David say they do not know who broke it. Only the guilty child was lying. Who broke the toy? A. Alex B. Barbara C. Claire D. David E. can't be sure.

12.

The diagram is made up of one circle and two semicircles. Which of the three regions | the black region, the white region and the large circle | has the longest perimeter?

27 A. the black region B. the circle C. the white region D. black and white are equal and longest E. all three perimeters are equal.

13. From my house the church spire is in the direction NNE. If I face in this

direction and then turn anticlockwise through 135 I can see the Town Hall clock. In which direction am I then facing? A. WSW B. due West C. SW D. due South E. SSE.

14. Samantha bought seven super strawberry swizzles and ten tongue twist-

ing to ees for $1.43. Sharanpal bought ve super strawberry swizzles and ten tongue twisting to ees for $1.25. How much is one tongue twisting to ee? A. 7p B. 8p C. 9p D. 10p E. 18p.

15. I am forty eight years, forty eight months, forty eight weeks, forty eight days and forty eight hours old. How old am I? A. 48 B. 50 C. 51 D. 52

16.

The numbers 1 to 12 are to be placed so that the sum of the four numbers in each of the six rows is the same. Where must the 7 go?

E. 53. 4

A

3

B

8

C A. at A

B. at B

C. at C

1 5

D

6

D. at D

9

E

E. at E

17. Three hedgehogs | Roland, Spike and Percival | have a leaf collecting

race. Roland collects twice as many as Percival, who collects one and a half times as many as Spike. (Spike is moulting, and so has fewer prickles for her to stick the leaves onto.) Between them they collect 198 leaves. How many did Spike manage to collect? A. 18 B. 22 C. 36 D. 44 E. 66.

28

18. The four digits 1, 2, 3, 4 are writtin in increasing order. You must insert

one plus sign and one minus sign between the 1 and the 2, or between the 2 and the 3, or between the 3 and the 4, to produce expressions with di erent answers. For example, 1 , 23 + 4 gives the answer , 18: How many di erent positive answers can be obtained in this way? A. 2 B. 3 C. 4 D. 5 E. 6.

19. Roger Rabbit has twice as many sisters as brothers. His sister Raquel

notices that 2=5 of her brothers and sisters are boys. How many Rabbit children are there in the family? A. 2 B. 4 C. 8 D. 16 E. 32.

20. The population of a new town in 1990 was 10; 000. It has since doubled

every year. If it kept on doubling every year for ten years, what would its population be in the year 2000? A. 100; 000 B. 200; 000 C. 1; 000; 000 D. 2; 000; 000 E. 10; 000; 000.

21. LMNO is a square. P is a point inside the square such that NOP is an equilateral triangle. How big is the angle PMN ? A. 75 B. 70 C. 60 D. 45

E. 30 .

22. If the perimeter of a rectangle is 16x + 18 and its width is 2x + 6, what is its length? A. 18x + 24 B. 7x + 6 C. 12x + 6 D. 6x + 3 E. 14x + 12.

23. In a group of fty girls each one is either blonde or brunette and is either

blue-eyed or brown-eyed. Fourteen are blue-eyed blondes, thirty one are brunettes and eighteen are brown-eyed. How many are brown-eyed brunettes? A. 5 B. 7 C. 9 D. 13 E. 18.

24. A bottle of Jungle Monster Crush (JMC) makes enough drink to ll sixty glasses when it is diluted in the ratio 1 part Crush to 4 parts water. How many glasses of drink would a bottle of JMC make if it is diluted in the ratio 1 part Crush to 5 parts water? A. 48 B. 60 C. 72 D. 75 E. 80.

29

25.

A 3 by 3 by 3 cube has three holes, each with a 1 by 1 cross section running from the centre of each face to the centre of the opposite face. What is the total surface area of the resulting solid?

PP  P PPP    P P PP P   PP P  PP PP    PP

PP 

PP  PP  PP

A. 24

B. 48

C. 72

D. 78

E. 84.

That completes the Skoliad Corner for this issue. Send me your contests, suggestions, and recommendations to improve this feature.

Introducing the new Associate Editor-in-Chief For those of you who do not know Clayton, here is a short pro le: Born: Haystack, Newfoundland Educated Haystack School School, Newfoundland Thornlea School, Newfoundland Memorial University of Newfoundland Queen's University, Kingston, Ontario Employment Random South School Board, Trinity Bay, Newfoundland Memorial University of Newfoundland Mathematical Mathematical Education Interests Commutative Algebra 1

You may have diculty nding Haystack on a map of Newfoundland | it is not lost | it no longer exists! 1

30

MATHEMATICAL MAYHEM Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. All material intended for inclusion in this section should be sent to the Mayhem Editor, Naoki Sato, Department of Mathematics, University of Toronto, Toronto, ON Canada M5S 1A1. The electronic address is [email protected]

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto). The rest of the sta consists of Richard Hoshino (University of Waterloo), Wai Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Editorial It gives me great pleasure to unveil the premiere issue of \Crux Mathematicorum with Mathematical Mayhem". This merger has been in the works for quite some time, and it has nally been successfully realized. For the bene t of those who have not heard of Mayhem, I will provide a brief description. Mayhem was founded in 1988 by two high school students Ravi Vakil and Patrick Surry, who wished to establish a journal speci cally oriented towards students, and totally operated by students. Although the journal has been passed down through many hands, and though it has not always been easy, this mandate has always been resolutely upheld; it has made Mayhem a unique and exceptional journal. And rest assured, we will still be running our share of \Crux with Mayhem". Our features include articles, olympiads, and a problems section. The material is generally focused towards contests and olympiads, and how to prepare for them, and the topics range from high school mathematics to undergraduate material. We cannot emphasize enough that we are a journal dedicated to mathematics students. I myself am a fourth-year student at the University of Toronto, and Cyrus Hsia (the Mayhem Assistant Editor) is a third-year student, also at the University of Toronto. Our fearless sta also consists of undergraduate and high school students. This brings me to my next point. After considerable discussion, \Mayhem" has decided to restrict itself to publishingsolutions only from students. The rationale behind this move is that the Crux problems already draw many solutions, and if the same people were to respond to our problems, which are considerably easier, it would simply overwhelm the section. We know that

31 there are many non-students who have contributed to the problems sections over the years, who have our full gratitude, and hope they understand our position. We are, however, prepared to make exceptions in, well, exceptional cases. However, we warmly welcome submissions for articles and problems from all people. Back issues are available; the information is inside the back cover. Any correspondence about Mayhem should be sent to Mathematical Mayhem, c/o Naoki Sato, Department of Mathematics, University of Toronto, M5S 1A1, or at the e-mail address . Subscriptions, however, should be sent to the Canadian Mathematical Society oces, as mentioned on the inside back cover. Well, I think that's about it. For people who have subscribed to Mayhem, welcome back, I know it's been a long wait. I look forward to working with Bruce Shawyer on our new project (I think he will bring a certain \discipline" to Mayhem, but we will resist it as much as possible. Don't tell him I said that.) Here's to a new year and a new era for Mayhem. Naoki Sato Mayhem Editor

Shreds and Slices Positive Matrices and Positive Eigenvalues Theorem. An n  n matrix M with positive real entries has at least one positive eigenvalue. Proof. Let S = f(x ; x ; : : : ; xn ) j x ; x ; : : : ; xn  0; x + x +    + xn = 1g; that is, S is the portion of the unit sphere in Rn with all coordinates nonnegative. v De ne the map f : S ! S by f (~v) = jM~ M~vj . Since M has all positive real entries, for any ~v 2 S , M~v also has all non-negative coordinates, so f is well-de ned, and does indeed map S into S . Note that S is a closed, simply connected set. Then by Brouwer's Fixed Point Theorem, there is a xed point of f ; that is, for some ~v 2 S , f (~v) = v ~v = jM~ M~vj ) M~v = ~v, for some positive value  (since  cannot be zero), namely  = jM~v j. This  is a positive eigenvalue of M . 1

2

1

2

2 1

2 2

2

32

Newton's Relations Given n reals a , a , : : : , an , let Sk be the sum of the products of the ai taken k at a time, and let Pk = ak + ak +    + akn . Consider the generating 1

2

1

functions

2

S + S x + S x +    + Snxn = 1 + (a + a +    + an )x + (a a + a a +    + an, an )x +    + a a    an xn = (1 + a x)(1 + a x)    (1 + anx) 0

1

2

2

1

2

1

1

1

3

2

1

2

1

and

2

2

P , P x +P x ,  = (1 , a x + a x ,    ) + (1 , a x + a x ,    ) +    + (1 , an x + an x ,    ) = 1 +1a x + 1 +1a x +    + 1 +1a x : 0

1

2

2

2 1

1

2

1

Their product is

2 2

2

2

2

2

n

2

(n , P x + P x ,    )(1 + S x + S x +    ) = (1 + a x)(1 + a x)    (1 + anx)   1 +1a x + 1 +1a x +    + 1 +1a x ; 1

2

2

1

1

2

2

2

1

n

2

since P = n and S = 1. We claim the expression is equal to n + (n , 1)S x + (n , 2)S x +    + Sn, xn, . To see this, consider the coecient of xk . Since the expression is symmetric, the coecient is some multiple of Sk . How many times does the term a a    ak xk appear? It must have appeared in the product (1 + a x)(1 + a x)    (1 + anx) as a a    akal, where k < l  n, before having the term al divided out. There are n , k choices for l, and hence the coecient is (n , k)Sk . Hence, 0

1

0

1

1

1

1

2

2

2

2

1

2

(n , P x + P x ,    )(1 + S x + S x +    ) = n + (n , 1)S x + (n , 2)S x +    + Sn, xn, ; 1

2

2

1

2

1

2

2

2

1

1

and equating coecients:

nS , P = (n , 1)S nS , S P + P = (n , 2)S nS , S P + S P , P = (n , 3)S  n , nSn, , Sn, P + Sn, P ,    + (,1) Pn, = Sn, ; 3

1

2

1

3

2

2

1

1

1

2

1

1

2

2

1

1

2

3

3

1

1

1

33 or

P ,S P , S P + 2S P , S P + S P , 3S  n , Pn, , S Pn, + S Pn, ,    + (,1) (n , 1)Sn, 2

1

1

2

2

2

1

3

= 0 = 0 = 0

1

= 0;

1

3

1

1

2

2

1

1

1

3

and

Pm , S Pm, + S Pm, ,    + (,1)nPm,n Sn = 0 1

1

2

2

for m  n: The last equation is the well-known recursion sequence for the Pi , and the previous equations (known as Newton's relations) can help pin down the values of P , P , : : : , Pn, , or vice-versa. Problem. If 1

2

1

x + y + z = 1; x + y + z = 2; x + y + z = 3; determine the value of x + y + z . 2

2

2

3

3

3

4

4

4

Solution. Newton's relations become

P , S = 1 , S = 0; P , S P + 2S = 2 , S + 2S = 0; P , S P + S P , 3S = 3 , 2S + S , 3S = 0; which imply that S = 1, S = ,1=2, and S = 1=6. Also, P , 3S + 2S , S = 0 ) P = 25=6. 1

1

1

2

1

1

3

1

2

2

2

1

2

3

1

1

2

3

1

2

2

3

3

4

1

4

Here is the last problem of the 1995 Japan Mathematical Olympiad, Final Round. Problem. Let 1  k  n be positive integers. Let a , a , : : : , ak be complex numbers satisfying 1

2

a + a +    + ak = n a + a +    + ak = n  ak + ak +    + akk = n , , , Show that (x + a )(x + a )    (x + an) = xk + n xk, + n xk, +    + nk . Solution. Given P = P, n =    = Pk = n, we must nd S , S , : ,: n: , Sk . We will prove that S = by induction. Clearly S = P = n = . 1

2

1

m

1

2

2 1

2 2

1

2

2

1

1

2

m

1

1

2

2

1

2

1

34

,

,

,



Now, for some m, assume S = n , S = n , : : : , Sm, = mn, . Then by the equations above, Pm , S Pm, + S Pm, ,   +(,1)m, Sm, P + m(,1)mSm = 0, or 1

2

1

1

1

2

1

2

2

1

1

1

1

n n  n  m , n , n 1 + n 2 ,    + (,1) n m , 1 + m(,1)mSm = 0 1

so that

m S =  n  ,  n  +  n  ,    =  n , 1 ; n m m,1 m,2 m,3 m,1 and further, n , 1 n n Sm = m m , 1 = m : So by induction, we are done.

Mathematically Correct Sayings [The following shred/slice appeared in the newsgroup rec.humor.funny.] After applying some simple algebra to some trite phrases and cliches, a new understanding can be reached of the secret to wealth and success. Here it goes. Knowledge is Power, Time is Money, and as everyone knows, Power is Work divided by Time. So, substituting algebraic equations for these time worn bits of wisdom, we get: K = P (1) T = M (2) P = W=T (3) Now, do a few simple substitutions. Put W=T in for P in equation (1), which yields: K = W=T (4) Put M in for T into equation (4), which yields: K = W=M (5) Now we've got something. Expanding back into English, we get: Knowledge equals Work divided by Money. What this MEANS is that:

35 1. The More You Know, the More Work You Do, and 2. The More You Know, the Less Money You Make. Solving for Money, we get:

M = W=K

(6)

Money equals Work divided by Knowledge. From equation (6) we see that Money approaches in nity as Knowledge approaches 0, regardless of the Work done. What THIS MEANS is: The More you Make, the Less you Know. Solving for Work, we get W =M K (7) Work equals Money times Knowledge From equation (7) we see that Work approaches 0 as Knowledge approaches 0. What THIS MEANS is: The stupid rich do little or no work. Working out the socioeconomic implications of this breakthrough is left as an exercise for the reader.

Contest Dates Here are some upcoming (or in some cases, already past) contest dates to mark on your calendar. Contest Grade Date Gauss Grades 7 & 8 Wednesday, May 14, 1997 Pascal Grade 9 Wednesday, February 19, 1997 Cayley Grade 10 Wednesday, February 19, 1997 Fermat Grade 11 Wednesday, February 19, 1997 Euclid Grade 12 Tuesday, April 15, 1997 Descartes Grades 12 & 13 Wednesday, April 16, 1997 CIMC Grades 10 & 11 Wednesday, April 16, 1997 AJHSME Grades 7 & 8 Thursday, November 21, 1996 AHSME High School Thursday, February 13, 1997 AIME High School Thursday, March 20, 1997 USAMO High School Thursday, May 1, 1997 AJHSME Grades 7 & 8 Thursday, November 20, 1997 COMC High School Wednesday, November 27, 1996 CMO High School Wednesday, March 26, 1997 APMO High School March, 1997 IMO High School July 18 { 31, 1997

36

A Journey to the Pole | Part I  Miguel Carrion  Alvarez

student, Universidad Complutense de Madrid Madrid, Spain For those of us who can not seem to get a strong grip on synthetic geometry, analytic geometry comes in handy. Even though polar coordinates can be superior to rectangular coordinates in some situations, they are systematically ignored by instructors and students alike. The purpose of this series is to introduce their uses with the idea that, as is always happening in mathematics, with a little ingenuity, the concepts central to polar coordinates can be applied elsewhere. This rst article uses polar coordinates in elementary geometry. De nition In polar coordinates, the position of a point P is determined by the distance r from a point O called the pole and the angle  between OP and a semi-in nite line called the polar axis. By convention, the polar axis is taken to be the positive x-axis, and the P transformation from polar to cartesian coordinates is given by x = r cos , y = r sin . The inverse change of coordinates is not r so straightforward; p The obvious expressions are r = x + y ,  = arctan( yx ), but the equation for  is not single-valued, even if  is re Polar axis stricted to [0; 2 ), and  is undeO ned at the origin. Fortunately, we need not worry about this: when handling curves in polar coordinates, the change from rectangular to polar coordinates is of little use, and it is convenient to allow r and  to take on all real values. With this provision, a point can be referred to by an in nite set of coordinate pairs: (r; ) = ((,1)nr;  + n ). Unless you want to do multiple integrals, this is not a problem, but rather something to exploit! Polar Curves Polar curves are usually written in the form r = r(), and unlike curves of the form y = y (x), they can be closed and need not be simple (they can intersect themselves). Implicit curves of the form f (r;) = 0 can be even more general. From a cartesian equation, the substitution x = r cos , y = r sin  yields a polar expression. Throughout this article, I have tried to avoid this whenever possible, and it turns out that it is always possible. 2

2

37 Some symmetries of the curves can be detected by checking the functions r() or f (r;) above for the following simple properties (this is not a complete list):

 The curve is symmetric about the pole if r() = r( + ) or f (,r;) = f (r;)  The curve has n-fold symmetry about the pole if r() = r( + n ) 2

 The curve is symmetric about the polar axis if r() = r(,)  The curve is symmetric about a line at an angle  to the polar axis if r() = r(2 , ) The following transformations are also useful:

 Any curve can be rotated through  by substituting  ,  for   The x- and y-axes can be permuted by substituting  ,  for  2

Example 1. The equation of a circle of radius a centered at the origin is

r = a.

Example 2. The equation of a line passing through the origin at an angle

 to the polar axis is  = .

Exercise 1. Find the equation of a line at an angle  to the polar axis passing at a distance d to the pole. Exercise 2. Identify the curve r = 2a cos . The Cosine Law More often than not, when working in polar coordinates, one uses nothing but trigonometry, and the cosine law is the starting point of many derivations. If you think about it, it comes closest to being a `vector addition rule' to use if you need to translate a curve, although this is best done in rectangular coordinates. I will not give a translation rule, because it is cumbersome and is of little use. Instead, I will use the cosine law to derive the equation of a circle of radius  centered at (R; ) (see gure). Applying the cosine law to side  of 4OCP , we have



2

= R + r , 2Rr cos( , ) = [r , R cos( , )] + R , R cos ( , ) =) [r , R cos( , )] =  , R sin ( , ): 2

2

2 2

2

2

2

2

2

2

38 From this equation, it is evident that if R  , then the curve is deP  ned for a limited range of  given by r , R  sin( , )  R , as we would C expect when the origin lies outside R the circle. The squared length of the   tangent from O, when sin( , ) = O  R2, is P = R cos ( , ) = R (1 ,  R2 ) = R ,  ; this is called the potence of the origin w.r.t. the circle. Note that this formula is correct, even when R <  and sin  = R has no solution. Incidentally, the solution to Exercise 2 can be obtained easily by noting that if the origin is on the circle, then R = , and 2

2

2

2

2

r , R cos( , ) = R cos( , ) =) r = 2R cos( , ): Example 3. Polar equation of the ellipse with one focus at the origin and the main axis at an angle  to the polar axis. The cosine law applied to side F 0P of triangle FF 0P gives

r

2

= =) =) =)

0

2c F

(2a , r)

P 2a , r F



r + 4c , 4rc cos( , ) a , ar = c , rc cos( , ) r[a , c cos( , )] = a , c b =a r = 1 , e cos(  , ) : 2

2

2

2

2

2

2

A Catalogue of Important Curves

The following curves are all important in their own right, but since their polar expressions are particularly simple, they make good examples of the use of polar coordinates.

39 Conic sections We already have the equation for the ellipse. The polar equation of the parabola is even easier to derive. In the gure, we have the focus at the origin, the axis at an angle  to the polar axis and a distance d from the focus to the directrix. From the gure on the previous page, we have

d + r cos  P & r  V F q

q

r = d + r cos( , ) =) r = 1 , cos(d  , ) : Exercise 3. In a similar way, derive the equation for the hyperbola, noting how both branches are handled. Hence, deduce that the general equation of the conic is r = ,e de , , where e is the eccentricity. This equation can be obtained immediately from the de nition of a conic as the locus of the points whose distances to a line (called the directrix) and a point (called the focus) are at a constant ratio e. The Cardioid P The cardioid is the trajectory of a point on a circle that rolls on another r  R circle of the same radius. So de 2#  P C ned, it is a special case of the epicyO " C " cloid, which is the curve described 2 by a point on a circle rolling on another circle with no restriction on the radii; the general equation of the epicycloid is best expressed in parametric form. 1

cos(

)

0

0

In the gure, the two circles have radius R. The condition that the one centered at C 0 rolls on the one centered at C implies that triangles OCP 0 and PC 0P 0 must be congruent.0 In triangles OCP 0 and PC 0P 0,  = 2R cos(=2). Similarly, in triangle OP P , r = 2 cos(=2). Putting all together, we have

r = 4R cos (=2) =) r = 2R(1 + cos ): 2

The cardioid is also a special case of Pascal's Limacon, of the equation

r = b + a cos . The cusp at O becomes a loop if b < a, and a smooth indentation if b > a. The limacon can be de ned as the locus of the feet

40 of perpendiculars dropped from the origin to tangent lines to a circle. The radius of the circle is b and the distance from O to its centre is a. The Lemniscate P The lemniscate is the locus of the at points such that the product of their r a=t  distances to two points 2a apart a O is a . In the gure, the cosine law gives 2

at

2 2

= =) =) =)

r a r r

2 4

4 2

+ a + 2ar cos ; a =t = r + a , 2ar cos  = (r + a ) , 4a r cos  = 2a r (2 cos  , 1) = 2a cos(2): 2

2

2

2

2 2

2

2

2

2

2

2

2

2

2

The lemniscate is a special case of Cassini's ovals, which are de ned in the same way, but with no restriction on the product of the distances. The Rose Loosely related to the lemniscate are the roses, of equation P a r = a cos[n( , )]  O with integer n. For odd n, the curve has n `leaves' and it is traced completely when  varies from 0 to  . For even n, the curve has 2n `leaves', and it is traced completely only when  varies from 0 to 2 (see the gure). More general curves can be obtained if n is rational or irrational. In the rst case there is an integer number of overlapping lobes and the curve is closed, but in the latter case the curve never closes, and in fact it is dense in the disc r  a. The Spirals Polar coordinates are particularly suited to spirals. The two most famous are the Archimedean Spiral r = a, which is the trajectory of a point whose angular and radial velocities are proportional, and the Logarithmic Spiral r = ea . In fact, any continuous monotonic function that goes to in nity as  goes to in nity de ned in a semi-in nite interval will give rise to a spiral, like r = a ln . A related feature of polar curves is the limit cycle, occurring when r() has a nite limit r at in nity. In that case, the curve winds around the origin in nitely many times, approaching the circle r = r . Recognizing a limit cycle makes it easier to sketch a polar curve. 0

0

41 Straight Lines We nish by deriving the equation of a straight line not passing through the origin. From the gure, we have d = r cos( , ), where d is the distance from the line to the origin and is the direction of the closest point. An alternative form is d = r sin( , ), where  is the direction of the line and 0 > , >  (see the gure).

r O 

d



Additional Problems Problem 1. Considering r( + ) and r( , ), derive an expression for a secant line to a conic. Passing to the limit ! 0, write an equation for the tangent line at  . Problem 2. PQ is a chord through the focus F of a conic, and the tangents at P , Q meet at T . Prove that T lies on the directrix corresponding to F and that FT ? PQ. Problem 3. Let s be the tangent line at the vertex of a parabola and t be the tangent at P . If r and s meet at Q, prove that FQ bisects the angle between FP and the axis of the parabola, and that FQ ? s. 0

0

0

IMO Report Richard Hoshino

student, University of Waterloo Waterloo, Ontario. After ten days of intensive training at the Fields Institute in Toronto, the 1996 Canadian IMO team travelled to Mumbai, India to participate in the 37th International Mathematical Olympiad. For the rst time in our team's history, every team member brought home a medal, with three silver and three bronze. This year's team members were: Sabin \Get me a donut" Cautis, Adrian \Da Chef" Chan, Byung Kyu \Spring Roll" Chun, Richard \YES! WE'VE GOT BAGELS!" Hoshino, Derek \Leggo my Eggo" Kisman, and Soroosh \Mr. Bean" Yazdani. Our team leaders were J.P. \Radishes" Grossman and Ravi \Oli" Vakil (no, he's not Italian). Special thanks go out to our coaches, Naoki \Dr. Cow" Sato and Georg \Where's my Ethanol" Gunther.

42 This year's paper was one of the most dicult ever, and thus, the cuto s for medals were among the lowest in history, 28 for gold, 20 for silver and 12 for bronze. Only one student, a Romanian, received a perfect score of 42. Our team's scores were as follows: CAN 1 Sabin Cautis 13 Bronze Medal CAN 2 Adrian Chan 14 Bronze Medal CAN 3 Byung Kyu Chun 18 Bronze Medal CAN 4 Richard Hoshino 22 Silver Medal CAN 5 Derek Kisman 22 Silver Medal CAN 6 Soroosh Yazdani 22 Silver Medal Some weird coincidences: all the silver medallists got the exact same score, are graduating and are headed to the University of Waterloo in September, and all the bronze medallists are eligible to return to Argentina for next year's IMO. Overall, Canada nished sixteenth out of seventy- ve countries, one of our highest rankings ever. A main reason for our success was our combined team score of 36 out of 42 on question #6, a problem that many countries answered very poorly (in fact, only two countries had more points on that problem than we did, and they both got 37 out of 42). Unfortunately, question #2 was answered very poorly by Canada, with only one 7, even though the problem was created by our own team leader, J.P. Grossman. We all owe special thanks to Dr. Graham Wright of the Canadian Mathematical Society, Dr. Bruce Shawyer of Memorial University and Dr. Richard Nowakowski of Dalhousie University for their hard work and organization in making our trip possible and Dr. Ed Barbeau of the University of Toronto for all his commitment and dedication to training all the IMO team hopefuls with his year-long correspondence program. Overall, the experience was memorable for all of us, although we could have done without the cockroaches in our rooms. Best of luck to all the students who will be working hard to make the 1997 IMO team, which will be held in Chapadmalal, Argentina near Mar del Plata.

Mayhem Problems A new year brings new changes and new problem editors. Cyrus Hsia now takes over the helm as Mayhem Advanced Problems Editor, with Richard Hoshino lling his spot as the Mayhem High School Problems Editor, and veteran Ravi Vakil maintains his post as Mayhem Challenge Board Problems Editor. Note that all correspondence should be sent to the appropriate editor | see the relevant section. In this issue, you will nd only problems | the next issue will feature only solutions. We intend to have problems and solutions in alternate issues.

43 We warmly welcome proposals for problems and solutions. With the new schedule of eight issues per year, we request that solutions be submitted by 1 June 1997, for publication in the issue 5 months ahead; that is, issue 6. We also request that only students submit solutions (see editorial), but we will consider particularly elegant or insightful solutions for others. Since this rule is only being implemented now, you will see solutions from many people in the next few months, as we clear out the old problems from Mayhem.

High School Problems Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario, Canada. L3S 3E8 H217. Let a , a , a , a , a be a ve-term geometric sequence satisfying the inequality 0 < a < a < a < a < a < 100, where each term is an integer. How many of these ve-term geometric sequences are there? (For example, the sequence 3, 6, 12, 24, 48 is a sequence of this type). H218. A Star Trek logo is inscribed inside a A circle with centre O and radius 1, as shown. Points        A, B, and C are selected on the circle so that AB =  B AC and arc BC is minor (that is ABOC is not a    BB  O B convex quadrilateral). The area of gure ABOC is  equal to sin m , where 0 < m < 90 and m is an in-  SSBB   SB teger. Furthermore, the length of arc AB (shaded as  S B shown) is equal to a=b, where a and b are relatively B C prime integers. Let p = a + b + m. 1

2

3

1

4

5

2

3

4

5

(i) If p = 360, and m is composite, determine all possible values for m. (ii) If m and p are both prime, determine the value of p.

H219. Consider the in nite sum a + a + a + a + ; S = 10 10 10 10 where the sequence fan g is de ned by a = a = 1, and the recurrence p relation an = 20an, + 12an, for all positive integers n  2. If S can a 0

1

0

2

2

3

4

6

0

1

be expressed in the form pb , where a and b are relatively prime positive integers, determine the ordered pair (a; b). 1

2

44

H220. Let S be the sum of the elements of the set f1; 2; 3; : : : ; (2p)n , 1g:

Let T be the sum of the elements of this set whose representation in base 2p consists only of digits from 0 to p , 1. Prove that 2n  TS = (p , 1)=(2p , 1).

Advanced Problems Editor: Cyrus Hsia, 21 Van Allen Road, Scarborough, Ontario, Canada. M1G 1C3 A193. If f (x;y) is a convex function in x for each xed y, and a convex function in y for each xed x, is f (x;y ) necessarily a convex function in x and y ? A194. Let H be the orthocentre (point where the altitudes meet) of a triangle ABC . Show that if AH : BH : CH = BC : CA : AB , then the triangle is equilateral. A195. Compute tan 20 tan 40 tan 60 tan 80. A196. Show that r + ra + rb + rc  4K , where r, ra, rb, rc, and K are the inradius, exradii, and area respectively of a triangle ABC . 2

2

2

2

Challenge Board Problems Editor: Ravi Vakil, Department of Mathematics, One Oxford Street, Cambridge, MA, USA. 02138-2901 C70. Prove that the group of automorphisms of the dodecahedron is S , the symmetric group on ve letters, and that the rotation group of the dodecahedron (the subgroup of automorphisms preserving orientation) is A . C71. Let L , L , L , L be four general lines in the plane. Let pij be the intersection of lines Li and Lj . Prove that the circumcircles of the four triangles p p p , p p p , p p p , p p p are concurrent. C72. A nite group G acts on a nite set X transitively. (In other words, for any x; y 2 X , there is a g 2 G with g  x = y .) Prove that there is an element of G whose action on X has no xed points. 5

5

1

12

23

31

2

23

3

34

4

42

34

41

13

41

12

24

45

PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 "11" or A4 sheets of paper. These may be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later than 1 September 1997. They may also be sent by email to [email protected]. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication. 1 2

2201. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABCD is a convex quadrilateral, and O is the intersection of its diagonals. Let L; M; N be the midpoints of DB; BC; CA respectively. Suppose that AL; OM; DN are concurrent. Show that either AD k BC

or

[ABCD] = 2[OBC ];

where [F ] denotes the area of gure F . 2202. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Suppose than n  3. Let A ; : : : ; An be a convex n-gon (as usual with interior angles A ; : : : ; An ). Determine the greatest constant Cn such that 1

1

n 1 n X X 1 :  C n A  , Ak k k k =1

Determine when equality occurs.

=1

46

2203. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let ABCD be a quadrilateral with incircle I . Denote by P , Q, R and S , the points of tangency of sides AB , BC , CD and DA, respectively with I . Determine all possible values of \(PR;QS ) such that ABCD is cyclic.  2204. Proposed by Sefket Arslanagic, Berlin, p Germany. For triangle ABC such that R(a + b) = c ab, prove that 3 a: r < 10

Here, a, b, c, R, and r are the three sides, the circumradius and the inradius of 4ABC . 2205. Proposed by Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA. Find the least positive integer n such that the expression

sinn A sinn B sinn C +2

+1

has a maximum which is a rational number (where A, B , C are the angles of a variable triangle). 2206. Proposed by Heinz-Jurgen Sei ert, Berlin, Germany. Let a and b denote distinct positive real numbers. (a) Show that if 0 < p < 1, p 6= , then 1 2

1 ,apb ,p + a ,p bp < 4p(1 , p)pab + (1 , 4p(1 , p)) a + b : 2 2 1

1

(b) Use (a) to deduce Polya's  Inequality:

a , b < 1 2pab + a + b  : log a , log b 3 2

Note: \log" is, of course, the natural logarithm. 2207. Proposed by Bill Sands, University of Calgary, Calgary, Alberta. Let p be a prime. Find all solutions in positive integers of the equation:

2 + 3 = 5:

a b

p

47

2208. Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. 1. Find a set of positive integers fx; y;z; a; b; c; kg such that y z = a +k z x = b +k x y = c +k 2

2

2

2

2

2

2

2

2

2

2

2

2. Show how to obtain an in nite number of distinct sets of positive integers satisfying these equations.

2209. Proposed by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain. Let ABCD be a cyclic quadrilateral having perpendicular diagonals crossing at P . Project P onto the sides of the quadrilateral. 1. Prove that the quadrilateral obtained by joining these four projections is inscribable and circumscribable. 2. Prove that the circle which passes through these four projections also passes through the mid-points of the sides of the given quadrilateral.

2210?. Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon,

Madrid, Spain. Given a = 1, the sequence fan g (n = 1; 2; : : : ) is given recursively by 0

n  n   n  n a , a + a , : : :  n n , n , n n,1 n,2 b n c ab n2 c = 0: 1

2

2

Which terms have value 0? 2211. Proposed by Bill Sands, University of Calgary, Calgary, Alberta. Several people go to a pizza restaurant. Each person who is \hungry" wants to eat either 6 or 7 slices of pizza. Everyone else wants to eat only 2 or 3 slices of pizza each. Each pizza in the restaurant has 12 slices. It turns out that four pizzas are not sucient to satisfy everyone, but that with ve pizzas, there would be some pizza left over. How many people went to the restaurant, and how many of these were \hungry"?

48

2212. Proposed by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Let S = f1; 2; : : : ; ng where n  3. (a) In how many ways can three integers x, y , z (not necessarily distinct) be chosen from S such that x + y = z ? (Note that x + y = z and y + x = z are considered to be the same solution.) (b) What is the answer to (a) if x, y , z must be distinct? 2213. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

A generalization of problem 2095 [1995: 344, 1996: 373]. Suppose that the function f (u) has a second derivative in the interval (a; b), and that f (u)  0 for all u 2 (a; b). Prove that 1. (y , z )f (x) + (z , x)f (y ) + (x , y )f (z ) > 0 for all x; y;z 2 (a; b),

z 0 and (m; n) = 1. Since the reverse of the G.P. a; ar; ar is another G.P. ar ; ar; a, we may also assume that jrj  1 and so 0 < m  jnj. From a(1 + r + r ) = 111 we get a(m + mn + n ) = 111m . Since clearly (m + mn + n ; m ) = 1 we have m ja. Letting a = km where k is an integer we then get k(m + mn + n ) = 111 which implies kj111. Since m + mn + n > 0 and 111 = 3  37, we have k = 1; 3; 37, or 111. Note that m + mn + n = m + jnj(m + jnj)  m . Case [1] If k = 1, then m + mn + n = 111 =) m  111 =) m  10. When m = 1, a = 1 and from n + n = 110 we get n = 10; ,11. Thus r = 10, ,11 and we obtain the solutions: 111 = 1 + 10 + 100 = 1 , 11 + 121: (2) For 2  m  9 it is easily checked that the resulting quadratic equation in n has no integer solutions. When m = 10, a = 100 and from n + 10n = 11 we get n = 1, ,11. Since m  jnj, n = ,11 and r = , yielding the solution: 111 = 100 , 110 + 121: (3) 2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

11 10

Case [2] If k = 3, then m + mn + n = 37 =) m  37 =) m  6. Quick checkings reveal that there are no solutions for m = 1, 2, 5, 6. When m = 3, a = 27 and from n + 3n = 28 we get n = 4, ,7. Thus r = ; , yielding the solutions: 111 = 27 + 36 + 48 = 27 , 63 + 147: (4) 2

2

2

4 3

7 3

2

54 When m = 4, a = 48 and from n + 4n = 21 we get n = 3, ,7. Since m  jnj, n = ,7 and r = , yielding the solution: 2

7 4

111 = 48 , 84 + 147:

(5)

Case [3] If k = 37, then m + mn + n = 3 =) m  3 =) m = 1 =) a = 37 and from n + n = 2 we get n = 1, ,2. Thus r = 1 or ,2 2

2

2

2

yielding the solutions:

111 = 37 + 37 + 37 = 37 , 74 + 148:

(6)

Case [4] If k = 111, then m + mn + n = 1 =) m  1 =) m = 1 =) a = 111, and from n + n = 0 we get n = ,1 as n 6= 0. Thus r = ,1 2

2

2

2

and we get the solution

111 = 111 , 111 + 111: (7) Reversing the summand in (2) , (7) and noting that two of them are \symmetric", we obtain seventeen solutions in all:

111 = = = = = =

111 + 0 + 0 1 , 11 + 121 121 , 110 + 100 27 , 63 + 147 147 , 84 + 48 148 , 74 + 37

= = = = = =

1 + 10 + 100 121 , 11 + 1 27 + 36 + 48 147 , 63 + 27 37 + 37 + 37 111 , 111 + 111:

= = = = =

100 + 10 + 1 100 , 110 + 121 48 + 36 + 27 48 , 84 + 147 37 , 74 + 148

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain, (who assumed r 6= 0 and found sixteen solutions); F.J. FLANIGAN, San Jose State University, San Jose, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; (both of them found all seventeen solutions). There were also two incomplete and twenty-three incorrect solutions submitted! Among these twenty-three, thirteen submissions claimed six solutions; six submissions claimed ve solutions; two submissions claimed six or nine solutions; one submission claimed two solutions, and one submission claimed one solution only. Most of the errors were the result of assuming by mistake that a(1 + r + r ) = 111 =) 1 + r + r must be an integer. 2

2

2105. [1996: 34] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK Find all values of  for which the inequality 2(x + y + z ) + 3(1 + 3)xyz  (1 + )(x + y + z )(yz + zx + xy) 3

3

3

holds for all positive real numbers x; y;z .

55 Solution by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. On setting x = y = 1 and z = 0, we obtain 4  (1 + )2 and thus nd that  must be  1. We now show that the inequality holds for all   1. First, if  = 1, the inequality reduces to

x + y + z + 6xyz  (x + y + z)(yz + zx + xy); which is equivalent to the special case n = 1 of the known Schur inequality xn(x , y)(x , z) + yn(y , z)(y , x) + zn(z , x)(z , y)  0; true for all real n, and which has come up many times in this journal. The rest will follow by showing that for all  < 1, (1 + )(x + y + z)(yz + zx + xy ) , 3(1 + 3)xyz  (1 + 1)(x + y + z)(yz + zx + xy) , 3(1 + 3)xyz: (1) 3

3

3

[Editor's note: rewrite the original inequality as

2(x + y + z )  (1 + )(x + y + z )(yz + zx + xy) , 3(1 + 3)xyz; then (1) says that the right hand side is largest when  = 1, so doing the case  = 1 would be enough.] But (1) can be written (1 , )[(x + y + z )(yz + zx + xy) , 9xyz]  0 which [after cancelling the positive factor 1 , ] is a known elementary in3

3

3

equality, equivalent to Cauchy's inequality

  (x + y + z ) x1 + y1 + z1  9:

Also solved by CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; HEINZ JURGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and HOE TECK WEE, Singapore. There were three incorrect solutions sent in.

2106. [1996: 34] Proposed by Yang Kechang, Yueyang University, Hunan, China. A quadrilateral has sides a; b;c; d (in that order) and area F . Prove that 2a + 5b + 8c , d  4F: 2

When does equality hold?

2

2

2

56 Solution by Federico Ardila, student, Massachusetts Institute of Technology, Cambridge, Massachusetts, USA. Let ABCD be the quadrilateral with AB = a, BC = b, CD = c, and DA = d. We can assume, without loss of generality, that AC = 1. Therefore, we can locate the quadrilateral in a system of Cartesian coordinates where

A = (0; 0); B = (p; q); C = (1; 0); D = (r; s): We assume that ABCD is simple so that its area is well-de ned. If ABCD

is not convex we can make it convex and keep the side lengths the same while increasing the area. This means that we will be done if we can show that the result is true for convex quadrilaterals. It's also clear from this that if the result is true for convex quadrilaterals, then equality cannot hold for non-convex quadrilaterals. Therefore, assume q < 0 and s > 0. Now note that

2(p + q ) + 5((p , 1) + q ) 7p , 10p + 5 + 7q 7 ,p ,  , + 5 + 7q 7q + ;

2a + 5b = = = 2a + 5b  2

2

2

and

2

2

2

2

2

2

2

5 2 7 10 7

2

25 7

2

(2)

8c , d = 8((r , 1) + s ) , (r + s ) = 7r , 16r + 8 + 7s = 7 ,r ,  , + 8 + 7s 8c , d  7s , : Combining (??) and (??), we get 2a + 5b + 8c , d  7, q + 7s + ,  = 7q + + 7s +  ,   ,    = 7 jqj , + 2jq j + 7 jsj , + 2jsj  2,jqj + jsj = 4(ABC ) + 4(CDA): 2a + 5b + 8c , d  4F; 2

2

2

2

2

2

2

2

2

2

8 7 8 7

64 7

2

(3)

2 7

1 7

2

2

1 7

2

1 7

2

2

2

2

2

2

2

2

2

2

2

1 7

2

(4)

2

as we wished to prove (where ABC and CDA refer to the areas of the two triangles ABC and CDA respectively). For equality to hold (when A = (0; 0) and C = (1; 0)), it must hold in steps (??), (??), and (??). Therefore p = , r = , q = , , and s = . Thus, in general, equality holds if and only if ABCD is directly similar to quadrilateral A B C D , where 5 7

8 7

1 7

1 7

0

0

0

0

A = (0; 0); B = ( ; , ); C = (1; 0); D = ( ; ): 0

0

5 7

1 7

0

0

8 7

1 7

57 Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck,   Y,  Ferris State University, Big Rapids, Michigan, Austria; VACLAV KONECN USA; KEE-WAI LAU, Hong Kong; and the proposer.  Y makes the observation that the quadrilateral is cyclic and KONECN that \ABD = 135 . The proposer makes the early observation that the maximum area for a quadrilateral with xed sides occurs when it is cyclic and uses properties of cyclic quadrilaterals in the proof. He also generalizes the result to an inequality which JANOUS uses in his proof and which appears in the Addenda to the Monograph \Recent Advances in Geometric Inequalities" by D. S. Mitrinovic et al. in I. Journal of Ningbo University 4, No. 2 (Dec. 1991), 79{145.

2107. [1996: 34] Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. Triangle ABC is not isosceles nor equilateral, and has sides a; b; c. D and E are points of BA and CA or their productions so that BD = CE = a. D and E are points of CB and AB or their productions so that CD = AE = b. Show that D E k D E . Solution by Florian Herzig, student, Perchtoldsdorf, Austria. Let S be the intersection of AB and D E . [Editor's comment by Chris Fisher. Even though there seem to be two choices for each Di and Ei , no solver had any trouble choosing the positions that make the result correct; furthermore, it must have been \obvious" to eveyone but me that AB is not parallel to D E , so that S exists. Alas, perhaps I need stronger glasses.] Then CS is the bisector of \ACB , since CE = CB and CA = CD . Therefore 1

1

1

1

2

2

2

2

1

1

2

2

2

2

1

1

1

2

D S = BD , BS = a = BSb , AS = a ; SE AE , AS b BS = a . It then follows that D E kD E , since since AS b E S = CE = a = D S : SD CD b SE 1

1

2

2

1

1

1

2

2

1

2

2

1

2

Also solved by FEDERICO ARDILA, student, MassachusettsInstitute of Technology, Cambridge, Massachusetts, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; JOHN G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus  Y,  Ferris State University, Big Rapids, Michigan, USA; tria; VACLAV KONECN P. PENNING, Delft, the Netherlands; JOEL SCHLOSBERG, student, Hunter

58 College High School, New York NY, USA; GEORGE TSAPAKIDIS, Agrinio, Greece; MELITIS D. VASILIOU. Elefsis, Greece; and the proposer. Janous adds the observation that D E and D E are not only parallel, but their lengths are in the ratios a : b (as is clear from the featured solution). 1

1

2

2

2108. [1996: 34] Proposed by Vedula N. Murty, Andhra University, Visakhapatnam, India. Prove that

s a + b + c  1 3 (b + c) (c + a) (a + b) ; 3 4 abc where a; b; c > 0. Equality holds if a = b = c. 2

2

2

Solution by Florian Herzig, student, Perchtoldsdorf, Austria, (modi ed slightly by the editor). By the arithmetic-geometric mean inequality we have

p a b + ab + b c + bc + c a + ca  6 6 a b c = 6abc; 2

2

2

2

2

2

6

6 6

which implies

9(a b + ab + b c + bc + c a + ca + 2abc)  8(a b + ab + b c + bc + c a + ca + 3abc); 2

2

2

or

2

2

2

2

2

2

2

2

2

9(a + b)(b + c)(c + a)  8(a + b + c)(ab + bc + ca) = 4(a + b + c)(a(b + c) + b(c + a) + c(a + b)): Using the arithmetic-geometric mean inequality again, we then have

3 (a + b)(b + c)(c + a) 4  (a + b + c)  a(b + c) + b(c 3+ a) + c(a + b)  (a + b + c)pabc(a + b)(b + c)(c + a)

(1)

From (1) it follows immediately that

s

1 3 (a + b) (b + c) (c + a)  a + b + c : 4 abc 3 Clearly, equality holds if a = b = c. [Ed. In fact, if equality holds, then from (1) we have a(b + c) = b(c + a) = c(a + b). The rst equality implies 2

2

2

59

a = b and the second one implies b = c. Thus, equality holds in the given inequality if and only if a = b = c. This was observed by about half of the

solvers.] Also solved by FEDERICO ARDILA, student, MassachusettsInstitute of Technology, Cambridge, Massachusetts, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; HAN PING DAVIN CHOR, Student, Cambridge, MA, USA; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; TIM CROSS, King Edward's School, Birmingham, England; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; KEE-WAI LAU, Hong  Kong; JUAN-BOSCO ROMERO MARQUEZ, Universidad de Valladolid, Valladolid, Spain; JOEL SCHLOSBERG, student, Hunter College High School,  New York NY, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. Janous commented that upon the transformation

a ! a1 ; b ! 1b ; c ! 1c ;

the given inequality can be shown to be equivalent to

r ab + bc + ca 3

s

 3 (a + b)(b +8 c)(c + a)

which is known in the literature as Carlson's inequality (cf. eg. P.S. Bullen, D.S. Mitrinovic and P.M. Vasic, \Means and Their Inequalities", Dordrechf, 1988. An anonymous reader commented that in this form, the inequality was problem 3 of the 1992 Austrian-Polish Mathematics Competition and has appeared in Crux before (see [1994:97; 1995:336-337]). Several solvers showed that the given inequality is equivalent to various other trigonometric inequalities involving a triangle XY Z , for example

 Z  3p3 cos 2 cos 2 cos 2  8 p3 3 sin X + sin Y + sin Z  2 ; X 

or

Y 

etc. These inequalities can be found in \Geometric Inequalities" by O. Bottema et al.

2109. [1996: 34] Proposed by Victor Oxman, Haifa, Israel.

In the plane are given a triangle and a circle passing through two of the vertices of the triangle and also through the incentre of the triangle. (The incentre and the centre of the circle are not given.) Construct, using only an unmarked ruler, the incentre.

60 Solution by P. Penning, Delft, the Netherlands. Let the triangle be ABC , and , the circle passing through B; C and the incentre. The angles of the triangle are denoted by the symbol for the corresponding vertex A; B or C . A

B

B0

..

X

C0 S

C ,

ANALYSIS: Let point S be the intersection of the circumcircle and the angular bisector through A. The arcs SB and SC of the circumcircle are now equal and so are the chords SB and SC . Introduce X on AS such that SX = SB = SC: \BSA = \BCA = C ; 4XBS is isosceles with SX = SB, so \XBS = 90 , C=2: \CBS = \CAS = A=2; \XBC = 90 , C=2 , A=2 = B=2: So BX is the angular bisector at vertex B , and X must be the incentre. The circle , apparently has the point S as centre. [It does, see Roger A. Johnson, Modern Geometry, 292]. [Editor's note: If either AB or AC is tangent to ,, then they both are and AB = AC . Suppose AB is tangent to ,. Then \SBA =  , so \BSA + \SAB =  . Since SC = SB; \BCS = \SBC = \SAC = \SAB . Therefore, \ACS = \ACB + \BCS = \BSA + \SAB =  and AC is tangent to ,. In addition, tangents to a circle from an exterior point are equal, so AB = AC:] So if AB 6= AC , neither line is tangent to ,. Let B 0 and C 0 be the other intersections of AB , respectively AC , with ,. There is mirror-symmetry with respect to the line AS : , remains ,; AB re ects into AC and AC re ects into AB ; B and C 0 are mirror-images and so are C and B 0 . The side BC becomes B0C 0; as a consequence they must intersect on the mirror-line AS. CONSTRUCTION: Find the other two intersections, B 0 and C 0 , of AB and AC with the circle ,. The intersection of BC and B 0 C 0 is M . The incentre is the inter2

2

2

61 section of AM and ,. COMMENT: The construction fails if ABC is isosceles, with AB = AC . In that case , touches both AB and AC in B and C respectively. M is now the midpoint of BC , but that cannot be found with unmarked ruler. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,  UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; VACLAV   KONECNY, Ferris State University, Big Rapids, Michigan, USA; KEE-WAI LAU, Hong Kong; and the proposer.

2110. [1996: 35] Proposed by Jordi Dou, Barcelona, Spain.

Let S be the curved Reuleaux triangle whose sides AB , BC and CA are arcs of unit circles centred at C , A and B respectively. Choose at random (and uniformly) a point M in the interior and let C (M ) be a chord of S for which M is the midpoint. Find the length ` such that the probability that C (M ) > ` is 1=2. Solution by the proposer. Let  be the locus of the mid-point M of segments of constant length  , whose ends S and S move on the boundary of S . For the points M inside  the chords bisected by M are greater than  . (?) The area contained between S and  is   . It is sucient to show that 1

2

2

4

 ` = [S] : 4 2 2

Since [S ] =  ,

p3

2

2

4

, we will have

p3 ! 12  , `= ' 0:67:  Brief proof of the assertion (?) (Special Case of Holditch's Theorem) The ends S , S of the chord of constant length move along the contour of the closed curve S . The mid-point M describes the curve . Let S = (x ; y ); S = (x ; y ), and M = ( x1 x2 ; y1 y2 ). Suppose that x , y , x , y are functions of t such that for t  t  t , we have S , S describing S. Z t1 Z t1 Z t1 [S] = y dx = y dx = 12 (y dx + y dx ) t0 Zt0 t1 Zt0t1 1 [] = y dx = 4 (y + y )(dx + dx ): 1

1

1

1

2

1

2

1

2

2

+ 2

2

2

+ 2

0

1

1

2

1

Then

t0

1

2

2

t0

1

1

2

1

1

2

Z t1 1 [S] , [] = 4 (y , y )(dx , dx ): t0 2

1

2

1

2

2

62 Substitute X = x , x , Y = y , y , giving 2

1

2

1 Z t1

[S] , [] = 4

t0

1

Y dX = 4  ; 2

since (X;Y ) describes a circle of radius  .

2111. [1996: 35] Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. Does there exist a function f : N ,! N (where N is the set of positive integers) satisfying the three conditions: (i) f (1996) = 1; (ii) for all primes p, every prime occurs in the sequence f (p), f (2p), f (3p); : : : ; f (kp); : : : in nitely often; and (iii) f (f (n)) = 1 for all n 2 N ? I. Solution by Shawn Godin, St. Joseph Scollard Hall, North Bay, On-

tario.

Yes, a function does exist that satis es the three conditions. It is:  condition  holds; f (x) = p1i ifotherwise ;

Q

where condition  is: if the prime factorization of x is x = pei i , there exists a power ei such that ei > 2 and ei > ej for all j 6= i. For example, 109850 has condition , since 109850 = 2  5  13 and the power of 13 is bigger than 2 and bigger than all other powers in the factorization; thus f (109850) = 13. Now  f satis es condition (i) since f (1996) = f (2  499) = 1;  f satis es condition (ii) because for any two primes p and q , f (xi) = q for every xi = q ip, i = 3; 4; : : : ;  f satis es condition (iii) since for all n either f (n) = 1 or f (n) = pi for some prime pi, and in either case f (f (n)) = 1. 2

3

2

II. Solution by Chris Wildhagen, Rotterdam, the Netherlands. For each n 2 N let qn be the nth prime and b(n) be the number of 1's in the binary representation of n. De ne f : N ! N as follows: q m = pn with p prime and n  2; f (m) = 1b n ifelse : (

)

63 Clearly f satis es the three given conditions. Also solved by FEDERICO ARDILA, student, Massachusetts Institute of Technology, Cambridge, Massachusetts, USA; MANSUR BOASE, student, St. Paul's School, London, England; CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; KEITH EKBLAW, Walla Walla, Washington, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; DAVID E. MANES, State University of New York, Oneonta, NY, USA; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; and the proposer. There were three incorrect solutions sent in. Most solvers gave a variation of Solution I.

2112. [1996: 35] Proposed by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. Find a four-digit base-ten number abcd (with a 6= 0) which is equal to aa + bb + cc + dd. Solution by Cyrus Hsia, student, University of Toronto, Toronto, Ontario (modi ed slightly by the editor). We rst stipulate that 0 = 1. Let m = abcd, s = aa + bb + cc + dd and assume that m = s. Clearly, 10  m < 10 . If x  6 for any x 2 fa; b; c; dg then s  6 > 10 which is a contradiction. So a; b;c; d  5. If x < 5 for all x 2 fa; b; c; dg, then s  4  4 = 1024 and furthermore a = b = c = d = 4 is the only combination for which s  10 . However, in this case, s = 1024 6= 4444 = m. Hence x = 5 for some x 2 fa; b; c; dg. We cannot have more than one 5 or else s  2  5 = 6250 would imply that some digit of m is at least 6. Hence, we have exactly one 5. Since s > 5 = 3125 and s  5 + 3  4 = 3893 < 4000, we must have a = 3. Thus, s = 5 +3 +xx +yy = 3152+xx+yy where x; y 2 fa; b; c; dg. Without loss of generality, we may assume that 0  y  x  4. If x = 0, then y = 0 and s = 3154 which has no 0 among its digits. If x = 1, then y = 0; 1 and s = 3154 while m has no 4 among its digits. If x = 2, then s = 3156 + y y and it is easily veri ed that s has no 2 among its digits for y = 0; 1; 2. If x = 3, then s = 3179 + y y and it is easily veri ed that s has no 5 among its digits for y = 0; 1; 2; 3. If x = 4, then s = 3408 + y y and, again, it is readily checked that s has no 5 among its digits when y = 0; 1; 2; 4. However, when y = 3, s = 3435 which is clearly a solution. To summarize, m = 3435 is the only solution. 0

3

6

4

4

4

3

5

5

5

5

3

4

64 Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; FEDERICO ARDILA, student, Massachusetts Institute of Technology, Cambridge, Massachusetts, USA; MANSUR BOASE, student, St. Paul's School, London, England; CHRISTOPHER J. BRADLEY, Clifton College,  OCHOA, Logro~no, Spain; Bristol, UK; MIGUEL ANGEL CABEZON THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; TIM CROSS, King Edward's School, Birmingham, England; JEFFREY K. FLOYD, Newnan, Georgia, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.   Y,  KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV KONECN Ferris State University, Big Rapids, Michigan, USA; LUKE LAMOTHE, student, St. Joseph Scollard Hall S.S., North Bay, Ontario, KATHLEEN E. LEWIS, SUNY Oswego. Oswego, NY, USA; DAVID E. MANES, State University of New York, Oneonta, NY, USA; JOHN GRANT McLOUGHLIN, Okanagan University College, Kelowna British Columbia; P. PENNING, Delft, the Netherlands; CORY PYE, student, Memorial University of Newfoundland, St. John's, Newfoundland; JOEL SCHLOSBERG, student, Hunter College High School, New York NY, USA; ROBERT P. SEALY, Mount Allison University,  Sackville, New Brunswick; HEINZ-JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. Of the twenty-six solvers (including the proposer), nine of them only gave the answer 3435. About half of all the solvers claimed, with or without proof, that 3435 is the only solution. Chronis, Hess, and Janous found the answer by computer search. Hess remarked that no other solutions were found for the present problem and the corresponding problem on 5{digit integers. Janous investigated the corresponding n{digit problem of nding nX , all n{digit integers an, an, : : :a a which equal aakk . He showed that k a necessary condition is n  10. For n > 1, he conducted an extensive, but not exhaustive, computer search, which revealed no solutions other than the one found by all the solvers! He made a guess that 1 and 3435 are the only integers with the desired property. Can any reader prove or disprove this? 1

1

2

1

0

=0

Crux Mathematicorum

Founding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell Editors emeriti / Redacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem

Founding Editors / Redacteurs-fondateurs: Patrick Surry & Ravi Vakil Editors emeriti / Redacteurs-emeriti: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

65

THE ACADEMY CORNER No. 9

Bruce Shawyer All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 In this issue, we feature a university undergraduate mathematics competition. We invite your solutions, especially from university students. Please send me your nice solutions.

Undergraduate Mathematics Competition September 18, 1996 Answer as many questions as you can. Complete solutions carry more credit than scattered comments about many problems.

1. If n is any integer, show that n5 , n is divisible by 5. 2. A line l with slope m = 2 cuts the parabola y 2 = 8x to form a chord. Find the equation of l if the midpoint of the chord lies on x = 4. 3. Show that three tangents can be drawn from the origin to the curve given by y = x3 , 13x2 + 10x , 36.

n2 n2  2   + 1 + : : : + nn = 2nn , for all positive

4. Prove that 0 integers n. 5. Show that

Z =2

sin13 x : dx = 13 13 4 sin x + cos x 0 6. In triangle ABC , \B = 3 \A.

If the sides opposite to the angles A; B; C have lengths a; b; c, respectively, prove that ac2 = (b , a)2 (b + a).

66

THE OLYMPIAD CORNER No. 180

R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. We begin this number of the Corner with the problems of the 3rd Mathematical Olympiad of the Republic of China (Taiwan). The contest was written April 14 and 15, 1994. Many thanks go to Richard Nowakowski, Canadian Team Leader to the IMO in Hong Kong for collecting these problems and many others.

3rd MATHEMATICAL OLYMPIAD OF THE

REPUBLIC OF CHINA (Taiwan) First Day | April 14, 1994

1. Let ABCD be a quadrilateral with AD = BC and \A + \B = 120. Three equilateral triangles 4ACP , 4DCQ and 4DBR are drawn

on AC , DC and DB away from AB . Prove that the three new vertices P , Q and R are collinear. 2. Let a, b, c be positive real numbers, be a real number. Suppose that

f ( ) = abc(a + b + c ); g( ) = a +2(b + c , a) + b +2(a , b + c) + c +2(a + b , c): Determine jf ( ) , g ( )j. 3. Let a be a positive integer such that (51994 , 1) j a. Show that the expression of the number a in the base 5 contains at least 1994 digits di erent from zero.

Second Day | April 15, 1994

4. Prove that there are in nitely many positive integers n having the following property: for every arithmetic progression a1 ; a2 ; : : : ; an of integers with n terms, both the mean and standard deviation of the set fa1; a2; : : : ; ang are integers. (Note: For any set fx1; x2; : : : ; xn g of real numbers, the mean of the set is de ned to be the number x = x1 + x2 +n    + xn

67 and the standard deviation of the set is de ned to be the number

sP

(xi , x)2 :

n

5. Let X = f0; a; b; cg and M (X ) = ff j f : X ! X g be the set of all functions from X into itself. Here, an addition table of X is given as follows:

 0 a b c 0 0 a b c a a 0 c b b b c 0 a c c b a 0 (1) If S = ff 2 M (X ) j f (x  y  x) = f (x)  f (y )  f (x); 8x;y 2 X g, determine the number of elements of S . (2) If I = ff 2 M (X ) j f (x  x) = f (x)  f (x); 8x 2 X g, determine the number of elements of I . 6. For ,1  x  1 de ne h p n  p ni Tn (x) = 21n x + 1 , x2 + x , 1 , x2 (1) Prove that, for ,1  x  1, Tn (x) is a monic polynomial of degree n in the x-variable and the maximum value of Tn (x) is 2n1,1 . (2) Suppose that p(x) = xn +an,1 xn,1 +  +a1 x+a0 is a monic polynomial with real coecients such that for all x in ,1  x  1, p(x) > , 2n1,1 . Prove that there exists x in ,1  x  1 such that p(x )  2n1,1 . Last issue we gave Five Klamkin Quickies. Here we give his \Quick" solutions plus another 5 problems. Many thanks to Murray S. Klamkin, the University of Alberta.

ANOTHER FIVE KLAMKIN QUICKIES October 21, 1996

6. Determine the four roots of the equation x + 16x , 12 = 0. 7. Prove that the smallest regular n-gon which can be inscribed in a 4

given regular n-gon is one whose vertices are the midpoints of the sides of the given regular n-gon. 8. If 311995 divides a2 + b2, prove that 311996 divides ab. 9. Determine the minimum value of

S

=

p pac (

(

b, d,

+ 1)2 + 2(

+ 1)2 + 2(

c a

2)2 + (

2)2 + (

+ 3)2 +

p pbd

+ 3)2 +

c, a,

+ 1)2 + 2(

(

(

+ 1)2 + 2(

2)2 + (

d

+ 3)2 ) +

b

2)2 + (

+ 3)2

68 where a, b, c, d are any real numbers. 10. A set of 500 real numbers is such that any number in the set is greater than one- fth the sum of all the other numbers in the set. Determine the least number of negative numbers in the set. We will give the solutions to these in the next issue so you can have some fun looking for the answers. Next we give his solutions to the ve Quickies we gave last issue.

FIVE KLAMKIN QUICKIES October 21, 1996

1. For x; y;z> 0, prove thatx 1

1

(i) 1 + (x + 1)  x 1 + x(x + 2) y , (ii) [(x + y )(x + z )] [(y + z )(y + x)] [(z + x)(z + y )]z  [4xy ]x [4yz ]y [4zx]z . Solution. Both inequalities will follow by a judicious application of the weighted arithmetic-geometric mean inequality (W{A.M.{G.M.) which for three weights is  

uavbwc  aua++bvb ++ccw where a; b; c; u; v;w  0.

a+b+c

;

(i) The inequality can be rewritten in the more attractive form

x  x+1 ; 1 + x1  1 + x +1 1



and which now follows by the W{A.M.{G.M.

x ( 1 + x ,1 + 1  )x+1  x+1 1 1 x 1+ x  = 1+ x+1 : 1+x



form

(ii) Also, the inequality here can be rewritten in the more attractive

 2x z+x  2y x+y  2z y+z  1: z +x x+y y+z

But this follows by applying the W{A.M.{G.M. to

1=

  X [z + x] z 2+x x  [z + x]:

X

2. If ABCD is a quadrilateral inscribed in a circle, prove that the four lines joining each vertex to the nine point centre of the triangle formed by the other three vertices are concurrent.

69 Solution. The given result still holds if we replace the nine point centres by either the orthocentres or the centroids. A vector representation is particularly a propos here, since (with the circumcentre O as an origin and F denoting the vector from O to any point F ) the orthocentre Ha, the nine point centre Na, the centroid Ga of 4BCD are given simply by Ha = B + C + D, Na = (B + C + D)=2, Ga = (B + C + D)=3, respectively, and similarly for the other three triangles. Since the proofs for each of the three cases are practically identical, we just give the one for the orthocentres. The vector equation of the line La joining A to Ha is given by La = A + a [B + C + D , A] where a is a real parameter. By letting a = 1=2, one point on the line is [A + B + C + D]=2 and similarly this point is on the other three lines. For the nine point centres, the point of concurrency will be 2[A + B + C + D]=3, while for the centroids, the point of concurrency will be 3[A + B + C + D]=4. 3. How many six digit perfect squares are there each having the property that if each digit is increased by one, the resulting number is also a perfect square? Solution. If the six digit square is given by

m2 = a  105 + b  104 + c  103 + d  102 + e  10 + f; then

n2 = (a+1)105 +(b+1)104 +(c+1)103 +(d+1)102 +(e+1)10+(d+1); so that

n2 , m2 = 111; 111 = (111)(1; 001) = (3  37)(7  11  13):

Hence,

n + m = di and n , m = 111; 111=di where di is one of the divisors of 111; 111. Since 111; 111 is a product of ve primes it has 32 di erent divisors. But since we must have di > 111; 111=di, there are at most 16 solutions given by the form lm = 12 (di , 111; 111=di). Then since m2 is a six digit number, we must have p 632:46  200 10 < 2m < 2; 000: On checking the various divisors, there are four solutions. One of them corresponds to di = 3  13  37 = 1; 443 so that m = 12 (1; 443 , 7  11) = 683 and m2 = 466; 489. Then, 466; 489 + 111; 111 = 577; 600 = 7602 . The others are given by the table

di

3  7  37 = 777 3  11  37 = 1; 221 7  11  13 = 1; 001

m

m2 317 100; 489 565 319; 225 445 198; 025

n2 211; 600 430; 336 309; 136

n

460 656 556

70

4. Let viwi, i = 1; 2; 3; 4, denote four cevians of a tetrahedron v1v2v3v4 which are concurrent at an interior point P of the tetrahedron. Prove that pw1 + pw2 + pw3 + pw4  max viwi  longest edge: Solution. We choose an origin, o, outside of the space of the tetrahedron and use the set of 4 linearly independent vectors Vi = ovi as a basis. Also the vector from o to any point q will be denoted by Q. The interior pointPp is then given by P = x1 V1 + x2 V2 + x3 V3 + x4 V4 where xi > 0 and i xi = 1. It now follows that Wi = P1,,xxi iVi (for other properties of concurrent cevians via vectors, see [1987: 274{275]) and then that

P , xiVi xi(P , Vi) X Vj , Vi pwi = 1 , x , P = 1 , x = xi xj 1 , x ; i i i j P , Vi X Vj , Vi P , xiVi viwi = 1 , x , Vi = 1 , x = xj 1 , x : i i i j

Summing

X X Vj , Vi X pwi = xi xj 1 , x = xi(viwi)  max i viwi; i i i j i and with equality only if vi wi is constant. Also, X  xj  max jVr , Vij = max viwi  r jVr , Vi j: j 6=i 1 , xi r X

Finally,

X

pwi  max r;s jVr , Vs j: i viwi  max

Comment: In a similar fashion, it can be shown that the result generalizes to n-dimensional simplexes. The results for triangles are due to Paul Erdos, } Amer. Math. Monthly, Problem 3746, 1937, p. 400; Problem 3848, 1940, p. 575. 5. Determine the radius r of a circle inscribed in a given quadrilateral if the lengths of successive tangents from the vertices of the quadrilateral to the circle are a; a; b; b; c; c; d; d, respectively. Solution. Let 2A, 2B , 2C , 2D denote the angles between successive pairs of radii vectors to the points of tangency and let r be the inradius. Then

r = tana A = tanb B = tanc C = tand D :

71 Also, since A + B + C + D =  , we have tan(A + B ) = tan(C + D) = 0; or so that Finally,

tan A + tan B tan C + tan D 1 , tan A tan B + 1 , tan C tan D = 0;

r(a + b) + r(c + d) = 0: r2 , ab r2 , cd + cda + dab r2 = abc +a bcd +b+c+d :

Now we turn our attention to solutions by our readers to problems given in the September 1995 number of the Corner where we gave the 16th Austrian Polish Mathematics Competition [1995: 221{222].

16TH AUSTRIAN POLISH MATHEMATICS

COMPETITION

First Day | June 30, 1993

Time: 4.5 hours (individual competition)

1. Determine all natural numbers x; y  1 such that 2x , 3y = 7.

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;  Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Amengual's solution. First of all we note that both numbers x and y must be even. Suppose to the contrary that one of the numbers is odd. If x is odd, the number 2x + 1 (which we have from the factorization x A + 1 = (A + 1)(Axx,1 ,y Ax,2 +    , A + 1) that 2x + 1) is a multiple of 3. Consequently 2 , 3  ,1 mod 3, while 7  1 mod 3, and the given equation is invalid mod 3. If y is odd, we will use the modular technique as in the previous case, but this time modulo 8. We have 32  1 mod 8. It follows that 32k+1  3 mod 8 for k = 0; 1; 2; : : : . Consequently 3y + 7  0 mod 8 and since 2x  3y + 7 we must have x2 2.yIf x = 1, we have 2 , 3y = 7, which is impossible. If x = 2, we have 2 , 3 = 7, which is also impossible. Suppose that the numbers x and y are even. So x = 2l, y = 2m, with l and m natural numbers. The given equation can then be written in the form (2l + 3m )(2l , 3m ) = 7, where 2l + 3m and 2l , 3m are natural numbers, which implies that 2l + 3m = 7 and 2l , 3m = 1. These two equations determine the values of l, m, namely l = 2, m = 1 for which we have x = 4, y = 2.

72 The only two natural numbers x; y  1 such that 2x , 3y = 7 are therefore x = 4 and y = 2. 5. Determine all real solutions x, y, z of the system of equations:

x3 + y = 3x + 4; 2y3 + z = 6y + 6; 3z3 + x = 9z + 8:

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; and by Panos E. Tsaoussoglou, Athens, Greece. We give the solution of Tsaoussoglou. From x3 + y = 3x + 4, we have

x3 , 1 , 1 , 3x = 2 , y;

or

(x , 2)(x + 1)2 = 2 , y: From 2y 3 , 2 , 2 , 6y = 2 , z , we have 2(y , 2)(y + 1)2 = (2 , z ); and 3z 3 , 3 , 3 , 9z = 2 , x gives 3(z , 2)(z + 1)2 = (2 , x); so that

and

(1) (2) (3)

(x , 2)(x + 1)2 = ,(y , 2); 2(y , 2)(y + 1)2 = ,(z , 2); 3(z , 2)(z + 1)2 = ,(x , 2);

  (x , 2)(y , 2)(z , 2) (x + 1)2(y + 1)2(z + 1)2 + 61 = 0:

As the last factor is always positive for real x; y;z , we have

(x , 2)(y , 2)(z , 2) = 0: This gives at least one of x = 2, y = 2, z = 2. In conjunction with (1), (2) and (3) this gives the unique solution x = y = z = 2. 6. Show: For all real numbers a; b  0 the following chain of inequalities is valid

pa + pb !2 2

p3 a2 b + p3 ab2 + b a +  4 v u p3 a2 + p3 b2 !3 p u a + ab + b t  :  3

2

73 Also, for all three inequalities determine the cases of equality.  Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; and by Panos E. Tsaoussoglou, Athens, Greece. We give Tsaossoglou's solution. p! p p p p

pa + b 2 3 a2(p3 a + 3 b) + 3 b2(p3 a + 3 b) 1.  2 4 is equivalent to p p p p (pa + b)2  ( 3 a2 + 3 b2 )( p3 a + 3 b);

which holds by the Cauchy inequality. p Let p6 a = A, 6 b = B , (A3 + B 3 )2  (A4 + B 4 )(A2 + B 2 ). p p p 2. 3(a + b) + 3 3 ab( p3 a + 3 b)  4(a + ab + b); or equivalently or or

p p p a + 3 3 a2b + 3 3 ab2 + b  2(a + 2 ab + b); p p ( p3 a + 3 b)3  2(pa + b)2;  A2 + B2 3  A3 + B3 2  ; 2 2

(with A, B as above), a known inequality.

u p3 a2 + p3 b2 !3 pab + b v u a + t 3.  . 3 2 With A and B as above this is equivalent to  A6 + A3B3 + B6 2  A4 + B4 3  : 3

For this it is enough to prove that or

2

 A4 + B4 3  A6 + A3B3 + B6 2 ,  0; 2 3

9(A4 + B 4)3 , 8(A6 + A3 B 3 + B 6 )2 = A12 , 16A 9 B 3 + 27A8B 4 , 24A6 B 6 + 27A4 B 8 , 16A3B9 + B 12 = (A , B )4 A8 + 4A7B + 10A6B 2 + 4A5 B 3 , 2A4B 4 +4A3B 5 + 10A2 B 6 + 4AB 7 + B 8  (A , B)4(A3B3(A , B)2)  0:

74

9. Let 4ABC be equilateral. On side AB produced, we choose a point P such that A lies between P and B. We now denote a as the length of sides of 4ABC ; r1 as the radius of incircle of 4PAC ; and r2 as the exradius of 4PBC with respect to side BC . Determine the sum r1 + r2 as a function of a alone. Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;  and by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina. We give Amengual's solution and comment. Looking at the gure, we see that \T1 O1 R = 60 since it is the supplement of \T1 AR = 120 (as an exterior angle for 4ABC ). Hence, \AO1 R = 30 . Similarly, we obtain \BO2 S = 30 . T20 C T10 O1 r1 P

have

and

T1 A

r2 R

O2

S B

T2

Since tangents drawn to a circle from an external point are equal, we

T1T2 = T1A + AB + BT2 = RA + AB + SB = r1 tan 30 + a + r2 tan 30 = r1p+ r2 + a; 3

T10 T20 = T10 C + CT20 = CR + CS = (a , RA) + (a , SB) = 2a , r1p+3r2 :

Since common external tangents to two circles are equal, T1 T2 = T10 T20 . Hence,

r1p+ r2 + a = 2a , r1p+ r2 ;

whence we nd that

3

p3 a r1 + r2 = 2 :

3

Comment. This problem is identical to problem 2.1.11, page 25, of H. Fukagawa and D. Pedoe, Japanese Temple Geometry Problems, (The Charles Babbage Research Centre, 1989).

75 We next give the solution to one problem of the VII Nordic Mathematical Contest. 2. [1995: 223] VII Nordic Mathematical Contest A hexagon is inscribed in a circle with radius r. Two of its sides have length 1, two have length 2 and the last two have length 3. Prove that r is a root of the equation

2r3 , 7r , 3 = 0:

Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain. Equal chords subtend equal angles at the centre of a circle; if each of sides of length i subtends an angle i (i = 1; 2; 3) at the centre of the given circle, then

2 1 + 2 2 + 2 3 = 360;

whence

1 + 2 = 90 , 3 ; 2

2

2

    cos 21 + 22 = cos 90 , 23 = sin 23 :

and

Next we apply the addition formula for the cosine:

cos 21 cos 22 , sin 21 sin 22 = sin 23 ;

(1)

where (see gures)

p4r2 , 1 1 = 2 1 1 sin 2 = r ; cos 2 = 2r ; p2 sin 22 = 1r ; cos 22 = r r, 1 ; sin 23 = 3=r2 :

r 1 2

A

3

2

1

2

2

r

2

3=2

1

B

r

C

We substitute these expressions into (1) and obtain, after multiplying both sides by 2r2 , p p

4r2 , 1  r2 , 1 , 1 = 3r:

76 Now write it in the form

p

and square, obtaining

(4r2 , 1)(r2 , 1) = 3r + 1;

(4r2 , 1)(r2 , 1) = 9r2 + 6r + 1;

which is equivalent to Since r 6= 0, we have

which was to be shown.

r(2r3 , 7r , 3) = 0: 2r3 , 7r , 3 = 0;

To nish this number of the Corner we turn to two problems from the

32nd Ukrainian Mathematical Olympiad given in the October 1995 number of the Corner [1995 : 266].

32nd UKRAINIAN MATHEMATICAL OLYMPIAD March 1992 | Selected Problems

that

2. (8) There are real numbers a, b, c, such that a  b  c > 0. Prove a2 , b2 + c2 , b2 + a2 , c2  3a , 4b + c: c a b

 Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; Panos E. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give the solution of Arslanagic. From a  b  c > 0 we have Now we get

and

a + b  2; 0 < b + c  2 and a + c  1: c a b a2 , b2  2(a , b); because a  b; c c2 , b2  2(c , b); because c  b; a a2 , c2  a , c; because a  c: b

After addition of these inequalities, we have

a2 , b2 + c2 , b2 + a2 , c2  2(a , b) + 2(c , b) + (a , c); c a b

77 that is,

a2 , b2 + c2 , b2 + a2 , c2  3a , 4b + c: c a b The equality holds if and only if a = b = c > 0. 5. (10) Prove that there are no real numbers x, y, z, such that x2 + 4yz + 2z = 0; x + 2xy + 2z2 = 0; 2xz + y 2 + y + 1 = 0:

 Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; by D.J. Smeenk, Zaltbommel, the Netherlands; by Panos E. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's solution. Label the three given equations (1), (2) and (3), (in that order), respectively. If x = 0 or z = 0, then from (3), y 2 + y + 1 = 0, which has no real solutions. Hence we may assume that xz 6= 0. From (1) and (2), we get x2 2=2 ,2z(2y + 1) and 2z2 = ,x(2y + 1), which, when multiplied gives 2 2x z = 2xz(2y + 1) or xz = (2y + 1)2. Substituting into (3) we get

2(2y + 1)2 + y 2 + y + 1 = 0

or which has no solution.

3y2 + 3y + 1 = 0;

That completes the Corner for this issue. Olympiad season is upon us. Send me your Olympiads, as well as your nice solutions.

78

BOOK REVIEWS Edited by ANDY LIU Leningrad Mathematical Olympiads 1987{1991, by Dmitry Fomin and Alexey Kirichenko, published by MathPro Press, Westford, MA, USA, 1994. (Contests in Mathematics Series, vol. 1.), paperbound, 202 + xxii pages, ISBN 0{9626401{4{X, US$24. Reviewed by Jozsef  Pelik an,  Eotvos Lorand University, Budapest, Hungary. The regulations of the International Mathematical Olympiad (IMO) stipulate that a country can send a team to the IMO consisting of at most six students. Still, there was a city in the 1965 IMO which participated with nine students: ve in the Russian team, two in the Israeli team, and one in each of the USA and German teams. This city was Leningrad, now known by its old name St. Petersburg again. If a city has such an output of talented young mathematicians, you can imagine that the city olympiad organized there should be of an exceptionally high quality. And so it is. Let me brie y describe the system of the Leningrad Olympiads. Until 1989 Soviet schools comprised grades 1{10, and the competition was held for 5{10 graders. Since that time the Russian schools comprise grades 1{11, and the competition is held for 6{11 graders. The competition is organized in four rounds. The rst round is at the school level in December and January and the second round held at the regional level (that is, in the 22 regions of Leningrad) in February, some 10,000 to 12,000 students taking part in the second round. Both of these rounds are written examinations. It is important to emphasize this, because the last two rounds | quite unusually for a mathematics competition | are oral examinations. This requires of course a huge number of able and devoted jurors, and few countries | let alone cities | can produce the number of mathematicians necessary for this task. In the third round (called \main round") about 90{130 students participate in each grade. In the nal round (called somewhat misleadingly \elimination round" | this term normally means the initial phase of a competition, not the nal one) which is held only for 9{11 (earlier 8{10) graders some 100 students take part altogether, although in 1991 there were only 34. The main round is held in February or March, and the elimination round in March. In the period 1962{1983 and also in 1991 the elimination round was used only to pick the city team at the All-Union Olympiad (this might explain the strange name), while in the period 1984{1990 the result of the competition was decided by the elimination round (not by the main round). A few words about the earlier history of the Leningrad (and other) Olympiads in the Soviet Union. The Leningrad Olympiad started in 1934 (the Moscow Olympiad one year later). The All-Russia Olympiad started

79 in 1961 and the All-Union (i.e. Soviet Union) Olympiad in 1967. These are respectable ages, although there are a few national mathematics contests which are much older (notably the Hungarian Eotvos | later Kurschak | competitions which started in 1894). But I fully agree with the authors that the Leningrad competition is quite unique in being an oral one. On the other hand I must contest another statement of the authors, namely that the Leningrad competition would be quite unique in having only new and original problems. In fact the majority of the competitions I know of have new and original problems. Besides the main part containing the problems and solutions the book contains useful appendices: statistical data about the number of participants, the names of each year's winners, a glossary and the names of the authors of each problem. Also, there are valuable comments on the book by the publisher, Stanley Rabinowitz, and a very good evaluation of the reasons of the excellence of Soviet mathematics by Mark Saul. To give a sample of the problems (and solutions) I give you two examples from the book. Both come from the 1991 Olympiad and were given to grade 11 students. Problem A black pawn is placed in the top right square of an 8  8 chess-board. One may place a white pawn on any empty square of the board, having repainted all pawns in adjacent squares so that black pawns become white, and vice versa. (Two squares are called adjacent if they have a common vertex.) Can one place pawns in this manner so that all 64 squares of the board would be lled with white pawns? Solution Connect the centres of each pair of adjacent free squares by a segment. Then, when we place the next white pawn on some square of the board, we erase all segments from the centre of this square. Thus, the number of erased segments coincides each time with the number of free adjacent squares for the square on which each white pawn is placed. To make a certain pawn be white at the end, it must be placed on a square having an even number of free adjacent squares because the number of repaintings for each pawn is equal to the number of pawns that will be placed on adjacent squares after this one. So, we conclude that each time we have to erase an even number of segments. But this is impossible because the initial number of segments is odd (we have lost at the beginning three segments caused by the black pawn in the corner) and the nal number is even | it is equal to zero. Thus, we see that in the nal position there is at least one black pawn. The next problem has a rather unusual solution. Problem One may perform the following two operations on a natural number: (a) multiply it by any natural number; (b) delete zeros in its decimal representation.

80 Prove that for any natural number n, one can perform a sequence of these operations that will transform n to a one-digit number. Solution We use the following fact: Lemma For any integer n that is not divisible by 2 or 5 one can nd a number consisting of digits 1 only that is a multiple of n: Proof Consider n numbers 1; 11; : : : ; 11 : : : 1 : If one of these numbers is a multiple of n, we are done. Otherwise, their remainders modulo n can have n , 1 possible values 1; : : : ; n , 1 ; and therefore (by the pigeonhole principle) at least two of those numbers have the same remainders modulo n: Then their di erence is divisible by n and looks like 11 : : : 1  10k : But n is coprime with 10, and we conclude that the rst factor is divisible by n: Now let n be an arbitrary natural number. Multiplying n by 2 and 5 and deleting zeros, we can transform n to a number that is coprime with 10. Then multiplying the result by the appropriate number, we can obtain a number containing only digits 1 in its decimal representation (we use the assertion of the lemma). Now the chain of the following operations leads to the desired result: (a) Multiplying by 82, we obtain the number 911 : : : 102 : (b) Delete zero in the last obtained number and multiply it by 9. This gives the number 8200 : : : 08 ; which is transformed into 828. (c) 828  25 = 20700 : (d) 27  4 = 108 : (e) 18  5 = 90 ; and we can obtain the single-digit number 9. All in all this book is very well written, full of interesting problems and I warmly recommend it to anyone interested in mathematical competitions, or just in nice problems.

81

Folding the Regular Heptagon Robert Geretschlager, Bundesrealgymnasium, Graz, Austria Introduction Ever since Greek antiquity, mathematicians have been considering constructions that can be done with straight-edge and compass only, the so-called Euclidean constructions. A number of famous problems, such as squaring the circle, trisecting angles and doubling the cube, were unsolvable for the Greeks, and later shown to be theoretically unsolvable by Euclidean methods. The reason for this is that only such problems that can be reduced algebraically to combinations of linear and quadratic equations are solvable in this sense. We now know that these three problems, as well as many others, cannot be represented by combinations of such equations. One speci c problem the Greeks attempted to solve in this way was the construction of regular n{gons for small n. They were successful in nding constructions for n = 3, 4, 5, 6, 8, 10 and 12, but not for n = 7, 9 or 11. Since 7 is the smallest n for which no construction could be found, it was of special interest why this particular problem should prove so stubborn. As it turned out, the construction of the regular heptagon by Euclidean methods is impossible for the same reason that angle trisection and doubling the cube are, in that each of these problems requires the graphic solution of an irreducible cubic equation in its algebraic representation. As shown in \Euclidean Constructions and the Geometry of Origami" ([1]), all cubic equations can be solved graphically using elementary methods of origami1 . This is especially interesting in light of the fact that regular n{gons are commonly used in the development of origami folding bases. A heptagon could conceivably nd use in developing models of insects for instance, since six legs + one head = seven corners. In this article, I present a theoretically precise method of folding the regular heptagon from a square, derived from the results established in the above-mentioned article. The folding method is presented in standard origami notation, and the mathematical section is cross-referenced to the appropriate diagrams.

The Cubic Equation

The seven corners of a regular heptagon can be thought of as the seven solutions of the equation z7 , 1 = 0 (1) 1 Origami is, of course, the art of paper folding. For readers not yet familiar with this ancient art, but interested in becoming so, there is a large amount of introductory literature easily available. I was personally introduced to origami by the books of Robert Harbin ([2]). A ne introduction to the geometry of origami is the classic \Geometric Exercises in Paper Folding" by T. Sundara Row ([3]).

82 in the complex plane. This implies that the unit circle is the circumcircle of the heptagon, and that one corner of the heptagon is the point z1 = 1 on the real axis (Fig. 1.1.). Since one solution of (1) is known, the other six are the roots of z7 , 1 = z6 + z5 + z4 + z3 + z2 + z1 + 1 = 0: (2)

z,1 For any speci c z satisfying this equation, the conjugate z is also a solution,

since the real axis is an axis of symmetry of the regular heptagon. Also, since

jzj = jzj = 1;

we have z = z . Therefore we can de ne 1

 = z + 1z = z + z = 2  Re z: z

3

r

iy

6

z

2

r

z

4

r

c z

5

b

r

O

z =1 x

a

r

1

r

z

r

z Fig. 1.1. r

7

6

z1 = 1 + i  0 z2 = cos 27 + i  sin 27 z3 = cos 47 + i  sin 47 6 z4 = cos 7 + i10sin 67 10 z5 = z4 = cos 8127+ i  sin 8712 z7 = z2 4= cos 7 + i  sin 7 z6 = z3 =2 cos 7 + i  sin 7 a = cos 7 = Re z2 = Re z7 b = cos 7 = Re z3 = Re z6 c = cos 67 = Re z4 = Re z5 Dividing by z 3 , we see that equation (2) is equivalent to z3 + z2 + z + 1 + z1 + z12 + z13 = 0

since 0 is not a root, and since

3 1  = z + z = z3 + 3z + z3 + z13  1 1 3 = z + z 3 + 3 z + z = z 3 + z13 + 3 ()  3 , 3 = z3 + z13 3



(3)

83 and

2 1  = z + z = z2 + 2 + z12 ()  2 , 2 = z2 + z12 ; 2



substituting yields

    1  1 1 3 2 z + z3 + z + z2 + z + z + 1 = 0 ()  3 , 3 +  2 , 2 +  + 1 = 0 ()  3 +  2 , 2 , 1 = 0:

From (3) ,we see that each root of the equation

 3 +  2 , 2 , 1 = 0

(4) is real, and is equal to twice the common real component of two conjugate complex solutions of (1). It is therefore possible to nd the six complex roots of (1) in the complex plane by nding the roots of (4), taking half their values, nding the straight lines parallel to the imaginary axis and at precisely these distances from it, and nally nding the points of intersection of these parallel lines with the unit circle. We shall now proceed to utilize these steps in folding the regular heptagon.

A Step-by-step Description of the Folding Process As is usually the case in origami, we assume a square of paper to be given. We consider the edge-to-edge folds in step 1 as the x{ and y {axes of a system of cartesian coordinates, and the edge-length of the given square as four units. The mid-point of the square is then the origin M (0; 0), and the end-points of the folds have the coordinates (,2; 0) and (2; 0), and (0; ,2) and (0; 2) respectively. For readers not familiar with origami notation, it should be mentioned that dashed lines represent so-called \valley" folds (folding up), and dot-dashed lines represent so-called \mountain" folds (folding down). Thin lines represent visible creases in the paper generated by previous folds. As shown in [1], the solutions of the cubic equation

x3 + px2 + qx + r = 0

are the slopes of the common tangents of the parabolas p1 and p2 de ned by the foci     and directrices

F1 , p2 + 2r ; q2

l1 : x = , p2 , r2

and

and

F2 0; 12

l2 : y = , 21

84 respectively. The solutions of (4) can therefore be obtained by nding the common tangents of the parabolas with foci

F1(,1; ,1) and directrices

l1 : x = 0

 1 F2 0; 2

and

l2 : y = , 12

and

respectively. Since the slope of the common tangents is not altered by translating the parabolas parallel to the y {axis we can, for convenience, use and

  F1 ,1; , 21 l1 : x = 0

and

F2(0; 1)

l2 : y = 0: This is precisely what is done in steps 2 to 5. F1 is the point A, and F2 is the point B . The fold in step 4 is then the only common tangent of the parabolas and

with positive slope, and thus twice the real component of the solutions of (1) which lie to the right of the imaginary axis and are not equal to 1. In other words, the slope of this fold is 2  cos 27 . Step 4, by the way, is the only step that cannot be replaced by a straight-edge and compass construction. In steps 6 to 8, the unit-length is then transferred in such a way that point E in step 8 has y {coordinate ,2  cos 27 . Since the distance from M to point 1 in step 9 is 2 units, the distances from M to points 2 and 7 are also 2 units, and so points 7, 1 and 2 are three consecutive corners of the regular heptagon. (We assume that point 1 with coordinates (0; ,2) is the rst corner, and continue from there.) Step 10 thus yields two sides of the heptagon, and steps 11 to 13 yield the remaining sides of the heptagon by making use of its radial symmetry, until nally step 14 shows us the completed regular heptagon. The folding process is shown at the end of the article.

Conclusion Unlike other regular n{gons with small n, the regular heptagon is not very common in popular culture or graphics. Apart from the seven-sided star one comes across in astrology, the heptagon does not seem to show up much in public, unlike its close relatives. We come across the octagon at many a street corner, and the pentagon and hexagon can be seen on most soccer balls, just to name a few. I do not know if this (relatively) easy generation of the regular heptagon will lead to its mass popularization, but an ardent Heptagonist can certainly dream. It should be mentioned that a similar folding method for the regular heptagon is described in the article \Draw of a Regular Heptagon by the

85 Folding" by Benedetto Scimemi ([4]) in the relatively hard to nd Proceedings of the First International Meeting of Origami Science and Technology. (I have only recently gained access to a copy myself.) This volume o ers a great many ideas for further research for anyone interested in the geometry of origami, and is certainly worth searching for.

References [1] R. Geretschlager, Euclidean Constructionsand the Geometry of Origami, Mathematics Magazine, 68 No. 5 December 1995, pp. 357-371 [2] R. Harbin, Origami, The Art of Paper-Folding, Vols. 1-4, Hodder Paperbacks, Norwich (1968) [3] T. Sundara Row, Geometric Exercises in Paper Folding, Dover Publications, Inc., Mineola, NY (1966) reprint of 1905 edition [4] B. Scimemi, Draw of a Regular Heptagon by the Folding, Proceedings of the First International Meeting of Origami Science and Technology, Ferarra, Italy (1989)

The Folding Process

1.

2.

Fold and unfold twice.

Fold back twice.

86

3.

4. B

q

A

A

q

Fold such that A and B come to lie on the creases.

Fold and unfold, making a crease mark at point A (bisecting the side).

5.

q

6.

q

q

B

C

q

q

D

A

Unfold everything.

Fold C to D.

87

7.

8.

E

q

Fold and unfold both layers at crease, then unfold everything.

9.

Fold horizontally through E , then unfold.

10.

M

7

q

q

2

q

q

1 q

Fold through M , such that 1 lies on crease, resulting in 2 and 7 (M is the mid-point of the heptagon, 1, 2 and 7 are corners).

q

q

q

q

q

q

Fold back twice, so that the marked points come to lie on one another; resulting folds are rst two sides of the heptagon.

88

11.

12.

M

q

2 q

Fold through M and 2.

13.

Fold back lower layers using edges of upper layer as guidelines; resulting folds are two more sides of the heptagon; open up fold from step 12 and repeat 11 and 12 on left side.

14. 5

q

q

4

q

Fold back nal edge of the heptagon through 4 and 5.

The nished heptagon.

89

THE SKOLIAD CORNER No. 20

R.E. Woodrow This issue we give the problems of the Mathematical Association National Mathematics contest, written November 18, 1994. The contest was written by about 30,000 students in the United Kingdom. My thanks go to Tony Gardiner, School of Mathematics, University of Birmingham for sending me the contest.

THE MATHEMATICAL ASSOCIATION NATIONAL MATHEMATICS CONTEST 1994 Friday, November 18, 1994 | Time: 90 minutes

A. 1

1. The average of x and 8x is 18. What is the value of x?

1 2

B. 2

C. 4

D. 4 12

E. 9.

2. Which of the following does not have six lines of symmetry?

A. g 2a

B. g 2b

C. g 2c

D. g 2d

E. g 2e.

3. I write out the numbers from 1 up to 30 in words. If N denotes the number of times I write the letter \n", M denotes the number of times I write the letter \m", and C denotes the number of times I write the letter \c", then N + M + C equals A. 27 B. 28 C. 29 D. 30 E. 31. 4. Which of the following ve numbers has a prime factor in common with exactly one of the other four numbers? A. 91 B. 52 C. 39 D. 35 E. 24. 5. ABCD is a quadrilateral with AB = AD = 25 cm, CB = CD = 52 cm and DB = 40 cm. How long in AC in cm? A. 32:5 B. 48 C. 52 D. 60 E. 63.

90

6. The number of pounds of pickled peppers that Peter Piper purchased for $59 is equal to the number of pounds Peter would pay for two hundred and thirty six pounds of peppers. How much would he pay for twenty pounds of pickled peppers? A. $5 B. $10 C. $20 D. $40 E. $80. 7. Which expression has the smallest value when x = ,0:5? ,

A. 2 =x 1

B. x1

C. x12

D. 2x

E. p,1 x .

8. Over an average lifetime in the UK, roughly how many times does a person's heart beat? A. 4  107 B. 5  107 C. 2  108 D. 3  109 E. 2  1010. 9. What is the sum of the reciprocals of the rst six triangular numbers

1, 3, 6, 10, etc.? A. 10 B. 127

C. 56

10. A rope 15 m long and 5 cm in diameter is coiled in a at spiral as shown. What is the best estimate for the diameter of the \circle" (in cm)? A. 10

B. 100

11. If a b =

A. 95

(

B. 100

C. 150

q

D. 200

ab+a+b+1) , then 19 94 equals a

C. 208

E. 49 21 .

D. 23

D. 1882

12. The diagram shows a semicircle with radius 1 cm and with centre O. If C is an arbitrary point on the semicircle, which of the following statements may be false?

E. 300. E. 1994.

C A

O B A. \ACB is a right angle. B. 4OAC is isosceles. C. the area of 4ABC is  1 cm2 D. 4AOC is equal in area to 4OBC E. AO2 + OB 2 = AC 2 + BC 2.

91

13. A giant marrow in my garden weighed 50 pounds and was 98% water. Suppose that during a rainy day it absorbed water so that it became 99% water. What would its new weight be (in pounds)? A. 50:01 B. 50:5 C. 98 D. 99 E. 100. 14. A solid cuboid has edges of lengths a, b, c. What is its surface area? A. (a + b + c)2 , (a2 + b2 + c2 ) B. abc C. 2(a2 + b2 + c2 ) 2 D. (a + b + c) E. ab + bc + ca. 15. Given two copies of an isosceles right-angled triangle ABC , squares BDEF and PQRS are inscribed in different ways as shown. What is the ratio area PQRS ? area BDEF A. 89 B. 23 C. 1 A. 0

16. What is the last digit of 1994 B. 2

C. 4

A

@ @ @E F @ @ @C B

D.

A

D

@ @R @ ,@ S , @Q @ ,@ B @, @ C

3

E. 98 .

q2

P

?

(1995+1996+1997+1998+1999+2000)

D. 6

E. 8.

17. When two dice are thrown the probability that the total score is a multiple of 2 is 21 . For how many other values of n is it true that, when two dice are thrown, the probability that the total score is a multiple of n is equal to n1 ? A. 1 B. 2 C. 3 D. 4 E. 5. 18. How many digits are there in the smallest number which is composed entirely of ves (eg. 5555) and which is divisibly by 99? A. 9 B. 10 C. 18 D. 36 E. 45. 19. The price of a secondhand car is displayed (in pounds) on four cards on the windscreen. Each card shows one digit. If the card with the thousands digit blew o in the wind, the apparent price of the car would drop to one forty-ninth of the intended value. What number is on that card? A. 5 B. 6 C. 7 D. 8 E. 9.

92

20. The graph of y , x against y + x is as shown. y,x 6 O

   

-

y+x

The same scale has been used on each axis. Which of the following shows the graph of y against x? y

y6

6 -

O

x

A. g 20a

O

   

y

-

x

B. g 20b

21.pWhich is smallest? p

A. 5 + 6 7

B. 7 + 6 5

O

y

6  

x

C. g 20c

p

C. 6 + 5 7

y

6

A A OAA

-

x

D. g 20d

p

D. 7 + 5 6

6 -

O

x

E. g 20e.

p

E. 6 + 7 5.

22. A train leaves London at 0600 and arrives in Newcastle at 0930. Another train leaves Newcastle at 0700 and arrives in London at 0930. If both used the same route and each travelled at a constant speed, at what time would they meet? A. 0757 21 B. 0802 21 C. 0807 21 D. 0827 21 E. more information required. 23. The triangle ABC has a right-angle at A. The hypotenuse BC is trisected at M and at N so that BM = MN = NC . If AM = x and AN = y, thenpMN is equal to p 2 2 q p 2 2 A. x+2 y B. (y 2,x ) C. (y 2 , x2 ) D. (x 3+y ) E. x2 +5 y2 . 24. If Susan drives to work at x mph she will be one minute late; if she drives at y mph she will be one minute early. How far does she drive to work (in miles)? A. 30(yx B. y2,yxx C. yx,+xy D. x+2 y E. 60(xy+,yx) . y,x)

93

25. The octagonal gure is obtained by tting eight congruent isosceles trapezia together. If the three shorter sides of each trapezium have length 1, how long is each outer edge? p

A. 1 + 2

B.

p

p

1+ 2 2

p

p

,L , L , , aa,

@ @ @ @ @!!

!!@ @ @ @ @

a , a , , L , L,

C. 2

D. 2

E. 1 + 2 , 2.

Last number we gave the 1994 Nat West U.K. Junior Mathematical Challenge. It was written Tuesday, April 26, 1994. Here are the answers. 1. 455 2. B 3. E 4. A 5. A 6. B 7. B 8. D 9. D 10. A 11. B 12. E 13. A 14. B 15. E 16. E 17. C 18. C 19. D 20. E 21. A 22. D 23. D 24. C 25. C That completes the Skoliad Corner for this issue. Send me your suitable contests and solutions. Also send me any suggestionsfor improvement of this feature.

94

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. All material intended for inclusion in this section should be sent to the Mayhem Editor, Naoki Sato, Department of Mathematics, University of Toronto, Toronto, ON Canada M5S 1A1. The electronic address is [email protected] The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto). The rest of the sta consists of Richard Hoshino (University of Waterloo), Wai Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Matrix Exponentials: An Introduction Donny Cheung

student, University of Waterloo Waterloo, Ontario. tion,

Let us start with the well-known power series for the exponential func-

1 k 3 4 2 X ex = 1 + x + x2! + x3! + x4! +    = xk! ; k=0 which works for all x 2 C . Likewise, we will de ne the exponential function

for matrices (with complex entries) as

1 k 3 4 2 X eM = I + M + M2! + M3! + M4! +    = Mk! k=0 for all n  n matrices M . Matrix Exponentials Really Do Exist

First of all, we must show that this sum actually converges. Otherwise, the answer we get would be meaningless, and that would be bad. To start us o , we will de ne the norm of an n  n matrix, which we will denote jAj, as follows: 0 1 n @ X jaij jA : jAj = 1max in j =1

That is, for each row, take the sum of the absolute values of the components in that row, and the norm will be the largest of these row sums.

95 Lemma 1 For n  n matrices A and B ,

jABj  jAjjBj: Proof We label the entries of A and B in the standard way, aij and bij , respectively. n X For any row in A, de ne the row sum as jaij j. Since (AB )ij = j =1 n X aik bkj , the row sum of row i in AB is k=1 n X n X n n X n X X n aik bkj  X j a b j = jaikj jbkj j ik kj j =1 k=1 j =1 k0=1 k =1 j =1 1 n n n X X @jaikj jbkj jA  X (jaikj jBj) = k=1 j =1 k=1 n X = jB j jaikj  jAjjB j: k=1

Now, jAB j, the maximum row sum, is still a row sum, thus jAB j  jAjjB j.  Lemma 2 For n  n matrices A and B ,

jA + Bj  jAj + jBj:

The proof is left as an exercise to the reader. Theorem 1 The sum eM converges for all n  n matrices M . Proof Using the two lemmas, we get

so that

2 3 4 eM = I + M + M2! + M3! + M4! +    ;

eM = jI + M + M 2 + M 3 + M 4 +    j 2! 3! 4! 2 3 4 j M j j M j  1 + jM j + + + jM j +    = ejM j:

2!

3!

4!

Since jM j is a real number, ejM j is nite, and jeM j is bounded. Thus, eM converges. 

96 Exercises

1. Prove Lemma 2.

Some Linear Algebra Lingo Recall that two n  n matrices A and B are similar if

B = C ,1AC for some invertible n  n matrix C . (It is usually easier, when verifying that two matrices are similar, to show that CB = AC instead.) Recall also that a diagonal matrix is a matrix with all its non-zero entries along the main diagonal. (But entries along the main diagonal are not necessarily non-zero). For example: 0 1

BB @

5

0

0

,

0

0

0

0

0

0

0

0

0

4

0 0

,

C C A:

1

If a matrix A is similar to a diagonal matrix, we say that A is diagonalizable.

Calculating Matrix Exponentials Suppose D is a diagonal matrix, with diagonal elements 1, 2, : : : ,

n. We have

0  BB   D B B@    n 0

1

0

=

and

eD

0 0

2

. . .

. . .

0

0

..

. . .

.

I D D  0  21    BB BB B@ 0 e1  BB e2    BB @    en +

+

2!

0

1

Dk

0

=

0 0

2

. . .

. . .

0

0

..

. . .

.

1 C C C ; C A

+

1 + 2! +

0

1+

. . .

0

0

0

0

=

0 k  k  B  B B B @    k

2

=

1+

=

1 C C C ; C A

0

. . .

. . .

0

0

..

.

. . .



0

. . .

1 CC CC : A

22

2 + 2! +

0

   ..

0 0

.



1+

n

. . .

2n

+ 2! +



1 C C C C C A

97

Jordan Matrices A Jordan block matrix is an n  n matrix of the form 0 BB  BB BB @ 1

0

0

1

. . .

. . .

. . .

0

0

0

0

0

0

 

0

0

0

0

..

. . .

. . .

.

1 C C C C C C A

    with 's all down the main diagonal and 1's directly above them (except for the rst column), and 0's everywhere else. Examples include 0, 1   B C , B C @ A: , , 1

0

3

(5),

5

1

0

5

1

0

, and

0

0

1

0

3

0

0

0

0

3

1

0

3

A Jordan matrix is an n  n matrix with Jordan blocks down the diagonal. An example containing our three examples above:

0 BB BB BB BB @

5 0 0

0

,

3

0

0

0

0

0

1

0

0

0

0

1

0

0

0

1

,

0

0

0

0

0

5

1

0

0

5

,

0

3

0

,

0

0

3

0

0

0

0

0

0

0

0

0

0

0

0

3

1 C C C C C : C C C A

Notice that all diagonal matrices are also Jordan matrices. Exercises

01.Verify that   1k BB  CC  BB CC BB CC @   A   0 k ,kk, ,kk, : : : , k k,n ,kk, : : : ,n,k k,n BB k  n, BB BB B@ ::: k ::: 1

0

0

0

0

1

0

0

. . .

. . .

. . .

. . .

. . .

0

0

0

0

0

0

..

.

1

0

1

1

0

=

2

2

1

1

1

2

. . .

. . .

. . .

0

0

0

0

0

0

.

..

. . .

0

2. Compute eB when B is a Jordan block matrix. 3. Compute eJ when J is a Jordan matrix.

+1 +2

,kk,n , knk,n n, 1

. . .

,kk, k

1

1

+1

1 C C C C : C C C A

98

So why did we just go through all that? Here's the reason: Theorem 2 For any n  n matrix A, there exists a Jordan matrix J which is similar, that is, C ,1 AC = J for some C . J is known as a Jordan canonical form of A. Some matrices have more than one Jordan canonical form. The proof is beyond the scope of this article, but can be found in any good advanced linear algebra textbook. But we will prove the following: Theorem 3 If J is a Jordan canonical form of n  n matrix A, with A = CJC ,1, then eA = C (eJ )C ,1. Proof

Ak = (CJC ,1)k = (|CJC ,1)(CJC{z,1)    (CJC ,1)} k = CJ (C ,1 C )J (C ,1C )J (C ,1C )    (C ,1 C )JC ,1 = CJ kC ,1: A2 Now, eA = I + A + +    2! 2 = CIC ,1 + CJC ,1 + C J2! C ,1 +      2 = C I + J + J2! +    C ,1 = C (eJ )C ,1: Thus, eA = C (eJ )C ,1 .  The process of nding a Jordan canonical form J of a matrix A is also beyond the scope of this article. Once again, I refer you to a good textbook. However, we have shown that it is possible to calculate eA for any n  n matrix A. And now for something less theoretical...

Here, we discuss an interesting application of matrix exponentials: rstorder systems of linear di erential equations. By that, we mean systems of the form:

x01(t) = a11x1(t) + a12x2(t) +    + a1nxn(t); x02(t) = a21x1(t) + a22x2(t) +    + a2nxn(t); .. .

x0n(t) = an1x1(t) + an2x2(t) +    + annxn(t):

99 Letting

0 BB x(t) = B B@

we de ne the derivative x(t) as

0 BB 0 x (t) = B B@

x1(t) 1 x2(t) C C; .. C C . A xn(t) x010 (t) 1 x2(t) C C: .. C C . A x0n (t)

We will also let A be the matrix of coecients for the system. Now we can rewrite the system of equations as x0 (t) = Ax(t). Exercises 1. Verify that x(t) = eAt c is a solution to x0 (t) = Ax(t) for every constant vector c.

Further Exploration This is where I become too lazy to show you all the neat stu you can do with matrix exponentials, and where I encourage you, the reader, to explore on your own. Exercises 1. We don't have to stop at matrix exponentials. We can de ne matrix equivalents for functions like sin(x) and cos(x) in a very similar fashion. What other types of functions can we extend to the matrices. 2. Prove that sin(A) sin(A) + cos(A) cos(A) = I for any arbitrary n  n matrix A. 3. When is eA+B = eA eB ? The fact that matrix multiplication isn't commutative causes some problems. 4. As a corollary to the above problem, show that e2A = eA eA . 5. Prove or disprove: there exist a 2  2 matrix A with real entries such that   [1996 Putnam, B4]

sin A = 10 1996 1 :

100 Open problems

1. As we have de ned eA , it is very easy to de ne xA for x 2 R. Can we de ne B A for n  n, matrices A and B? What would their properties  C BC B be? Would A = A ? 2. Can we de ne a log function? Would it have the similar properties to the real log function? 3. Can we solve other types of vector di erential equations with matrix exponentials or (more generally) matrix functions like sin and cos?

Mayhem Problems A new year brings new changes and new problem editors. Cyrus Hsia now takes over the helm as Mayhem Advanced Problems Editor, with Richard Hoshino lling his spot as the Mayhem High School Problems Editor, and veteran Ravi Vakil maintains his post as Mayhem Challenge Board Problems Editor. Note that all correspondence should be sent to the appropriate editor | see the relevant section. In this issue, you will nd only solutions | the next issue will feature only problems. We intend to have problems and solutions in alternate issues. We warmly welcome proposals for problems and solutions. With the new schedule of eight issues per year, we request that solutions from the previous issue be submitted by 1 June 1997, for publication in the issue 5 months ahead; that is, issue 6. We also request that only students submit solutions (see editorial [1997: 30]), but we will consider particularly elegant or insightful solutions for others. Since this rule is only being implemented now, you will see solutions from many people in the next few months, as we clear out the old problems from Mayhem.

101

High School Problems | Solutions Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario, Canada. L3S 3E8 [email protected] H205. A circular billiard table is given with a cue ball at the circumference. It is shot at an angle of to the line from the ball to the centre of the table. For what angles will the ball come back to this point, assuming the ball keeps going inde nitely? Solution by Samuel Wong, Mary Ward Catholic Secondary School, Toronto.

A

2

A

1



O A

3

We let the ball start at A1 as shown. Then \OA1 A2 = \A1 A2 O = (equal radii) and \A1 OA2 = 180 , 2 (angles sum to 180). Since the angle of incidence equals the angle of re ection, we have \A3 A2 O = \A1 A2 O = = \A2A3O by equal radii so \A3OA2 = \A2OA1 = 180 , 2 . Thus, the angles at the centre are always equal. The cue ball returns to A1 i

k(180 , 2 ) = 360c =) k(90 , ) = 180c; for some positive integers c and k. Since must be real, 90 , must be either a rational or irrational number. Case I: If 90 , is rational, it can be expressed in the form mn , where m and n are positive, relatively prime integers. Thus k  mn = 180c. We may take k = 180n and then m = c, so for all rational, the cue ball will return to A1 . Case II: If 90 , is irrational, let x = 90 , . Then kx = 180 c =) 180 c x = k . But x is irrational, so we have a contradiction.  ALVAREZ,  Also solved by MIGUEL CARRION Universidad Complutense de Madrid, Spain.

102

H206. For what values of n is an n-digit natural number uniquely determined from the sum and product of its digits?  Solutionby Miguel Carrion  Alvarez, Universidad Complutense de Madrid, Spain. The sum and product of the digits are two conditions, and so only two digits can be determined in general. This means n = 1 or 2, but when n = 2, the order cannot be determined (for example, consider 12 and 21), so the answer is n = 1. H207. Is there a natural number n, such that (n) = p, where p is an odd prime number? Solution by Bob Prielipp, University of Wisconsin-Oshkosh, WI. We shall show that there is NO positive integer n such that (n) is an odd prime number. This is an immediate consequence of the fact that (1) = 1 = (2) and the result established below. Theorem: If n is a positive integer, n  3, then 2 divides (n). Proof: Let n be a positive integer, n  3. Case I: n = 2a , where a is an integer, a  2. Then (n) = 2a,1 , where a , 1 is a positive integer, so 2 divides (n). Case II: n has an odd prime factor p. If the prime factorization of n is pa11 pa22    pakk , where pk is odd, then (n) = pa11,1pa22,1    pakk,1(p1 , 1)(p2 , 1)    (pk , 1): Because pk is odd, pk , 1 is even so, 2 divides (n)

 ALVAREZ,  Also solved by MIGUEL CARRION Universidad Complutense de Madrid, Spain, SAMUEL WONG, Mary Ward Catholic Secondary School, Toronto.

Advanced Problems | Solutions Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada. M1G 1C3 [email protected]

 k A185. Let bn be the highest power of 3 dividing 3n , 0  n  3k. 3k X 1. Calculate n=0 bn

Solution by Wai Ling Yee, student, University of Waterloo, Waterloo, Ontario. The exponent of the highest power of 3 dividing t! is

t  t   t  3 + 32 + 33 +    :

103

3k 3k ! Thus, the exponent of the highest power of 3 dividing = n n!(3k , n)! is $ % $ % $ %! k X 3k , n , 3k , n : j =1

3j

3j

3j

3k , n = 3k,j , an integer, 3j $ k% $ % $ k % 3 , n , 3 ,n 3j 3j 3j can only take two possible values: 0 and 1. When 3j j n, the value is 0. Otherwise, it is 1. Therefore, the value of bn is 3k divided by the highest power of 3 dividing n. Consider the numbers divisible by 3j ; 0  j  k , 1. Ignoring 0, we see that 3k,j of the numbers are divisible by 3j . These numbers can be written as 3j  (3m); 3j  (3m + 1), or 3j  (3m + 2). Therefore, there are 2  3k,j,1 numbers in f1; 2; : : : ; 3k g for which 3j is the highest power of 3

n

Since j + 3

dividing them.

1

3k X 1

Each of the 2  3k,j ,1 numbers contributes k,j to the sum 3 b for a n =0 n total of 23 . Therefore, k ,1 3k X 1 = 3X 1 + 2 = 2 k + 2: 3 n=0 bn n=1 bn

Also solved by Edward Wang, Wilfred Laurier University, Waterloo, On-

tario.

A186. Let an be the sequence de ned by a1 = 1, and 2 2 an+1 = 1 + a1 +n   + an ; n  1: Prove that every an is an integer.

Solution An April Fool's joke. It turns out that a1 , a2 , : : : , a43 are all integers, but a44 is not an integer. [Ed: can you prove this?]

104 n X A187. Let Sk(n) be the polynomial in n, such that Sk(n) = ik for i=1 all positive integers n, e.g. S0 (n) = n, S1 (n) = (n2 + n)=2. Prove that for k  1, n(n + 1) j Sk (n). Furthermore, prove that for k odd, k  3, n2(n + 1)2 j Sk (n).

Solution by Wai Ling Yee, student, University of Waterloo, Waterloo, Ontario. n(n + 1) . Assume that n(n +1) divides Note that n(n +1) j S1(n) = 2 all Sk (n) up to k = t. Consider: t+2 (n + 1) t+2 (n , 1 + 1)

=

=

=



1

t t

+

+

. . .

t+2 (1 + 1)

!

t t

=

t+2 + 2

t

=

t+2 + 1

t

nt+1 +

+2

t !

1 + 2

n,

(

+2

2

n,

t t

nt +    + t

!

+2 2

(

+2

!

n

+1

n,

1)

;

+1

t

;

1) + 1

t+1 + (n , 2)

t

+2 2

!

n,

(

2)

t+

;

2) + 1

!

t+1 + 2

t

t+1 + 1

t

!

!

!

1

(

+1

1

!

+1

+ 2

+2

t+1 + (n , 1)

+ 2 1

+2

t

!

t

t+2 + (n , 2)

 t+2 (2 + 1)

+2

t+2 + (n , 1) +

t+2 (n , 2 + 1)

t

nt+2 +

+2 2 +2 2

! !

t 2 ++ t 1 ++

t t t t

!

+2 +1

!

+2 +1

;

2+1

:

1+1

Adding all of the equations together, we have:

tX +1 X n n  t+2  X t +2 t +2 i = i + ji t + 2 , i i=2 i=1 i=0 j =1

nX +1

if and only if

  t  t+2  X (n + 1)t+2 , n , 1 = t +1 2 St+1 (n) + Si(n): i=1 t + 2 , i By the induction hypothesis, we need only to prove that n(n + 1) divides (n + 1)t+2 , n , 1 = n(n + 1)t+1 , which is true.

105 By induction, n(n + 1) j Sk (n) for k  1.

Now, S3 (n) = n (n4+1) ; therefore n2 (n + 1)2 divides S3(n). Assume that n2 (n +1)2 divides Sk (n) for all odd k up to k = t , 1, t even. Consider: 2

t+2 (n , 1)

n, ,

(

1

t+2

1)

=

=

2

nt+2 ,

t

n,

t+2 ,

(

+2 1

=

(2

,

t+2

1)

t+2 (1 , 1)

=

=

t+2 ,

t

t+2 , 1

t

2

!

t

!

1

+ 2

+ 2

!

1)

t t

nt ,    ,

t+1 +

t

+2

!

2

(

+2

!

n

+1

;

+1

t

n,

1)

;

1) + 1

t+1 + (n , 2)

n,

t

+2 2

!

n,

(

2)

t

;

2) + 1

2

t+1 +

t

t+1 + 1

t

!

!

!

(

+ 1

1

n,

n,

+2

+2 2

(

(

+1

+ 2

1

+ 2

+2

,    , tt . . .

!

1

t+2 , (n , 2)

t

nt+1 +

t

1)

,    , tt

t+2 (n , 2 , 1)

!

+2 2 +2 2

!

t t ! t ,, t t t ,,

2

1

!

+2 +1

!

+2 +1

;

2+1

:

1+1

Subtracting the sum of all the equations in the above group from the sum of all the equations in the rst group, we have:

    (n + 1)t+2 + nt+2 , 1 = 2 t +1 2 St+1 (n) + t +3 2 St,1 (n) t + 2 t + 2  +    + t , 1 S3 (n) + t + 1 S1 (n) : By the induction hypothesis, we have to prove only that n2(n + 1)2 divides

(n + 1)t+2 + nt+2 ,

t + 2 2 t + 1 (n + n) , 1:

106

t + 2 t + 2 2 t +2 2 t + 1 (n + n) , 1 = (n + 1 , 1)  , t+ 1 (n + n) , 1 2 (n + 1)1 + 1 = (n + 1)t+2 ,    + t +t 2 (n + 1)2 , tt + +1 t + 2 , t + 1 (n2 + n) , 1 t + 2 t + 2 t +2 2 = (n + 1) ,    + t (n + 1) , t + 1 (n2 + 2n + 1) t + 2 t + 2 t +2 2 = (n + 1) ,    + t (n + 1) , t + 1 (n + 1)2: t + 2 2 Therefore (n + 1)2 divides (n + 1)t+2 + nt+2 , t + 1 (n + n) , 1. t + 2 t +2 (n + 1) , t + 1 n , 1 t + 2 t + 2 t + 2 t +2 2 = n + + t n + t + 1 n +1 , t + 1 n, 1 t + 2 t +2 = n +    + t n2: ,  Therefore n2 divides (n + 1)t+2 + nt+2 , t+2 (n2 + n) , 1. nt+2 ,

t+1

Therefore n (n + 1) divides St+1 (n). By induction, n2(n + 1)2 divides Sk (n) for odd k  3. 2

2

Challenge Board Problems | Solutions Editor: Ravi Vakil, Department of Mathematics, One Oxford Street, Cambridge, MA, USA. 02138-2901 [email protected] We begin by dredging up an old favourite of ours. Well, maybe not so old { a solution appeared in [Mayhem 8: 4, 25: 1996]. We have since received a new solution that is worth printing. C64. The numbers x1, x2 , : : : , xn are such that x1 + x2 +    + xn = 0 and x21 + x22 +    + x2n = 1. Prove that there are two numbers among them whose product is no greater than ,1=n. (1991 Tournament of Towns)

107 tario.

Solution by Naoki Sato, student, University of Toronto, Toronto, OnChoose i such that x1  x2      xi  0  xi+1      xn . Then

x21 + x22 + : : : + x2i  x1  x1 + x1  x2 +    + x1  xi = x1 (x1 + x2 +    + xi) = ,x1 (xi+1 +    + xn )  ,(n , i)x1  xn:

Similarly,

x2i+1 +    + x2n  xn(xi+1 +    + xn) = ,xn(x1 + x2 +    + xi)  ,ix1  xn : Thus 1 = x21 +    + x2n  ,nx1  xn .

(As always, please send us new solutions to old problems.) C68. Proposed by Vin de Silva, student, Oxford University, Oxford, England. Let M be an n  n orthogonal matrix. (In other words, the rows are n vectors in n-space that are of length 1 and mutually perpendicular.) Let A be the k  k matrix in the upper-left corner of M , and let B be the (n , k)  (n , k) matrix in the lower-right corner of M . Prove that det A = det B . Solution by Eric Wepsic, D.E.Shaw and Co., New York, NY, USA. Let ci be the ith column of the orthogonal matrix Q, and let ri be the th i row. Let xi be the ith basis vector. Let M1 be the top k  k submatrix of Q, and let M2 be the bottom (n , k)  (n , k) submatrix. Then

det M1 = c1 ^ c2 ^    ^ ck ^ xk+1 ^    ^ xn = Qx1 ^ Qx2 ^    ^ Qxk ^ xk+1 ^    ^ xn : As Qt has determinant 1, this is equal to:

QtQx1 ^ Qt Qx2 ^    ^ QtQxk ^ Qtxk+1 ^    ^ Qt xn = x1 ^ x2 ^    ^ xk ^ rk+1 ^    ^ rn = det M2:

Solution by Sam Vandervelde, University of Chicago, Illinois, USA. Let Q1 be the matrix whose rst k columns are the rst k columns of Q, and whose remaining n , k columns are the last n , k columns of the identity matrix. Let Q2 be the same sort of thing with the rst k columns identical to I , and the last n , k columns those of Qt . Then an easy calculation (using Qt Q = I , QtI = Qt ) shows that QtQ1 = Q2. Taking determinants of both sides yields the desired result. The two solutions presented were in some sense identical, although the perspectives were di erent.

108

C69. Let Q be a polyhedron in R3. Let ~n1, : : : , ~nk and A1, : : : , Ak be the normal vectors and areas of the faces of Q respectively. Prove that k X Ai~ni = 0: i=1

 Solutionby Miguel Carri on  Alvarez, student, Universidad Complutense de Madrid, Spain. H P We have ki=1 Ai~ni = d~s, where d~s is the element of area (pointing outwards) and the integral is over the faces of the polyhedron. Then

 I I  I I I dsx ; dsy ; dsz = ~ux  d~s; u~ y  d~s; ~uz  d~s ; H for elds ~ux , ~uy , ~uz . We recall the divergence theorem: A~  d~s = R (constant ~  ~ui = 0. Thus . But ~ux , ~uy , ~uz are constant elds, so r HVd~sr~= (0A~); dV 0; 0). I

d~s =

I

Comments. 1. If  is the pressure eld in a uid, the total force on a (three-dimensional) object is I Z

F = ,  d~s = , (r~ )dV:

(Prove this!) In this case, we have constant pressure. What this result says is that a polyhedron sitting in a uid with constant pressure from all sides (and no other forces) will not move. 2. This is true in all dimensions. 3. Many other interesting results follow from this one, including the following variant of the Pythagorean theorem. Consider a tetrahedron ABCD with AB, AC , AD mutually perpendicular. Then

(ABC )2 + (ACD)2 + (ADB )2 = (BCD)2

where the brackets denote the area of the triangle. This is also true in other dimensions. If you can think of any other interesting consequences (including other variants of Pythagoras), please send them in!

109

PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later than 1 October 1997. They may also be sent by email to [email protected]. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication.

2214. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let n  2 be a natural number. Show that there exists a constant C = C (n) such that for all real x1 ; : : : ; xn  0 we have v u n uY xk  t (xk + C ):

n p X

k=1

k=1

Determine the minimum C (n) for some values of n. [For example, C (2) = 1.] 2215?. Proposed by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. Let P be a point inside a triangle ABC . It is known how to determine P such that PA + PB + PC is a minimum (known as Fermat's Problem for Torricelli). Determine P such that PA + PB + PC is a maximum.

110

2216. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Suppose that   1 is a natural number. 1. Determine the set of all 's such that the diophantine equation x + y2 = z2 has in nitely many solutions. 2.? For any such , determine all solutions of this equation. 2217. Proposed by Bill Sands, University of Calgary, Calgary, Alberta. (a) Prove that for every suciently large positive integer n, there are arithmetic progressions a1 ; a2 ; a3 and b1; b2; b3 of positive integers such that n = a1 b1 + a2 b2 + a3 b3. (b) What happens if we require a1 = b1 = 1? (This is a variation of problem 3 of the 1995/96 Alberta High School Mathematics Competition, Part II, which will appear in a future Skoliad Corner.) 2218. Proposed by Victor Oxman, University of Haifa, Haifa, Israel. Suppose that a, b, c are positive real numbers and that abc = (a + b , c)(b + c , a)(c + a , b):

Clearly a = b = c is a solution. Determine all others. 2219. Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. Show that there are an in nite number of solutions of the simultaneous equations:

x2 , 1 = (u + 1)(v , 1) y2 , 1 = (u , 1)(v + 1) with x; y;u; v positive integers and x 6= y .

2220. Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Let V be the set of an icosahedron's twelve vertices, which can be partitioned into four classes of three vertices, each one in such a way that the three selected vertices of each class belong to the same face. How many ways can this be done?

111  2221. Proposed by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina. Find all members of the sequence an = 32n,1 + 2n,1 ; (n 2 N) which are the squares of any positive integer. 2222. Proposed by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. Find the value of the continued root:

v s u r u t4 + 27 4 + 29 4 + 31q4 + 33p: : ::

NOTE: This was inspired by the problems in chapter 26 \Ramanujan, In nity and the Majesty of the Quattuordecillion", pp 193{195, in \Keys to In nity" by Cli ord A. Pickover, John Wiley and Sons, 1995. 2223. Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. We are given a bag with n identical bolts and n identical nuts, which are to be used to secure the n holes of a gadget. The 2n pieces are drawn from the bag at random one by one. Throughout the draw, bolts and nuts are screwed together in the holes, but if the number of bolts exceeds the number of available nuts, the bolt is put into a hole until one obtains a nut, whereas if the number of nuts exceed the number of bolts, the nuts are piled up, one on top of the other, until one obtains a bolt. Let L denote the discrete random variable which measures the height of the pile of nuts. Find E [L] + E [L]2 . 2224. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Point P lies inside triangle ABC . Triangle BCD is erected outwardly on side BC such that \BCD = \ACP and \CBD = \ABC . Prove that if the area of quadrilateral PBDC is equal to the area of triangle ABC , then triangles ACP and BCD are similar. 2225. Proposed by Kenneth Kam Chiu Ko, Mississauga, Ontario. (a) For any positive integer n, prove that there exists a unique n-digit number N such that: (i) N is formed with only digits 1 and 2; and (ii) N is divisible by 2n . (b) Can digits \1" and \2" in (a) be replaced by any other digits?

112

SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. We have identi ed \anonymous" in the solutions of problems 2064,  2065, 2067, 2068, 2069, 2071 and 2077 as MARIA MERCEDES SANCHEZ BENITO, I.B. Luis Bu~nuel, Madrid, Spain. The name of VICTOR OXMAN, University of Haifa, Haifa, Israel, should be added to the list of solvers of problem 2091. The name of ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick, should be added to the list of solvers to problem 2104.

2113. [1996: 35] Proposed by Marcin E. Kuczma, Warszawa, Poland. Prove the inequality

! X n ! X n ! n n ai bi ! X X ai bi  (ai + bi) i=1 i=1 i=1 i=1 ai + bi for any positive numbers a1 ; : : : ; an ; b1; : : : ; bn . I. Solution by Vedula N. Murty, Andhra University, Visakhapatnam, India. The given inequality follows from the easily veri ed identity:

! X n ! X n ! n n ai bi ! X X ai bi , (ai + bi) i=1 i=1 i=1 i=1 ai + bi X (aibj , aj bi)2 : = ( a + b )( a + b ) i i j j 1i AC by the less stringent condition \ABC < 90 . (That seems

117 to be sucient for the inequality in his rst proof: 90 > \ABC > \PBC implies sin \ABC > sin \PBC .) Neither he nor the other solvers explicitly stated that they had stronger results.] Solution I. Let E be the point where the internal bisector of \P of 4PBC meets side BC . Then BE = PB = sin \PCB > sin \ACB = AB = BD ; EC PC sin \PBC sin \ABC AC DC so that E lies between D and C . It follows that \BPD < \DPC . [Editor's further comment: The above argument clearly continues to work for some positions of P on the side AC even when \ABC  90 . A set of conditions on P and 4ABC under which \BPD < \DPC is implicit in Shirali's second argument.] Solution II. The set of points X in the plane of the triangle for which \BXD = \DXC is , the circle of Apollonius through A with respect to B XB = BD ; it is and C . [Equivalently, is the locus of points X for which XC DC the circle through A and D whose centre lies on BC ; inversion in this circle xes A and D, and interchanges B with C ; see, for example, H.S.M. Coxeter, Introduction to Geometry, x6.6 pages 88-89.] When AB > AC [so that BD > DC ], the set of points X for which \BXD < \DXC is the set of points interior to . In this case the centre of lies on the extension of BC beyond C , so the segment AC lies entirely within . It follows that for all points P on AC , we have \BPD < \DPC . [When AB < AC , the centre of lies on the part of BC beyond B , so that now we seek points P outside

; when \B  90 , the entire line segment AC lies outside , which agrees with the rst proof. When \B is obtuse, then some points of AC lie inside

{ for those points, the desired inequality no longer holds.] Also solved by FEDERICO ARDILA, student, MassachusettsInstitute of   UniTechnology, Cambridge, Massachusetts, USA; SEFKET ARSLANAGIC, versity of Sarajevo, Sarajevo, Bosnia and Herzegovina; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JORDI DOU, Barcelona, Spain;   Y,  Ferris State University, Big Rapids, Michigan, USA; VACLAV KONECN KEE-WAI LAU, Hong Kong; WALDEMAR POMPE, student, University of Warsaw, Poland; D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; GEORGE TSAPAKIDIS, Agrino, Greece; MELETIS VASILIOU, Elefsis, Greece; and the proposer.

2118. [1996: 76] Proposed by Paul Yiu, Florida Atlantic University, Boca Raton, Florida, USA. The primitive Pythagorean triangle with sides 2547 and 40004 and hypotenuse 40085 has area 50945094, which is an 8{digit number of the form abcdabcd. Find another primitive Pythagorean triangle whose area is of this form.

118 Solution by the proposer. There are three such primitive Pythagorean triangles:

m

n m2 , n2 2mn m2 + n2

(i) 146 137 (ii) 146 9 (iii) 105 32

2547 21235 10001

40004 2628 6720

40085 21397 12049

Area

50945094 27902790 33603360

8{digit numbers of the form abcdabcd can be factored as 10001  N = 73  137  N for an integer N in the range 1000  N < 10000. The sides of a primitive Pythagorean triangle are given by m2 , n2 , 2mn, m2 + n2, with relatively prime integers m and n of di erent parity. The area of the triangle is 4 = mn(m , n)(m + n). Note that the factors m, n, m + n, and m , n are pairwise relatively prime, and exactly one of them is even. These numbers are in the order (i) m + n > m > n > m , n, or (ii) m + n > m > m , n > n. [Clearly, none of m; n or m , n can be 10001r. Also m + n 6= 10001; otherwise, mn  (10000)(1) and 4 > 108 .] Each of 137 and 73 divides exactly one of the four numbers m; n;m  n. We list the twelve di erent types in the table below, in which h and k are positive integers. Type (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)

m+n 2(137h) + 73k 137h + 2(73)k 137h + 73k ,137h + 2(73)k 73k 73k 137h + 73k 2(137h) , 73k 137h 137h 137h 73k

m 137h + 73k 137h + 73k 73k 73k ,137h + 73k 1 (137h + 73k) 2 137h 137h 137h , 73k 1 (137h + 73k) 2 73k 137h

n

h k 137h ,137h + 73k 137h 1 (,137h + 73k) 2 73k 137h , 73k 73k 1 (137h , 73k) 2 137h , 73k ,137h + 73k 137 73

m,n conditions 73k k odd 137h h odd 73k , 137h h + k odd 137h h odd 73k , 2(137)h k odd 137h h  k odd 173h , 73k h + k odd 73k k odd 137h , 2(73k) h odd 73k h  k odd ,137h + 2(73k) h odd 2(137)h , 73k k odd

In Types (1) and (2), 4 > ((137h)(73k))2 = (hk)2(10001)2 > 108. In each of Types (3), (4), (5), (6), note that 137h > 100. This cannot be the smallest of the four numbers m; n; m  n, and must therefore be the third in order. The area of the triangle is greater than (137h)83  1. We need 3 therefore consider only h satisfying (137h)3 < 108; h < 10  3:388; in 137 other words, h = 1, 2, 3. In each of these types, the smallest number is a linear function ,p 137h + q 73k, where each of p and q is 1, 2 or 12 .

119 Note that ,p 137h + q 73k < h1 73 since the product of the two smallest measurements cannot exceed 10001. It follows that 0 < ,p 137h + q 73k < h1 73: From this,

p  137  h < k < p  137  h + 1 : q 73 q 73 qh In each of Types (7), (8), (9), (10), either n or m , n is 73k, so 4 > (73k)3. In other words, k  6. Also, from 0 < p 137h , q 73k < k1 137, we have q  73  k < h < q  73  k + 1 : p 137 p 137 pk p q

possible (

(3)

1

1

(1 2), (2 4)

(1 2)

(i)

(4)

1

1

(1 2), (2 4)

(1 2)

(ii)

(5)

2

1

(1 4), (2 8)

none

(6)

1 2

1 2

(1 2), (1 3), (2 4), (3 6)

(1 3)

(7)

1

1

(1 1)

none

(8)

1

1

(1 1)

(1 1)

(9)

1

2

(2 1)

none

(10)

1 2

1 2

(1 1), (2 1), (2 2),

(1 1)

(2 3), (3 5)

(3 5)

Type

; ; ; ; ; ; ; ; ;

h; k ; ; ; ; ; )

relevant (

; ;

;

h; k

; ;

)

triangle

m

= 178,

n

= 41

m

= 137,

n

= 64

; ; ; ; ; m n Type (11). We consider (h; k) in either of the parallelograms 137h , 73k = 0; 137h , 73k = 100; , 137h + 2(73k) = 0; ,137h + 2(73k) = 464; 137h , 73k = 0; 137h , 73k = 464; , 137h + 2(73k) = 0; ,137h + 2(73k) = 100: p 3 (Note: 464 = [ 108]). These are the points (h; k) = (1; 1); (2; 3); (t; t) for t = 2; : : : ; 8; and we need only consider (h;k) = (1; 1) (since h is odd and gcd (h;k) = 1). But this triangle, with m = 73, n = 64, is too small; it has area 5760576. Type (12). We consider (h; k) in either of the parallelograms ,137h + 73k = 0; ,137h + 73k = 100; 2(137)h , 73k = 0; 2(137)h , 73k = 464; (iii)

= 388,

= 23

120

,137h + 73k = 0; ,137h + 73k = 464; 2(137)h , 73k = 0; 2(137)h , 73k = 100:

These are the points

(h;k) = (1; 2); (1; 3); (2; 7); (3; 11); (2; 4); (3; 6); and we need only consider (h;k) = (1; 3); (2; 7) and (3,11). But the corre-

sponding triangles are all too big. Remark Enlarging the primitive triangle with m = 73; n = 64 by the factors 2, 3, 4, we obtain non-primitive Pythagorean triangles of areas 23042304, 51845184, and 92169216 respectively. Enlargements of the primitive triangles (i), (ii), (iii) all have areas exceeding 108: The non-primitive triangle from m = 137, n = 73 has area 13441344. Both the additional triangles were also found by TIM CROSS, King Edward's School, Birmingham, England; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; and RICHARD I. HESS, Rancho Palos Verdes, California, USA. The following readers each found one of the triangles: CHARLES ASHBACHER, Cedar Rapids, Iowa, USA; SAM BAETHGE, Science Academy, Austin, Texas, USA; MANSUR BOASE, student, St. Paul's School, London, England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; DAVID HANKIN, Hunter College Campus Schools, New York, NY, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; WALTHER JANOUS, Ursulinengymnasium,   Y,  Ferris State University, Big Rapids, Innsbruck, Austria; VACLAV KONECN Michigan, USA; DAVID E. MANES, State University of New York, Oneonta, NY, USA; P. PENNING, Delft, the Netherlands; CORY PYE, student, Memorial University of Newfoundland, St. John's, Newfoundland; JOEL SCHLOSBERG, student, Hunter College High School, New York NY, USA;  HEINZ-JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece. There was one incorrect submission.

2119. [1996: 76] Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. (a) Show that for any positive integer m  3, there is a permutation of m 1's, m 2's and m 3's such that (i) no block of consecutive terms of the permutation (other than the entire permutation) contains equal numbers of 1's, 2's and 3's; and

121 (ii) there is no block of m consecutive terms of the permutation which are all equal. (b) For m = 3, how many such permutations are there? Solution by P. Penning, Delft, the Netherlands. (b) I found two: aabbcbcca and abbcbccaa; where a; b; c is any permutation of 1; 2; 3. (In fact these are mirror-images of one another.) Thus there are 3!  2 = 12 such permutations of 1's, 2's and 3's. (a) To generate a solution for m > 3, start with three blocks of m equal terms, and to make it satisfy condition (ii) let the last term of each block change places with the rst term of the next block (and do the same with the rst and last block):

z m}|,2 { z m}|,2 { z m}|,2 { 1 22 : : : 2 3 2 33 : : : 3 1 3 11 : : : 1 2:

Note that for m = 3 this permutation violates condition (i). [It is easy to check that it works whenever m > 3.|Ed.] Also note that if the permutation is considered as a cycle then there are solutions only if m > 3. Also solved by FEDERICO ARDILA, student, Massachusetts Institute of Technology, Cambridge, Massachusetts, USA; MANSUR BOASE, student, St. Paul's School, London, England; HANS ENGELHAUPT, Franz{Ludwig{ Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; DAVID E. MANES, State University of New York, Oneonta, NY, USA; JOEL SCHLOSBERG, student, Hunter College High School, New York, NY, USA; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; and the proposer. Part (b) only was solved by CHARLES ASHBACHER, Cedar Rapids, Iowa, USA. Note that Penning's solution shows that for m > 3, condition (ii) can be strengthened just a teensy bit to: (ii)0 there is no block of m , 1 consecutive terms of the permutation which are all equal. However, there is an even better result. For any m  3 there is a permutation of m 1's, m 2's and m 3's which satis es condition (i) and also (ii)? there is no block of three consecutive terms of the permutation which are all equal. (Thanks to expert colleague James Currie for suggesting examples of such permutations, which readers might enjoy nding for themselves!)

122

2120. [1996: 76] Proposed by Marcin E. Kuczma, Warszawa, Poland. For

Let A1 A3 A5 and A2 A4 A6 be non-degenerate triangles in the plane.

i = 1; : : : ; 6 let `i be the perpendicular from Ai to line Ai,1Ai+1 (where of course A0 = A6 and A7 = A1 ). If `1 ; `3; `5 concur, prove that `2; `4; `6 also concur. Solution by Christopher J. Bradley, Clifton College, Bristol, UK. Let the vector position of the vertex Ai relative to the point of concurrency of `1; `3 ; `5 be a~i . Then a~1  (a~2 , a~6 ) = a~3  (a~4 , a~2 ) = a~5  (a~6 , a~4 ) = 0. In particular, the sum is also zero, so a~2  (a~1 , a~3 ) + a~4  (a~3 , a~5) + a~6  (a~5 , a~1) = 0: (1) Now suppose that the perpendicular from A2 to A1 A3 meets the perpendicular from A4 to A3 A5 at B , with position vector ~b. (These perpendiculars cannot be parallel.) Then (~b , a~2 )  (a~1 , a~3 ) = 0 and (~b , a~4 )  (a~3 , a~5 ) = 0:

Adding gives ~b  (a~1 , a~5 ) = a~2  (a~1 , a~3 ) + a~4 (a~3 , a~5 ), which, by (1), is equal to ,a~6  (a~5 , a~1 ). Hence (~b , a~6 )  (a~1 , a~5 ) = 0, which means that BA6 ? A1 A5 . Also solved by CLAUDIO ARCONCHER, Jundia, Brazil; FEDERICO ARDILA, student, Massachusetts Institute of Technology, Cambridge, Massachusetts, USA; JOHN G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; MELETIS VASILIOU, Elefsis, Greece; and the proposer. Problem 2120 can be found in many reference texts. Dan Pedoe provides two proofs in his Geometry, A comprehensive Course (Dover, 1998). Clerk Maxwell's remarkable treatment of the problem, reducing it to an observation about the radical axes of appropriate circle, appears in x28.4, page 115. Earlier, in x6.1, pages 35-37, Pedoe uses barycentric coordinates for a proof of a closely related problem involving reciprocal triangles. He comments that whole books have been written on the subject of reciprocal gures. Seimiya uses Steiner's theorem in his solutions (Jacob Steiner, Gesammelte Werke I, p. 189): A necessary and sucient condition in order for `1; `3; `5 to be concurrent is

A1A22 + A3A24 + A5A26 = A2A23 + A4A25 + A6A21:

Since (by permuting the subscripts) this is also a necessary and suf cient condition for `2 ; `4 ; `6 to concur, our problem follows immediately. Arconcher refers to this condition as Carnot's Theorem, but he provides no

123 reference. Ardila discovered the theorem for himself and showed that it follows fairly quickly from Pythagoras' Theorem, so maybe we should refer to it as the Pythagoras-Ardila-Steiner-Carnot Theorem. Seimiya added that the problem may also be found in Aref and Wernick, Problems and Solutions in Euclidean Geometry (Dover), p. 55 problems 2.32, He includes two of his interesting generalizations that have appeared in Japanese, one in 1928 and one in 1967. Finally, Heuver found the problem as Ex. 7 of x54 of George Salmon's A Treatise on Conic Sections, 6th ed. His solution exploits Salmon's imaginative use of pencils of lines in Cartesian coordinates.

2121. [1996: 76] Proposed by Krzysztof Chelminski,  Technische Hochschule Darmstadt, Germany; and Waldemar Pompe, student, University of Warsaw, Poland. Let k  2 be an integer. The sequence (xn ) is de ned by x0 = x1 = 1 and k xn+1 = xn + 1 for n  1: xn,1

(a) Prove that for each positive integer k  2 the sequence (xn ) is a sequence of integers. (b) If k = 2, show that xn+1 = 3xn , xn,1 for n  1. (c)? Note that for k = 2, part (a) follows immediately from (b). Is there an analogous recurrence relation to the one in (b), not necessarily linear, which would give an immediate proof of (a) for k  3? I Solution ((a) and (b) only) by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. (a) We immediately get x2 = 2 and x3 = 2k + 1. Now we use mathematical induction for the proof. Assume that x0 ; x1 ; : : : ; xn are all natural numbers. We must show that xn+1 2 N. First we note that since xn,2  xn = xkn,1 +1 it follows that xn,2 and xn,1 are relatively prime. Using xn = (xkn,1 + 1)=xn,2 we infer that k k k k xn+1 = xxn + 1 = (xn,1xk+ 1)x + xn,2 : n,1

n,2 n,1 Thus obviously xkn,2 divides N = (xkn,1 + 1)k + xkn,2 since xn is a natural number. Furthermore, modulo xn,1 we have: N  1 + xkn,2 = xn,3  xn,1  0:

That is, xn,1 also divides N and we are done.

124 (b) Now,

xn+1 = xxn + 1 () xn,1  xn+1 , x2n = 1: n,1 That is, the sequence fyng = fxn,1  xn+1 , x2n g is constant. Setting yn+1 = yn we have xn  xn+2 , x2n+1 = xn,1  xn+1 , x2n () xn (xn + xn+2) = xn+1(xn,1 + xn+1) xn + xn+2 = xn,1 + xn+1 : () xn+1 xn That is, the sequence fzn g = f(xn,1 + xn+1 )=xn g is constant. From z1 = 3 we get (xn,1 + xn+1 )=xn = 3; that is, xn+1 = 3xn , xn,1 for all n  1, 2

as claimed. II Solution by Christopher J. Bradley, Clifton College, Bristol, UK. We rst establish the following lemma: Lemma. Suppose a and b have highest common factor 1 and (ak + bk +1)=ab is an integer (where k is an integer, k  2), then (i) c = (bk + 1)=a is an integer; (ii) (bk + ck + 1)=bc is an integer; (iii) b and c have highest common factor 1. Proof. Let ak + bk + 1 = (a; b)ab where (a; b) is an integer. Then since a divides (a; b)ab , ak it also divides bk + 1; that is, (bk + 1)=a = c is an integer, which proves (i). Also

bk + ck + 1 = ac + ck = a + ck,1 = ak + (bk + 1)k,1 : bc bc b ak,1 b Now bk + 1 = ac has a as a factor, so ak + (bk + 1)k,1 is divisible by ak,1 and hence the numerator is divisible by ak,1 . Also ak + 1 is divisible by b (from part (i)), so multiplying out the numerator by the Binomial Theorem we see that the numerator is divisible by b. But a and b have highest common factor 1, so the numerator is divisible by ak,1 b. Hence k k (b;c) = b +bcc + 1 is an integer, which proves (ii). Since bk + ck + 1 = (b;c)bc, if b and c have a common factor h, then h divides bk + ck , (b; c)bc; that is, h divides 1. Hence b and c have highest common factor 1, which proves (iii), and the lemma is proved. We claim that the recurrence relation sought is

xn+1 = (xn,1; xn)xn , xkn,,11

()

125 with (xn,1 ; xn ) = (xkn + xkn,1 +1)=(xnxn,1 ), which is but one step away from xn+1 = (xkn + 1)=xn,1 . That (xn,1 ; xn ) is always an integer follows by induction, using the lemma, with the start of induction satis ed since x0 = x1 = 1 have highest common factor 1 and (x0; x1 ) = 3 is an integer. Relation () then establishes that fxn g is a sequence of integers. This gives us parts (c) and (a). The term (x2n + x2n,1 +1)=(xnxn,1 ) is known from work on alternate terms of the Fibonacci sequence to be equal to 3 for all n  2 and is also 3 for n = 1, which proves (b). Parts (a) and (b) together were also solved by FEDERICO ARDILA, student, Massachusetts Institute of Technology, Cambridge, Massachusetts, USA; FLORIAN HERZIG, student, Perchtoldsdorf,Austria; and the proposers.   University of Sarajevo, Part (b) alone was solved by SEFKET ARSLANAGIC, Sarajevo, Bosnia and Herzegovina; and TIM CROSS, King Edward's School, Birmingham, England. There was one incorrect solution. JANOUS comments that this was posed as a problem of the nal round of the third Austrian Mathematical Olympiad held in 1972, and refers the in terested reader to the book \Osterreichische Matematik-Olympiaden" 19701989, G. Baron & E. Windischbacher, Innsbruck 1990, problem 42.

2122.[1996: 77] Proposed by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. Little Sam is a unique child and his math marks show it. On four tests this year his scores out of 100 were all two-digit numbers made up of eight di erent non-zero digits. What's more, the average of these scores is the same as the average if each score is reversed (so 94 becomes 49, for example), and this average is an integer none of whose digits is equal to any of the digits in the scores. What is Sam's average? I. Solution by Mansur Boase, student, St. Paul's School, London, England. Let the four marks be 10a + b, 10c + d, 10e + f and 10g + h. Then 10a + b + 10c + d + 10e + f + 10g + h = 10b + a + 10d + c + 10f + e + 10h + g; so and

9(a + c + e + g ) = 9(b + d + f + h)

a + c + e + g = b + d + f + h:

(1)

The average of the four marks is therefore 10a + b + 10b + a + 10c + d + 10d + c + 10e + f + 10f + e + 10g + h + 10h + g 4 2 

= 11(a + b + c + d8+ e + f + g + h) :

(2)

126 The average must consist only of k's and 0's, the digits not equal to a, b, c, d, e, f , g or h. Thus (2) is

11(1 + 2 +    + 9 , k) = 11(45 , k) ; 8 8 and so 8 j (45 , k). Since k  9, the only solution for k is k = 5, and therefore Sam's average is 11(40)=8 = 55. An example of four such marks is 98; 76; 34; 12. II. Solution by Tara McCabe, student, Mount Allison University, Sackville, New Brunswick. [Editor's note: McCabe rst obtained equation (1) as above, using the same notation.] Since a; : : : ; h are all di erent and non-zero, their total must lie between 1 +    + 8 = 36 and 2 +    + 9 = 44 inclusively. From (1),

36 = 18  a + c + e + g  22 = 44 : 2 2

(3)

Letting the average be the two digit number xy ,

10a + b + 10c + d + 10e + f + 10g + h = 10x + y; 4

which by (1) means

11(a + c + e + g ) = 4(10x + y ): Consequently, a + c + e + g must be divisible by 4, and from (3), a + c + e + g = 20. Therefore 11(20) = 4(10x + y), and so 55 = 10x + y . Little Sam has an average of 55.

A natural addition to this problem is to try to nd Little Sam's four test scores ab; cd; ef and gh. We know that a + c + e + g = 20, and the problem becomes: how many sets of test scores are possible? First, choose four digits from f1; 2; 3; 4; 6; 7; 8; 9g to be a; c; e and g (0 and 5 are not allowed). Notice that for a sum of 20 to be possible, two,4digits must be chosen from f1; 2; 3; 4g and two from f6; 7; 8; 9g. There are 2 = 6 ways to choose two digits from f1; 2; 3; 4g. The sum S1 of these two digits is such that 3  S1  7, and the only sum that can occur twice is 5. Similarly, there are 6 ways to choose two digits from f6; 7; 8; 9g, the sum S2 of these two digits is such that 13  S2  17, and the only sum that can occur twice is 15. There are 2  2 = 4 combinations when S1 = 5 (and S2 = 15) and one combination for each other sum S1 . Therefore, there are eight ways to choose four digits from f1; 2; 3; 4; 6; 7; 8; 9g such that the sum is 20. Once a; c; e and g are chosen, b;d; f and h are simply the four remaining digits. There are 4! = 24 di erent ways to assign these four digits to be b; d; f and h. Therefore there are 8  24 = 192 di erent sets of four test scores possible for Little Sam. He's not such a unique child after all!

127 Also solved by SAM BAETHGE, Nordheim, Texas, USA; CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, UK; TIM CROSS, King Edward's School, Birmingham, UK; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; NOEL EVANS and CHARLES DIMINNIE, Angelo State University, San Angelo, Texas, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, New York, USA; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; JOHN GRANT MCLOUGHLIN, Okanagan University College, Kelowna, B.C.; P. PENNING, Delft, the Netherlands; CORY PYE, student, Memorial University of Newfoundland, St. John's, Newfoundland; JOEL SCHLOSBERG, student, Hunter College High School, New  York, NY, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. Three other solvers did not prove that 55 is the only possible answer. There was also one incorrect answer sent in. At the end of his solution, the proposer asks for the number of sets of test scores satisfying the problem, but only McCabe was able to read his mind and answer this too!

2123. [1996: 77] Proposed by Sydney Bulman{Fleming and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. It is known (e.g., exercise 23, page 78 of Kenneth H. Rosen's Elementary Number Theory and its Applications, Third Edition) that every natural number greater than 6 is the sum of two relatively prime integers, each greater than 1. Find all natural numbers which can be expressed as the sum of three pairwise relatively prime integers, each greater than 1. Solution by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. Since every number greater than 6 is the sum of two relatively prime integers, then each even number greater than 6 can be expressed as the sum of of two relatively prime odd numbers. If we add 2 to each of these expressions we see that every even number greater than 8 can be expressed as the sum of three pairwise relatively prime integers (two relatively prime odd integers and 2). All odd numbers can be expressed in the form 18N + k, where N is a nonnegative integer and k is an odd number less than 18. Note that: 18N + 1 = (6N , 3) + (6N , 1) + (6N + 5) for N  1 18N + 3 = (6N , 1) + (6N + 1) + (6N + 3) for N  1 18N + 5 = (6N , 1) + (6N + 1) + (6N + 5) for N  1

128 for N  1 for N  1 for N  1 for N  1 for N  0 for N  1 Clearly each of the pairs of terms in each expression is relatively prime, since if there is a number which divides each term in a pair it must divide the di erence. The restriction on N ensures that each term is greater than 1. Putting all this together shows that the numbers which can be so written are 10, 12, 14, 15, 16, and all the natural numbers greater than 17. [Ed: it is easily veri ed that all other integers do not have the desired property.] Also solved by RICHARD I. HESS, Rancho Palos Verdes, California, USA; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL JOSEPHY, Universidad de Costa Rica, San Jose, Costa Rica; KEE-WAI LAU, Hong Kong; DAVID E. MANES, State University of New York, Oneonta, NY, USA; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; and the proposers. There was 1 incorrect submission. Most solvers used mod 18 or mod 12 arithmetic to handle the odd values of n. MANES pointed out that both this problem and the problem in Rosen's book can be found in Sierpinski's book, \250 Problems in Elementary Number Theory", American Elsevier, New York, 1970, (problems 47 and 48) where they are both solved.

18N + 7 18N + 9 18N + 11 18N + 13 18N + 15 18N + 17

= = = = = =

(6N , 1) + (6N + 3) + (6N + 5) (6N + 1) + (6N + 3) + (6N + 5) (6N + 1) + (6N + 3) + (6N + 7) (6N + 1) + (6N + 5) + (6N + 7) (6N + 3) + (6N + 5) + (6N + 7) (6N + 1) + (6N + 7) + (6N + 9)

Crux Mathematicorum

Founding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell Editors emeriti / Redacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem

Founding Editors / Redacteurs-fondateurs: Patrick Surry & Ravi Vakil Editors emeriti / Redacteurs-emeriti: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

129

THE ACADEMY CORNER No. 10

Bruce Shawyer All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 This month, we present a university entrance scholarship examination paper from the 1940's. Thanks to Georg Gunther, Sir Wilfred Grenfell College, Corner Brook, Newfoundland, for providing this. We challenge today's university students to solve these problems | send me your nice solutions! 1. Find all the square roots of

p

1 , x + 22x , 15 , 8x2: 2. Find all the solutions of the system of equations:

x + y + z = 2;

x2 + y2 + z2

= 14; xyz = ,6: 3. Suppose that n is a positive integer and that Ck is the coecient of xk in the expansion of (1 + x)n . Show that n X , 1)! : (k + 1)Ck2 = (n +n!2)(n(2,n1)! k=0 4. (a) Suppose that a 6= 0 and c 6= 0, and that ax3 + bx + c has a factor of the form x2 + px + 1. Show that a2 , c2 = ab. (b) In this case, prove that ax3 + bx + c and cx3 + bx2 + a have a common quadratic factor. 5. Prove that all the circles in the family de ned by the equation

x2 + y2 , a(t2 + 2)x , 2aty , 3a2 = 0

(a xed, t variable) touch a xed straight line. 6. Find the equation of the locus of a point P which moves so that the tangents from P to the circle x2 + y 2 = r2 cut o a line segment of length 2r on the line x = r.

130 7. If the tangents at A, B and C to the circumcircle of triangle 4ABC meet the opposite sides at D, E and F , respectively, prove that D, E and F are collinear. 8. Find the locus of P which moves so that the polars of P , with respect to three non-intersecting circles, are concurrent. 9. Suppose that P is a point within the tetrahedron OABC . Prove that \AOB + \BOC + \COA is less than \APB + \BPC + \CPA. 10. Two unequal circles of radii R and r touch externally, and P and Q are the points of contact of a common tangent to the circles, respectively. Find the volume of the frustum of a cone generated by rotating PQ about the line joining the centres of the circle. 11. Prove that sin2 ( + )+sin2 ( + ) , 2 cos( , ) sin( + ) sin( + ) = sin2 ( , ):

12. Three points A, B and C are on level ground. B is east of A, C is N. 49 E. of A, and C is N. 11300 W. of B . Find the direction of C as seen from the mid-point of AB . 13. With each corner of a square of side r as a centre, four circles of radius r are drawn. Show that the area of the central curvilinear quadrilateral formed inside the square by the intersection of the four circles is

 p  1, 3+ :



r2

3

14. An observer on a boat is vertically beneath the centre of a bridge, which crosses a straight canal at right angles. Looking upwards, the observer sees that the angle subtended by the length of the bridge is 2 . The observer then rows a distance  along the middle of the canal, and then nds that the length of the bridge now subtends an angle of 2 . Show that the length of the bridge is

p

2

cot2 , cot2

:

131

THE OLYMPIAD CORNER No. 181

R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. We begin this number with two contests. Thanks go to Richard Nowakowski, Canadian Team Leader to the 35th IMO in Hong Kong, for collecting them and forwarding them to us.

SELECTED PROBLEMS FROM THE ISRAEL MATHEMATICAL OLYMPIADS, 1994 1. p and q are positive integers. f is a function de ned for positive p q

numbers and attains only positive values, such that f (xf (y )) = x y . Prove that q = p2 . 2. The sides of a polygon with 1994 sides are ai = p4 + i2, i = 1; 2; : : : ; 1994. Prove that its vertices are not all on integer mesh points. 3. A \standard triangle" in the plane is a ( lled) isosceles right triangle whose sides are parallel to the x and y axes. A nite family of standard triangles, containing at least three, is given. Every three of this family have a common point. Prove that there is a point common to all triangles in that family. 4. A shape c0 is called \a copy of the planar shape c" if the following conditions hold: (i) There are two planes  and  0 and a point P that does not belong to either of them. (ii) c 2  and c0 2  0 . (iii) A point X 0 satis es X 0 2 c0 i X 0 is the intersection of  0 with the line passing through X and P . Given a planar trapezoid, prove that there is a square which is a copy of this trapezoid. 5. Find all polynomials p(x), with real coecients, satisfying for all x.

(x , 1)2p(x) = (x , 3)2p(x + 2)

132

PROBLEMS FROM THE BI-NATIONAL ISRAEL-HUNGARY COMPETITION, 1994 1. a1; : : : ; ak ; ak+1 ; : : : ; an are positive numbers (k < n). Suppose that the values of ak+1 ; : : : ; an are P xed. How should one choose the values of a1 ; : : : ; ak in order to minimize i;j;i6=j aaji ? 2. Three given circles pass through a common point P and have the same radius. Their other points of pairwise intersections are A, B , C . The 3 circles are contained in the triangle A0 B 0 C 0 in such a way that each side of 4A0 B0 C 0 is tangent to two of the circles. Prove that the area of 4A0 B0 C 0 is at least 9 times the area of 4ABC . 3. m, n are two di erent natural numbers. Show that there exists a real number x, such that 13  fxng  23 and 13  fxmg  23 , where fag is the fractional part of a. 4. An \n-m society" is a group of n girls and m boys. Show that there exist numbers n0 and m0 such that every n0 -m0 society contains a subgroup of ve boys and ve girls in which all of the boys know all of the girls or none of the boys knows none of the girls. Last issue we gave ve more Klamkin Quickies. Next we give his \Quicky" solutions to these problems. Many thanks to Murray S. Klamkin, the University of Alberta, for creating the problems and solutions.

ANOTHER FIVE KLAMKIN QUICKIES October 21, 1996

6. Determine the four roots of the equation x4 + 16x , 12 = 0. Solution. Since

x4 + 16x , 12 = (x2 + 2)2 , 4(x , 2)2 = (x2 + 2x , 2)(x2 , 2x + 6) = 0; p p the four roots are ,1  3 and 1  i 5. 7. Prove that the smallest regular n-gon which can be inscribed in a given regular n-gon is one whose vertices are the midpoints of the sides of the given regular n-gon. Solution. The circumcircle of the inscribed regular n-gon must intersect each side of the given regular n-gon. The smallest that such a circle can be is the inscribed circle of the given n-gon, and it touches each of its sides at its midpoints.

133

8. If 311995 divides a2 + b2, prove that 311996 divides ab.

Solution. If one calculates 12 ; 22 ; : : : ; 302 mod 31 one nds that the sum of no two of these equals 0 mod 31. Hence, a = 31a1 and b = 31b1 so that 311993 divides a21 + b21. Then, a1 = 31a2 and b1 = 31b2. Continuing in this fashion (with p = 31), we must have a = p998m and b = p998 n so that ab is divisible by p1996. More generally, if a prime p = 4k + 3 divides a2 + b2 , then both a and b must be divisible by p. This follows from the result that \a natural n is the sum of squares of two relatively prime natural numbers if and only if n is divisible neither by 4 nor by a natural number of the form 4k + 3" (see J.W. Sierpinski, Elementary Theory of Numbers, Hafner, NY, 1964, p. 170).

9. Determine the minimum value of p

p

S = p(a + 1)2 + 2(b , 2)2 + (c + 3)2 + p (b + 1)2 + 2(c , 2)2 + (d + 3)2 ) + (c + 1)2 + 2(d , 2)2 + (a + 3)2 + (d + 1)2 + 2(a , 2)2 + (b + 3)2

where a, b, c, d are any real numbers. Solution. Applying Minkowski's inequality,

p p S  (4 + s)2 + 2(s , 8)2 + (s + 12)2 = 4s2 + 288 p where s = a + b + c + d. Consequently, min S = 12 2 and is taken on for a = b = c = d = 0.

10. A set of 500 real numbers is such that any number in the set is greater than one- fth the sum of all the other numbers in the set. Determine the least number of negative numbers in the set. Solution. Letting a1 ; a2 ; a3 ; : : : denote the numbers of the set and S the sum of all the numbers in the set, we have a1 > S ,5 a1 ; a2 > S ,5 a2 ; : : : ; a6 > S ,5 a6 :

Adding, we get 0 > S , a1 , a2 ,    , a6 so that if there were six or less negative numbers in the set, the right hand side of the inequality could be positive. Hence, there must be at least seven negative numbers. Comment. This problem where the \5" is replaced by \1" is due to Mark Kantrowitz, Carnegie{Mellon University.

134 First a solution to one of the 36th IMO problems:

2. [1995: 269] 36th IMO

Let a, b, and c be positive real numbers such that abc = 1. Prove that

1

a3(b + c)

+ b3 (c1+ a) + c3(a1+ b)  23 :

Solution by Panos E. Tsaoussoglou, Athens, Greece. By the Cauchy{Schwartz inequality

 1 1 1 [a(b + c) + b(c + a) + c(a + b)] a3 (b + c) + b3 (c + a) + c3(a + b) 

or

 1 1 1 2  a+b+c ;   2(ab + ac + bc) a3(b1+ c) + b3 (c1+ a) + c3(a1+ b) ac + bc) ;  (ab +(abc )2 2

or

1

because abc = 1. Also Therefore holds.

1

1

a3(b + c) + b3(c + a) + c3(a + b)

+ bc ;  ab + ac 2

ab + ac + bc  p3 a2b2c2 = 1: 3

1

1

1

3

a3(b + c) + b3(c + a) + c3(a + b)  2

135 Now we turn to some of the readers' solutions to problems proposed to the jury but not used at the 35th IMO in Hong Kong [1995: 299{300].

PROBLEMS PROPOSED BUT NOT USED AT THE 35th IMO IN HONG KONG Selected Problems

3. A semicircle , is drawn on one side of a straight line `. C and D are points on ,. The tangents to , at C and D meet ` at B and A respectively, with the center of the semicircle between them. Let E be the point of intersection of AC and BD, and F be the point on ` such that EF is perpendicular to `. Prove that EF bisects \CFD. Solutions by Toshio Seimiya, Kawasaki, Japan; and by D.J. Smeenk, Zaltbommel, the Netherlands. We give Seimiya's write-up. P

D C B



E T FQ O

A Let P be the intersection of AD and BC . Then \PCO = \PDO = 90, \CPO = \DPO and PC = PD. Let Q be the intersection of PE with AB . Then by Ceva's Theorem, we get BQ  AD  PC = BQ  AD = 1: QA DP CB QA CB Thus we get Since \BPO = \APO we get

`



BQ = BC : QA AD

PB = BO : PA AO We put \PAB = , \PBA = .

(1) (2)

136 Let T be the foot of the perpendicular from P to AB . Then from (1) and (2) we have

BC = BO cos = PB cos = PT : AD AO cos PA cos TA

(3)

From (1) and (3) we have

BQ = PT : QA TA Hence Q coincides with T so that P , E , F are collinear. [See page 136.] Because \PCO = \PDO = \PFO = 90 , P , C , F , O, D are concyclic. Hence \CFE = \CFP = \CDP = \DCP = \DFP = \DFE . Thus EF bisects \CFD.

P

D C B

E F

O

`

A

4. A circle ! is tangent to two parallel lines `1 and `2. A second circle !1 is tangent to `1 at A and to ! externally at C . A third circle !2 is tangent to `2 at B , to ! externally at D and to !1 externally at E . AD intersects BC at Q. Prove that Q is the circumcentre of triangle CDE. Solutions by Toshio Seimiya, Kawasaki, Japan; and by D.J. Smeenk, Zaltbommel, the Netherlands. We give Smeenk's solution. We denote the three circles as ! (O;R), !1 (O1; R1 ), !2 (O2; R2 ). Now let ! touch `1 at F and `2 at F 0 . Let the line through O2 parallel to `1 intersect FF 0 at G and the production of AO1 at H .

137

`1

F

L

A R1

R

O

Q E D R 2 O 2 N

K G

`2

O1

C R1

R2

F0

M B

Let the line through D parallel to `1 intersect FF 0 at K . Let the line through D parallel to FF 0 intersect `1 at L, `2 at M and GO2 at N . Now AF is a common tangent of ! and !1, so and

p AF = 2 RR1

(1)

p BF 0 = 2 RR2 = GO2:

(2)

It follows that

p p HO2 = j2 RR2 , 2 RR1 j; HO1 = 2R , R1 , R2: In right triangle O1 HO2 , p p (2 RR2 , 2 RR1 )2 + (2R , R1 , R2 )2 = (R1 + R2)2 : p After some reduction R = 2 R1 R2 . Next consider triangle GOO2 . p GO = R , R2; GO2 = 2 RR2; DO2 = R2; DO = R; KDkGO2: pRR R 2 R 2. We nd that GN = FL =  GO 2= R + R2 R + R2 With (1) we have

pRR 2 R AL = 2 RR1 , R + R 2 : 2 p

(3)

138 Furthermore DN =

R2 R2(R , R2) R + R2  GO = R + R2 and

2 DL = 2R , R2 , R2R(R+,RR2) = R2+RR : 2

2

(4)

Now AD2 = AL2 + DL2 . With (3) and (4),

pRR !2  2R2 2 p 2 R 2 RR1 , R + R 2 + R + R = 4RR1 = AE 2 : 1

2

1

2

So AD = AF . That means that AD touches ! at D and AD is a common tangent and the radical axis of ! and !2. In the same way BC is the radical axis of ! and !1 and Q is the radical point of ! , !1 and !2. So QC = QD = QE , as required.

5. A line ` does not meet a circle ! with center O. E is the point on ` such that OE is perpendicular to `. M is any point on ` other than E. The tangents from M to ! touch it at A and B . C is the point on MA such that EC is perpendicular to MA. D is the point on MB such that ED is perpendicular to MB . The line CD cuts OE at F . Prove that the location of F is independent of that of M . Solution by Toshio Seimiya, Kawasaki, Japan. As MA, MB are tangent to ! at A, B respectively, we get \OAM = \OBM = 90 and OM ? AB . Let N , P be the intersections of AB with OM and OE respectively. Since M , E , P , N lie on the circle with diameter MP we get ON  OM = OB2 = r2 where r is the radius of !. Hence P is a xed point. (P is the pole of `.) Let G be the foot of the perpendicular from E to AB . As \OBM = \OAM = \OEM = 90 , O, B , M , E , A are concyclic, so that by Simson's Theorem C , D, G are collinear.

139

! B

r

O N

P A

D

M

C

F E

G

`

Since A, C , E , G lie on the circle with diameter AE we get \EGF = \EGC = \EAC = \EAM: (1) As O, M , E , A are concyclic and OM is parallel to EG we have \EAM = \EDM = \DEG = \FEG: (2) From (1) and (2) we get \EGF = \FEG: (3) Since \EGP = 90 we get \FGP = \FPG: (4) From (3) and (4) we have EF = FG = FP . Thus F is the midpoint of EP . Hence F is a xed point.

140 Next, we give a counterexample to the rst problem of the set of problems proposed to the jury, but not used at the 35th IMO in Hong Kong given in the December 1995 number of the corner. 1. [1995: 334] Problems proposed but not used at the 35th IMO in Hong Kong. ABCD is a quadrilateral with BC parallel to AD. M is the midpoint of CD, P that of MA and Q that of MB . The lines DP and CQ meet at N . Prove that N is not outside triangle ABM . Counterexamples by Joanna Jaszunska,  student, Warsaw, Poland; and by Toshio Seimiya, Kawasaki, Japan. We give Jaszunska's  example.

B

C Q X

M

P N

t

A

D We draw a triangle ADM and denote the midpoint of MA by P . Let C be a point on the half-line DM such that M is the midpoint of CD. Let N be any point of the segment PD, inside triangle ADM . We construct a parallelogram MCBX such that MX and BC are parallel to AD and X lies on the segment CN . Let us denote the point where the diagonal MB of this quadrilateral meets CN by Q. Q is then the midpoint of MB . Connect points A and B . We have thus constructed a quadrilateral ABCD with BC parallel to AD, M is the midpoint of CD, P that of MA and Q that of MB . Lines DP and CQ meet at N . N is inside triangle ADM ; hence it is outside triangle ABM .

141 Next we look back to some further solutions to problems of the Sixth Irish Mathematical Olympiad given in [1995: 151{152] and for which some solutions were given in [1997: 9{13]. An envelope from Michael Selby arrived which I mis led. It contains solutions to problems 1, 2, and 4 of Day 1, and problems 1, 2, 3 and 4 of Day 2.

1. [1995: 152] Second Paper, Sixth Irish Mathematical Olympiad.

Given ve points P1 , P2 , P3 , P4 , P5 in the plane having integer coordinates, prove that there is at least one pair (Pi; Pj ) with i 6= j such that the line Pi Pj contains a point Q having integer coordinates and lying strictly between Pi and Pj . Solution by Michael Selby, University of Windsor, Windsor, Ontario. The points can be characterized according to the parity of their x and y coordinates. There are only four such classes: (even, even), (even, odd), (odd, even), (odd, odd). Since we are given ve such points, at least two must have the same parity of coordinates by the Pigeonhole Principle. Suppose they are Pi and Pj , Pi = (xi; yi), Pj = (xj ; yj ). Then xi + xj is even and yi + yj is even, since the xi , xj have the same parity and yi, yj have the same parity. Hence the midpoint   has integral coordinates.

Q = xi +2 xj ; yi +2 yj

2. [1995: 152] Second Paper, Sixth Irish Mathematical Olympiad.

Let a1 ; a2 ; : : : an , b1; b2 ; : : : bn be 2n real numbers, where a1 ; a2; : : : ; an are distinct, and suppose that there exists a real number such that the product

(ai + b1 )(ai + b2 ) : : : (ai + bn) has the value for all i (i = 1; 2; : : : ; n). Prove that there exists a real number such that the product

(a1 + bj )(a2 + bj ) : : : (an + bj ) has the value for all j (j = 1; 2; : : : ; n).

Solution by Michael Selby, University of Windsor, Windsor, Ontario. De ne

Pn(x) = (x + b1)(x + b2)    (x + bn) , : (1) Then Pn (ai) = 0 for i = 1; 2; : : : ; n. Therefore Pn (x) = (x , a1 )(x , a2 )    (x , an ) by the Factor Theorem.

142 Now (,1)n Pn (,x) = (x + a1 )(x + a2 )    (x + an ). So

(,1)n Pn (,bi) = (bi + a1 )(bi + a2 )    (bi + an ) = (,1)n+1 by (1): Hence (bi + a1 )(bi + a2 )    (bi + an ) = (,1)n+1 for i = 1; 2; : : : ; n. Thus, the result is true with = (,1)n+1 . That completes the Corner for this number. We are in high Olympiad season. Send me your nice solutions and contests.

Do you believe what occurs in print? The last sentence of the quoted passage, taken from The Daughters of Cain by Colin Dexter (Macmillan, 1994), contains two factual erors. What are they? `Have you heard of \Pythagorean Triplets"?' `We did Pythagoras Theorem at school.' `Exactly. The most famous of all the triplets, that is | \3, 4, 5" 32 + 42 = 52. Agreed?' `Agreed.' `But there are more spectacular examples than that. The Egyptians, for example, knew all about \5961, 6480, 8161".' Contributed by J.A. McCallum, Medicine Hat, Alberta.

143

BOOK REVIEWS Edited by ANDY LIU The Lighter Side of Mathematics, edited by Richard K. Guy and Robert E. Woodrow, Mathematical Association of America, Washington DC, 1994, ISBN 0-88385-516-X, 376+ pages, softcover, US $38.50, reviewed by Murray S. Klamkin, University of Alberta. This book is the proceedings of the Eugene Strens Memorial Conference on Recreational Mathematics and its History held at the University of Calgary in August 1986 to celebrate the founding of the Strens Collection which is now the most complete library of recreational mathematics in the world. I had been invited to attend this conferences but unfortunately had a previous committment. To make up for missing this conference, the next best thing was reviewing this proceedings book which on doing so made me realize what I had missed, for example, some very interesting talks plus getting together with the leading practitioners of recreational mathematics, some of whom were long time colleagues. One does not normally include a list of contents in a book review, but by doing so, it will give the reader a good indication of the wealth of recreational material here . So if you have any interest in recreational mathematics, this is a book for you. And even if you do not have such an interest, reading this book may give you one. Contents

Preface The Strens Collection Eugene Louis Charles Marie Strens Part 1: Tiling & Coloring

Frieze Patterns, Triangulated Polygons and Dichromatic Symmetry, H. S. M. Coxeter & J. F Rigby Is Engel's Enigma a Cubelike Puzzle? J. A. Eidswick Metamorphoses of Polygons, Branko Grunbaum SquaRecurves, E-Tours, Eddies, and Frenzies: Basic Families of Peano Curves on the Square Grid, Douglas M. McKenna Fun with Tessellations, John F. Rigby Escher: A Mathematician in Spite of Himself, D. Schattschneider Escheresch, Athelstan Spilhaus The Road Coloring Problem, Daniel Ullman Fourteen Proofs of a Result About Tiling a Rectangle, Stan Wagon Tiling R3 with Circles and Disks, J. B. Wilker

1 5 15 28 35 49 74 91 101 105 113 129

144 Part 2: Games & Puzzles

Introduction to Blockbusting and Domineering, Elwyn Berlekamp 137 A Generating Function for the Distribution of the Scores of all Possible Bowling Games, Curtis N. Cooper & Robert E. Kennedy 149 Is the Mean Bowling Score Awful? Curtis N. Cooper & Robert E. Kennedy 155 Recreation and Depth in Combinatorial Games, Aviezri Fraenkel 159 Recreational Games Displays Combinatorial Games, Aviezri S. Fraenkel 176 Combinatorial Toys, Kathy Jones 195 Rubik's Cube | application or illumination of group theory? Mogens Esrom Larsen 202 Golomb's Twelve Pentomino Problems, Andy Liu 206 A New Take-Away Game, Jim Propp 212 Confessions of a Puzzlesmith, Michael Stueben 222 Puzzles Old & New: Some Historical Notes, Jerry Slocum 229 Part 3: People & Pursuits

The Marvelous Arbelos, Leon Banko Cluster Pairs of an n-Dimensional Cube of Edge Length Two, I. Z. Bouwer & W. W. Cherno The Ancient English Art of Change Ringing, Kenneth J. Falconer The Strong Law of Small Numbers, Richard K. Guy Match Sticks in the Plane, Heiko Harborth Misunderstanding My Mazy Mazes May Make Me Miserable, Mogens Esrom Larsen Henry Ernest Dudeney: Britain's Greatest Puzzlist, Angela Newing From Recreational to Foundational Mathematics, Victor Pambuccian Alphamagic Squares, Lee C. F. Sallows Alphamagic Squares: Part II, Lee C. F. Sallows The Utility of Recreational Mathematics, David Singmaster The Development of Recreational Mathematics in Bulgaria, Jordan Stoyanov V , E + F = 2, Herbert Taylor Tracking Titanics, Samuel Yates

247

List of Conference Participants

363

254 261 265 281 289 294 302 305 326 340 346 353 355

Postscript: There is a typographical error in Andy Liu's article on page 206. The number of tetrominoes is ve and not four. This is corrected in the reprint of this article as Appendix C in the new edition of Golomb's \Polyominoes".

145

In Memoriam | Leon Banko We were saddened to learn of the recent death of Dr. Leon Banko , who has been a contributor to CRUX over many years. Sadly, he was not able to contribute recently, and some of our more recent readers may not know so much about him. We refer you to an excellent article in the March 1992 issue of the College Mathematics Journal, entitled A Conversation with Leon Banko , written by G.L. Alexanderson. One of Leon's long time friends, Dr. Clayton Dodge, has written the appreciation printed below. Leon Banko practiced dentistry for sixty years in Beverly Hills, California, until his retirement just a few years ago. His patients included many Hollywood personalities whose names are household words. Among his several other interests, such as piano, guitar, calculators, and computers, he lectured and wrote papers both on dentistry and mathematics. His specialty was geometry, and the gure he loved best was the arbelos, or shoemaker's knife, which consists of three semicircles having a common diameter line. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. It is said that the test of a mathematician is not what he himself has discovered, but what he inspired others to do. Leon discovered a third circle congruent to the twin circles of Archimedes and published that result in the September 1974 issue of Mathematics Magazine (\Are the Twin Circles of Archimedes Really Twins?", pp. 214-218.) This revelation motivated the discovery by Leon and by others of several other members of that family of circles. An article on those circles is in progress. Dr. Banko edited the Problem Department of the Pi Mu Epsilon Journal from 1968 to 1981, setting and maintaining a high standard of excellence in the more than 300 problems he included in its pages. Although the Journal has a relatively small circulation, its Problem Department grew to have a large number of regular contributors. He became acquainted with Crux Mathematicorum early in its history, when it was called Eureka, and made many contributions to its pages over the years, maintaining a close friendship with its founder and rst editor Leo Sauve. Like Leo, who started Crux to add some spice to his mathematical life of teaching basic post high school courses, Leon worked in mathematics for mental exercise and recreation, making friends with and earning the respect of many well known mathematicians. Leon and I became good friends, rst through correspondence regarding the Pi Mu Epsilon Journal Problem Department, and later through many personal meetings, including the August 1979 meeting of problemists in Ottawa, sponsored by Leo Sauve and Fred Maskell of Crux. Following the

146 formal sessions in Ottawa, seven of us drove to Quebec City for an enjoyable weekend of sightseeing and fellowship: Leo and Carmen Sauve, Leon and Francine Banko , Charles and Avetta Trigg, and I. Since his retirement from his dentistry, Leon has worked on the manuscript for a proposed book on the properties of the arbelos, carrying on a monumental task started by him and the late Victor Thebault. Much material has been collected for this project and much remains to be done on it. Indeed, he asked me to nish the job. At one time some years ago a schoolgirl wrote to Albert Einstein about a mathematical question she had. Apparently Einstein misinterpreted her question and gave an incorrect answer. Banko pointed out this error and in his mathematical museum he now has a copy of the Los Angeles Times with the front page headline \Local Dentist Proves Einstein Wrong." Leon developed many physical problems in his later years. He was a ghter and he won several physical battles. When I last visited him at his home in Los Angeles in October 1996, he was ghting liver cancer, but still working on the Thebault material, in spite of failing eyesight. On Sunday afternoon, February 16, 1997, the cancer overtook him and he died at his home at the age of 88. He was a gentleman, a scholar, and a true friend. Clayton W. Dodge University of Maine Orono, ME 04469-5752

Heronian Triangles with Associated Inradii in Arithmetic Progression Paul Yiu

Department of Mathematics, Florida Atlantic University

In memory of Dr. Leon Banko 1. The area of a triangle is given in terms of its sides a, b, c by the Heron formula p 4 = s(s , a)(s , b)(s , c); where s := 12 (a + b + c) is the semiperimeter. A triangle (a; b;c; 4) is called Heronian if its sides and area are all integers. L. Banko [1] has made an interesting observation about the Heronian triangle (13; 14; 15; 84). The

147 height on the side 14 being 12, this triangle can be decomposed into two Pythagorean components, namely, (5; 12; 13) and (9; 12; 15). The inradii of these Pythagorean triangles, and that of the Heronian triangle, are respectively 2, 3, 4, three consecutive integers! (See Figure 1). Noting that the sides of the Heronian triangle are themselves three consecutive integers, Banko remarked that \no other Heronian triangle can claim that distinction". Actually, apart from this, there are exactly two other Heronian triangles with three consecutive integers for the associated inradii. Each of these two Heronian triangles is decomposable into two Pythagorean components, namely,

(15; 20; 25; 150) = (9; 12; 15) [ (16; 12; 20); (2) (25; 39; 56; 420) = (20; 15; 25) [ (36; 15; 39): (3) The three inradii in these two cases are 3, 4, 5, and 5, 6, 7 respectively. The Heronian triangle (15; 20; 25; 150) in (2) has an interesting property that no other Heronian triangle shares. Here, if the smaller Pythagorean component (9; 12; 15) is excised from the larger one (16; 12; 20), another Heronian triangle results, namely,

(15; 20; 7; 42) = (16; 12; 20) n (9; 12; 15): (4) This has inradius 2. (See Figure 2). Note that the four inradii are consecutive integers! The same construction applied to Banko 's example (13; 14; 15; 84) gives (13; 15; 4; 24), with inradius 32 , not an integer. For the Heronian triangle (25; 39; 56; 420) in (3), this fourth inradius is 3, albeit not consecutive with the other three inradii 5, 6, and 7. In x 4 below, we shall show that, up to similarity, the con guration in Figure 2 is the only one with the four associated inradii in arithmetic progression. 13

15

12

12

15

20

q q q

4 3

q

q

q

2 5

q

q

9

Figure 1.

9

3

q

5 q

q

4 9

q

Figure 2.

q

2

q

7

2. Consider two right triangles with a common side y and opposite acute angles  and  juxtaposed to form a triangle ,(+). We shall assume  > , so that ,1 can be excised from ,2 to form another triangle ,(,). (See Figures 3 and 4). The inradius of a triangle with area 4 and semiperimeter s is given by r = 4s . (See, for example, Coxeter [2, p. 12]). For a right triangle

with legs a, b, and hypotenuse c, this is also given by the simpler formula

148

r = s,c = 12 (a+b,c). We shall determine the similarity classes of triangles

for which the inradii of the triangle and its two Pythagorean components are in arithmetic progression. The calculations can be made simple by making use of the fact that in a right triangle with an acute angle , the sides are in the ratio 2t : 1 , t2 : 1 + t2 , where t = tan 2 . Let t1 := tan 2 and t2 := tan 2 .

a1

x

,1 1

y

,2

a2

 ,(+) = ,1 [ ,2 x2

Figure 3.

y

a1

,1

a2

,(,)

  a03 ,(,) = ,2n,1

x1

Figure 4.

By choosing y = 2t1 t2 , we have, in Figures 3 and 4,

a1 = t2 (1 + t212); x1 = t2 (1 , t1 ); a3 = x1 + x2 = (t1 + t2 )(1 , t1 t2 );

a2 = t1 (1 + t222 ); x02 = t1 (1 , t2 ); a3 = x2 , x1 = (t1 , t2 )(1 + t1 t2 ): From these, we determine the inradii of the four triangles ,1 , ,2, ,(+) and ,(,) :

r1 = t1t2(1 , t1); r2 = t1t2(1 , tt2); (5) r+ = t1t2(1 , t1t2); r, = t1t2(1 , t21 ): Since r1 < r2 , r, < r2 , and r, < r+ , there are only three cases in

which three of these inradii can be in arithmetic progression: (i) r1, r2 , r+ are in A.P. if and only if t1 , t2 , t1 t2 are in A.P., that is, t1 + t1t2 = 2t2. From this, t2 = 2,t1t1 , and r1 : r2 : r+ = 2 , t1 : 2 : 2 + t1: (6) (ii) r, , r1 , r2 are in A.P. if and only if tt12 , t1 , t2 are in A.P., that is, t2 + t2 = 2t1 . From this, t2 = 2t21 , and t1 1+t1 r, : r1 : r2 = 1 : 1 + t1 : 1 + 2t1: (7) (iii) r1 , r, , r2 are in A.P. if and only if t1 , tt21 , t2 are in A.P., that is, t1 + t2 = 2tt12 . From this, t2 = 2,t21t1 , and r1 : r, : r2 = 2 , t1 : 2 : 2 + t1: (8)

149 In each of these cases, with t := t1 , the proportions of the sides of ,() are as follows. These triangles are all genuine for 0 < t < 1.

a1 : a2 : a3 (or a03 ) (i) r1 ; r2 ; r+ (2 , t)(1 + t2 ) : 2(2 , 2t + t2 ) : (3 , t)(1 , t)(2 + t) (ii) r, ; r1 ; r2 2t(1 + t)(1 + t2 ) : 1 + 2t + t2 + 4t4 : (1 , t)(1 + t + 2t3 ) (iii) r1 ; r, ; r2 t(2 , t)(1 + t2 ) : 4 , 4t + t2 + t4 : 2(1 , t)(2 , t + t3 ) A.P.

3.

Among the triangles constructed above with three associated inradii in A.P., the only cases in which the three sides also are in A.P. are tabulated below. In each case, we give the smallest Heronian triangle with three associated inradii in integers. (t1 ; t2 )

Triangle with decomposition

inradii

; 112 ) (13; 15; 14;84) = (5; 12; 13) [ (9; 12; 15) (r1; r2 ; r+ ) = (2; 3; 4) ; 3 ) (15; 20; 25;150) = (9; 12; 15) [ (16; 12; 20) (r1; r2 ; r+ ) = (3; 4; 5) ; 161) (15; 37; 26;156) = (35; 12; 37) n (9; 12; 15) (r1; r,; r2 ) = (3; 4; 5) ; 15 ) (39; 113; 76;570) = (112; 15; 113) n (36; 15; 39) (r,; r1; r2 ) = (5; 6; 7) Banko 's observation [1] on the Heronian triangle (13; 14; 15; 84) is case (1) in this table. 4. Finally, we consider the possibility for the four inradii r, , r1, r2, r+, to be in A.P. First, assume2 r1 ; r, ; r2 in A.P. By (8), r1 : r, : r2 = 2 , t1 : 2 : 2 + t1; indeed, t2 = 2,t1t1 . From (5), we have, after simpli cation, (1) (2) (3) (4)

(2 3 (1 2 (1 2 (1 5

r1 : r, : r2 : r+ = 2 , t1 : 2 : 2 + t1 : 2 + t1 + t21: Now, these four inradii are in A.P. if and only if 2 + t1 + t21 = 2 + 2t1 . This is clearly impossible for 0 < t1 < 1. It remains, therefore, to consider the possibility that r, , r1, r2 , r+ be in A.P. This requires, by (7), r1 : r2 = 1 + t1 : 1 + 2t1 , and also by (6), r1 : r2 = 21, t1 : 2. It follows that 1 + t1 : 1 + 2t1 = 2 , t1 : 2, from which t1 = 2 . Consequently, r1 : r2 = 3 : 4. Also, r1 : r+ = 3 : 5 from (6), and r, : r1 = 2 : 3 from (7). Thus, we have the con guration in Figure 2, in which the four inradii are in the ratio

r, : r1 : r2 : r+ = 2 : 3 : 4 : 5:

The author thanks the referee for valuable comments and suggestions.

References 1. L. Banko , An Heronian oddity, Crux Math. 8 (1982) p.206. 2. H.S.M. Coxeter, Introduction to Geometry, Wiley, New York, 1961.

150

THE SKOLIAD CORNER No. 21

R.E. Woodrow This number we give the problems of the Manitoba Mathematical Contest for 1995. This is a two hour contest aimed primarily at grade 12 students, and sponsored by the Actuaries Club of Winnipeg, The Manitoba Association of Mathematics Teachers, The Canadian Mathematical Society and The University of Manitoba. My thanks go to Diane and Roy Dowling, organizers of the contest for supplying us with it.

THE MANITOBA MATHEMATICAL CONTEST 1995 For Students in Grade 12

Wednesday, February 22, 1995 | Time: 2 hours

1. (a) If a and b are real numbers such that a + b = 3 and a2 + ab = 7

nd the value of a. (b) Noriko's average score on three tests was 84. Her score on the rst test was 90. Her score on the third test was 4 marks higher than her score on the second test. What was her score on the second test? 2. (a) Find two numbers which di er by 3 and whose squares di er by 63. (b) Find the real number which is a root of the equation

27(x , 1)3 + 8 = 0:

3. (a) Two circles lying in the same plane have the same centre. The radius of the larger circle is twice the radius of the smaller circle. The area of the region between the two circles is 7. What is the area of the smaller circle? (b) The area of a right triangle is 5. Also, the length of the hypotenuse of this triangle is 5. What are the lengths of the other two sides? 4. (a) The parabola2 whose equation is 8y = x2 meets the parabola whose equation is x = y at two points. What is the distance between these two points? (b) Solve the equation 3x3 + x2 , 12x , 4 = 0. 5.2 (a) Find the real number a such that a4 , 15a2 , 16 = 0 and 3 a + 4a , 25a , 100 = 0. p (b) Find all positive numbers x such that xx x = (xpx)x .

151

6. If x, y and z are real numbers prove that

,xjyj , yjxj ,yjzj , zjyj,xjzj , zjxj = 0:

7. x and y are integers between 10 and 100. y is the number obtained by reversing the digits of x. If x2 , y 2 = 495 nd x and y . 8. Three points P , Q and R lie on a circle. If PQ = 4 and \PRQ = 60 what is the radius of the circle? 9. Three points are located in the nite region between the x-axis and the graph of the equation 2x2 + 5y = 10. Prove that at least two of these points are within a distance 3 of each other. 10. Three circles pass through the origin. The centre of the rst circle lies in the rst quadrant, the centre of the second circle lies in the second quadrant, and the centre of the third circle lies in the third quadrant. If P is any point that is inside all three circles, show that P lies in the second quadrant. Last number we gave the problems of the Mathematical Association National Mathematics Contest 1994 from the United Kingdom. Here are the answers. 1. C 2. E 3. C 4. D 5. E 6. B 7. A 8. D 9. B 10. B 11. B 12. E 13. E 14. A 15. A 16. C 17. D 18. C 19. B 20. B 21. C 22. B 23. E 24. B 25. E That completes the Skoliad Corner for this issue. I need suitable contest materials and welcome your suggestions for the evolution of this feature.

152

MATHEMATICAL MAYHEM Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. All material intended for inclusion in this section should be sent to the Mayhem Editor, Naoki Sato, Department of Mathematics, University of Toronto, Toronto, ON Canada M5S 1A1. The electronic address is [email protected] The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto). The rest of the sta consists of Richard Hoshino (University of Waterloo), Wai Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

A Journey to the Pole | Part II  Miguel Carrion  Alvarez

student, Universidad Complutense de Madrid Madrid, Spain In this second (and last, for your relief) article, we look at some advanced topics like inversion or the applications of calculus to the theory of curves. Inversion Inversion is a transformation determined by a point called the centre of inversion O and an inversion ratio k2 . The image of a point P is a point P 0 such that P 0 is on line OP and jOP 0 j = k2 jOP j. It is evident k2 . Inversion is that a curve r = f () can be inverted by letting r =  f () a conformal transformation, meaning that the angles between intersecting curves are preserved. We will prove this in a later section. Exercise 1. Prove that the inverse curve of a straight line is itself if it passes through the origin or a circle through the origin if it does not.

de

Example 1. By inspection of the equations of the conic r = 1 , e cos( , ) and the limacon r = b + a cos , it is evident that the inverse of a conic about its focus is a limacon. I imagine that trying to prove this theory with synthetic geometry would result in a severe headache. This last result provides a di erent de nition of conics as loci if we invert the de nition of the limacon given above. Consider a circle or straight

153 line and a point O not on it. Draw a circle through O tangent to the circle or line. The diameter through O intersects the circle at P . The locus of all P 's is a conic with O at one focus. Tangent Lines We leave the realm of elementary geometry to enter calculus, where we will teach the same old dog new tricks. The rst is how to nd the tangent line to a polar curve. Our starting point will be the equation of the straight line d = r sin( , ). The tangent line at 0 is a rst-order approximation to the curve involvdr ing r(0 ) and . Di erentiating the equation of the straight line with

d 0 dr respect to  at 0 , we get 0 = sin( , 0 ) , r(0 ) cos( , 0 ), which d 0 r . This quantity can sometimes be useimplies that tan( , 0 ) = (dr=d ) 0 ful in itself, as  , 0 represents the angle between the radius vector and the tangent line. We will make use of it in the next example. The orientation  is determined from its tangent, and implies that

 , 0) + tan 0 tan( , 0 + 0) = 1tan( , tan( ,  ) tan  0

0

dr 0 : tan  = rdr(j 0) ,+ rd(j0) tan tan  d 0

0

0

The parameter d in the equation of the tangent line is given (after some trigonometric manipulations) by which gives

d = pr(0) tan(2  , 0) ; 1 + tan ( , 0 )

2 0) d = pr2( )r+((dr=d : )2 j 0

0

Example 2. Proof that inversion preserves angles. Let r = f () and r = g () be two curves that intersect at 0 . Their directions at 0 are 1 and 2, and the angle between them satis es

2 , 0) , tan(1 , 0) : tan(2 , 1 ) = tan(2 , 0 + 0 , 1) = 1tan( + tan( ,  ) tan( ,  ) 2

0

1

0

Now, the inverted curves, r0 = 1=f () and r0 = 1=g (), also intersect at 0 and their directions 01 and 02 satisfy

,f = tan(0 , 1) tan(01 , 0) = ,11=fdf  = (df=d ) 2 f

d

154 (and similarly for 02). Hence, tan(2 , 1) = tan(01 , 02 ) and we are done. Tangent lines through the origin An interesting special case is when r(0) = 0. In that case, tan  = tan ,0 or  = 0 . In words, if the curve crosses the origin for a given 0, the equation of the tangent line at the origin is  = 0 .

6

6

-r

-r

When sketching curves, a useful result is that if r() has an odd-order root at 0 , then the curve is smooth at the origin, but if the root is even-order, then there is a cusp (see the gure). Asymptotes In certain cases r tends to in nity for some nite value of , signalling the possibility of an asymptote. In handling asymptotes, it is convenient to consider s() = 1=r(), in which ,s j case tan( , 0 ) = 0

(ds=d)

d

O

0

ds j0 tan 0 , s(0 ) = d . ds d j0 + s(0 ) tan 0

and tan  There is a possible asymptote if s(0) = 0 and its equation is: s() = sin(d, 0) , or  = 0 if d = 0.

This is because, as in the case of tangent lines through the origin, the slope of the tangent line is tan  = tan 0 . The parameter d is given by

1 p d = lim !0 s2 () + (ds=d)2 :

If this does not diverge, there is an asymptote. When sketching curves, it is useful to know from which side of the asymptote the curve approaches in nity. This is achieved by studying the sign of s() , d1 sin( , 0 ), which tends to 0 at  = 0 . If it tends to 0+, the curve is closer to the origin than the asymptote (see the gure on the last page), and if it tends to 0,, the curve is farther from the origin than the asymptote. Exercise 2. Sketch the curve r = ln , its asymptote and the tangent line at the origin.

155

1 satis es lim r() = 1, but it has !0 1 , cos  1 no asymptotes since p =p 1 , which 2 2 (1 , cos 0) + (sin 0 ) 2(1 , cos 0 )

Example 3. The parabola r =

diverges at 0 . Arc Length Another application of calculus is the computation of curve lengths. Usually one would take the expression for the line element in cartesian coordinates, dl2 = dx2 + dy 2 and transform it to polar coordinates. To use only polar coordinates, one could apply Pythagoras' Theorem to (dr) and (rd). Although this gives the right answer, it is not rigorous. A rigorous argument that does not rely on rectangular coordinates follows. Applying the cosine rule to side PP 0 of POP 0 (see gure) we have

dl2 = r2( + d) + r2() , 2r()r( + d) cos(d):

Expanding each term in a Taylor series up to the second order in d, we get

dl2

=

" 2 2r # dr d 2 d + d + r d2 d  dr 1 d2r  1  2 +r , 2r r + d + 2 d2 1 , 2 d2 :

r2 + 2r dr

Keeping terms up to second order in d we have

 dr 2# dl2 = r2 + d d2 : "

We can thus write the expression for the length of a curve inspolar coordinates

 dr 2 2 as follows: l = r + d d = 0 s  ds 2 Z 1 1 2 s2 s + d d, where s = r . Z

0

dr P 0 r d dl

r , rd

O

d 

r

P

Exercise 3. Derive the polar expression for arc length from the cartesian expression dl2 = dx2 + dy 2. Curvature It seems tautological to say that curvature is an important feature of curves, but the fact is that a planar curve is uniquely determined (up to translations and rotations) if its curvature is known as a function of arc length. This is generally of little practical importance, since the resulting di erential equations can only be solved if you know the solution! We will give the formula for curvature in terms of s() = l=r() and some applications.

156 Curvature, , can be de ned as the rate of change of the direction of the tangent line per unit arc length. We have

 d( , )  d  d = q s2  1 + d :  = d = ds ds d ds 2 s2 + , d ,s , so that Now, tan( , ) = (ds=d)  ,s  d arctan d( , ) = (ds=d) d d ds 2 + s  d2 s2  , , d 1 d = 2  s 2 (ds=d) 1+ giving

(ds=d)

2 , ds 2 + s  d2s  3 2 , 2 s  = q , ds 2  41 + d2 , ds 2d 5 s + s2 + d 2  d2s 2 3 d 2 s+ 2 = q s , 2  64 2 ,dds 2 75 ds s + d s2 + d  d2s  s+ 2 = h , d2 i3=2 : 1 ds 1 + 2 d

In terms or r, we have

2 )2 , r(d2 r=d2 ) :  = r + 2([rdr=d 2 + (dr=d)2 ]3=2

Example 4. If curvature is zero we obtain the di erential equation s + s00 = 0, with general solution s = d cos( , 0 ), that is, the equation of a straight line. Exercise 4. Check that the curvature at each point of a lemniscate is proportional to the distance to the origin. (Hint: to simplify the algebra, divide numerator and denominator by r3 in the curvature formula above.)

157 Area Enclosed by Polar Curves

P0 P

Q0

d

max O

r + dr r Q

The last application of calculus is the calculation of areas. The natural surface element is the `triangle' de ned by a segment of curve and the radius vectors at the endpoints (OPP 0 in the gure). The area of this triangle is, to rst order in d, dA = 1 r2d.

2



If the origin does not lie inside the curve the equation r = r() will have more than one branch, as shown, and the sign of the `enclosed area' dA depends on the orientation given to the curve. In the gure the curve is traversed counterclockwise, and so the outer branch (PP 0) has positive sign (the positive sense of d coincides with the direction of the curve) and the inner branch (QQ0 ) has negative sign (the positive sense of d as opposed to the direction of the curve). The same applies when calculating the area enclosed by two intersecting curves. Example 5. As our last example, we will evaluate the area enclosed by the circle (r , R cos )2 = 2 , R2 sin2 . The area is given by

Z sin ==R 1 (r2 , r,2 )d; A= sin =,=R 2 + p where r = R cos   2 , R2 sin2 . We have Z Z q A = 12 (r+ + r, )(r+ , r, )d = 2(R cos ) 2 , R2 sin2 d:

Letting R sin  =  sin  and R cos d =  cos d, we have

A= as expected.

Z =2

=,=2

22 cos2 d = 2

158

A Pattern in Permutations John Linnell

student, University of Massachusetts Boston, Massachusetts, USA Let pn (k) be the number of permutations on n elements (say, n large anteaters) with exactly k xed points (i.e. a permutation which takes k elements to themselves), so for example, p3 (0) = 2, p3 (1) = 3, p3(2) = 0, and n X p3(3) = 1. It should be clear that pn(k) = n!, but may be not so k=0

n X

kpn(k) = n! (this was problem 1 on the 1987 IMO). It may n X be even more surprising to learn that for n  2, k2 pn(k) = 2n!. What obvious that

k=0

k=0

kind of pattern ensues? As my old analysis prof would no doubt say, \this is good exercise," and it is kind of fun to follow a trail like this and see where it leads. Based on the above results, for n  1 and t  0, let

Q(n; t) = n1!

n X

k=0

ktpn(k):

Then Q(n; 0) = Q(n; 1) = 1 for all n  1 and Q(n; 2) = 2 for all n  2 (not n  1, and we will see why soon). Our rst conjecture would probably be then that indeed each Q(n; t) is an integer. But we will have to see a little more before we can prove anything. Take n = 5. Then we can make the following table:

k p5 (k) kp5 (k) k2 p5 (k) k3 p5 (k) k4 p5 (k) k5 p5 (k) k6 p5 (k)

0 1 2 3 4 5 P

P =5!

44 45 20 10 0 1 120 1

0 45 40 30 0 5 120 1

0 45 80 90 0 25 240 2

0 45 160 270 0 125 600 5

Table 1.

0 45 320 810 0 625 1800 15

0 45 640 2430 0 3125 6240 52

0 45 1280 7290 0 15625 24240 202

159 Thus, Q(5; 0) = 1, Q(5; 1) = 1, Q(5; 2) = 2, Q(5; 3) = 5, and so on. So far so good. We can make a second table with the actual Q(n;t) values:

nnt 0 1 2 3 4 5 1 2 3 4 5 6

1 1 1 1 1 1

1 1 1 1 1 1

1 2 2 2 2 2

1 4 5 5 5 5

1 8 14 15 15 15

1 16 41 51 52 52

6 1 32 122 187 202 203

Table 2. Now we are getting somewhere. Notice how the rows seem to converge to a single sequence of integers, with one new term kicking in with each row. This sequence begins 1, 1, 2, 5, 15, 52, 203, : : : . I could not nd this sequence in any of my references, so I did what any enterprising student would do. I sent an e-mail to [email protected], with \lookup 1 1 2 5 15 52 203" in the body. For those not familiar, it is an on-line sequence server that tries to solve or match any sequence you might send it; I should also add that they ask that you send at most one request per hour. Soon enough, I had a response, which indicated that this was a sequence known as the Bell numbers, which satisfy n n X B(0) = 1; B(n + 1) = B(k): k=0 k

This recursion is striking, because if one looks at the second row of n n X table 2, it may remind one of the identity 2n = . Could it be? Yes, k=0 k in fact the same recursion that generates the Bell numbers is what generates successive rows of table 2. Now we really have something. Claim. For all n  1 and t  0, Q(n + 1; t + 1) =

t t X Q(n; i). i=0 i

n Proof. First, note that pn (k) = k pn,k (0) [why?]. Then Q(n + 1; t + 1) = (n +1 1)!

nX +1

kt+1pn+1(k) k=0 n + 1 n +1 X 1 t +1 = (n + 1)! k k pn+1,k(0) k=0

160 +1 1 nX n! kt+1 p = (nn ++ 1)! n+1,k (0) k=1 (k , 1)!(n + 1 , k)! k nX +1  n  kt k , 1 pn+1,k(0) = n1! k=1 n +1 X = n1! ktpn(k , 1) k=1 n X 1 = n! (k + 1)t pn (k) k=0 nX +1 X t t kipn(k) = n1! i k=1 i=0 !  t n X t 1 X i = k pn(k) i=0 i n! k=0

t t X = Q(n; i): i=0 i

So we have proven quite a bit actually, including: 1. Each Q(n; t) is an integer (i.e.,

n X

k=0

ktpn(k) is divisible by n!), and

2. For xed t, Q(n; t) eventually becomes B (t) for suciently high n. The claim looks complicated, but we know what we want to prove, and it turns out to be just a little algebraic manipulation. So in the end, we have a nice result from a simple observation.

161

IMO CORRESPONDENCE PROGRAM Canadian students wishing to participate in this program should rst contact Professor Edward J. Barbeau, Department of Mathematics, University of Toronto, Toronto, Ontario. Please note that there is a fee for participation in the program: $12. Please make the cheque payable to Edward J. Barbeau.

PROBLEM SET 1 Algebra 1. Solve the system of equations

x2 + 2yz = x; y2 + 2xz = z; z2 + 2xy = y: 2. Let m be a real number. Solve, for x, the equation jx2 , 1j + jx2 , 4j = mx: 3. Let fx1 ; x2 ; : : : ; xn ; : : : g be a sequence of nonzero real numbers. Show that the sequence is an arithmetic progression if and only if, for each integer n  2,

1 + 1 +    + 1 = n , 1: x1x2 x2x3 xn,1 xn x1xn

4. Suppose that x and y are two unequal positive real numbers. Let

 x2 + y2 1=2

g = (xy)1=2 h = x2+xyy : Which of the numbers r , a, a , g , g , h is largest and which is smallest? 5. Simplify x3 , 3x + (x2 , 1)p x2 , 4 , 2 p x3 , 3x + (x2 , 1) x2 , 4 + 2 to a fraction whose numerator and denominator are of the form upv with u and v each linear polynomials. For which values of x is the r=

2 a = x +2 y

equation valid?

162 6. Prove or disprove: if x and y are real numbers with y  0 and y(y + 1)  (x + 1)2, then y(y , 1)  x2. 7. X is a collection of objects upon which the operation of addition, subtraction and multiplication are de ned so as to satisfy the following axioms: (1) if x; y belong to X , then x + y and xy both belong to X ; (2) for all x; y in X , x + y = y + x; (3) for all x; y;z in X , x + (y + z ) = (x + y ) + z and x(yz ) = (xy )z ; (4) for all x; y; z in x, x(y + z ) = xy + xz ; (5) there is an element 0 such that 0+x = x+0 = x and for each x in X , there exists a unique element denoted by ,x for which x + (,x) = 0; (6) x , y = x + (,y ) for each pair x; y of elements of X ; (7) x3 , x = x + x + x = 0 for x in X . Note that these axioms do not rule out the possibility that the product of two non-zero elements of X may be zero, and so it may not be valid to cancel terms. On X , we de ne a relation  by the following condition: x  y if and only if x2y , xy2 , xy + x2 = 0. Prove that the following properties obtain: (i) x  x for each element x of X ; (ii) if x  y and y  x, then x = y ; (iii) if x  y and y  z , then x  z . 8. Let n be a positive integer and suppose that u and v are positive real numbers. Determine necessary and sucient conditions on u and v such that there exist real numbers a1 ; a2 ; : : : ; an satisfying

a1  a2      an  0 u = a1 + a2 +    + an v = a21 + a22 +    + a2n:

When such a representation is possible, determine the maximum and minimum values of a1 . 9. Suppose that x + y1 = y + 1z = z + x1 = t, where x; y;z are not all equal. Determine xyz . 10. Let a  0. The polynomial x3 , ax + 1 has three distinct real roots. For which values of a does the root u of least absolute value satisfy 1 2 a < u < a?

163 11. Determine the range of values of cd subject to the constraints ab = 1, ac + bd = 2, where a, b, c, d are real. 12. Find polynomials p(x) and q (x) with integer coecients such that

p3 + p5) p p p(p 2 + q(p2 + p3 + p5) = 2 + 3:

Mayhem Problems The Mayhem Problems editors are: Cyrus Hsia Mayhem Advanced Problems Editor, Richard Hoshino Mayhem High School Problems Editor, Ravi Vakil Mayhem Challenge Board Problems Editor. Note that all correspondence should be sent to the appropriate editor | see the relevant section. In this issue, you will nd only problems | the next issue will feature only solutions. We warmly welcome proposals for problems and solutions. With the new schedule of eight issues per year, we request that solutions to the new problems in this issue be submitted by 1 August 1997, for publication in the issue 5 months ahead; that is, issue 8. We also request that only students submit solutions (see editorial [1997: 30]), but we will consider particularly elegant or insightful solutions from others. Since this rule is only being implemented now, you will see solutions from many people in the next few months, as we clear out the old problems from Mayhem.

164

High School Problems Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario, Canada. L3S 3E8 There is a correction for H220; the expression 2n  TS should be 2n  TS . H221. Let P = 195 + 6605 + 13165. It is known that 25 is one of the forty-eight positive divisors of P . Determine the largest divisor of P that is less than 10; 000. H222. McGregor becomes very bored one day and decides to write down a three digit number ABC , and the six permutations of its digits. To his surprise, he nds that ABC is divisible by 2, ACB is divisible by 3, BAC is divisible by 4, BCA is divisible by 5, CAB is divisible by 6, and CBA is a divisor of 1995. Determine ABC . H223. There are n black marbles and two red marbles in a jar. One by one, marbles are drawn at random out of the jar. Jeanette wins as soon as two black marbles are drawn, and Fraserette wins as soon as two red marbles are drawn. The game continues until one of the two wins. Let J (n) and F (n) be the two probabilities that Jeanette and Fraserette win, respectively. 1. Determine the value of F (1) + F (2) +    + F (3992). 2. As n approaches in nity, what does J (2)  J (3)  J (4)      J (n) approach?

H224. Consider square ABCD with side length 1. Select a point M exterior to the square so that \AMB is 90 . Let a = AM and b = BM . Now, determine the point N exterior to the square so that CN = a and DN = b. Find, as a function of a and b, the length of line segment MN .

D

N

A a b

C

B

M

165

Advanced Problems Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada. M1G 1C3 A197. Calculate

Z

2 , 2

sin(2N + 1) d; sin 

where N is a non-negative integer. A198. Given positive real numbers a, b, and c such that a + b + c = 1, show that aa bb cc + ab bcca + ac bacb  1. A199. Let P be a point inside triangle ABC . Let A0 , B0, and C 0 be the re ections of P through the sides BC , AC , and AB respectively. For what points P are the six points A, B , C , A0 , B 0 , and C 0 concyclic? A200. Given positive integers n and k, for 0  i  k , 1, let

n Sn;k;i = : j i (mod k) j Do there exist positive integers n, k > 2, such that Sn;k;0, Sn;k;1, : : : , Sn;k;k,1 are all equal? X

Challenge Board Problems Editor: Ravi Vakil, Department of Mathematics, One Oxford Street, Cambridge, MA, USA. 02138-2901 There are no new Challenge Board Problems this month | we reprint those from issue 1 this year [1997: 44]' C70. Prove that the group of automorphisms of the dodecahedron is S5, the symmetric group on ve letters, and that the rotation group of the dodecahedron (the subgroup of automorphisms preserving orientation) is A5 . C71. Let L1, L2, L3, L4 be four general lines in the plane. Let pij be the intersection of lines Li and Lj . Prove that the circumcircles of the four triangles p12 p23p31 , p23 p34p42 , p34 p41p13 , p41p12 p24 are concurrent. C72. A nite group G acts on a nite set X transitively. (In other words, for any x; y 2 X , there is a g 2 G with g  x = y .) Prove that there is an element of G whose action on X has no xed points.

166

PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later than 1 November 1997. They may also be sent by email to [email protected]. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication.

2226.

Proposed by K.R.S. Sastry, Dodballapur, India. An old man willed that, upon his death, his three sons would receive the u'th, v 'th, w'th parts of his herd of camels respectively. He had uvw , 1 camels in the herd when he died. Obviously, their sophisticated calculator could not divide uvw , 1 exactly into u, v or w parts. They approached a distinguished CRUX problem solver for help, who rode over on his camel, which he added to the herd and then ful lled the old man's wishes, and took the one camel that remained, which was, of course, his own. Dear CRUX reader, how many camels were there in the herd? 2227. Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Evaluate " 1 ,2k #

Y X

k 2k p k=0 (2p)

:

where the product is extended over all prime numbers.

167

2228. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Let A be the set of all real numbers from the interval (0; 1) whose decimal representation consists only of 1's and 7's; that is, let A=

(X 1

k=1

) ak : a 2 f1; 7g : 10k k

Let B be the set of all reals that cannot be expressed as nite sums of members of A. Find sup B . 2229. Proposed by Kenneth Kam Chiu Ko, Mississauga, Ontario. (a) Let m be any positive integer greater than 2, such that x2  1 (mod m) whenever (x; m) = 1. Let n be a positive integer. If mjn +1, prove that the sum of all divisors of n is divisible by m. (b)? Find all possible values of m

2230. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Triangles BCD and ACE are constructed outwardly on sides BC and CA of triangle ABC such that AE = BD and \BDC + \AEC = 180. The point F is chosen to lie on the segment AB so that Prove that

AF = DC : FB CE

EF FD DE CD + CE = BC = AC :

2231. Proposed by Herbert Gulicher, Westfalische Wilhelms-Universitat, Munster, Germany. In quadrilateral P1P2 P3 P4 , suppose that the diagonals intersect at the point M 6= Pi (i = 1; 2; 3; 4). Let \MP1 P4 = 1, \MP3 P4 = 2, \MP1 P2 = 1 and \MP3 P2 = 2 . Prove that 1 M j = cot 1  cot 1 ; 13 := jjPMP 3 j cot 2  cot 2 where the +(,) sign holds if the line segment P1 P3 is located inside (outside)

the quadrilateral.

168  2232. Proposed by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina. Find all solutions of the inequality:

n2 + n , 5
1. Show that the constant Cp is best possible in all cases. 2234. Proposed by Victor Oxman, University of Haifa, Haifa, Israel. Given triangle ABC , its centroid G and its incentre I , construct, using only an unmarked straightedge, its orthocentre H . 2235. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Triangle ABC has angle \CAB = 90 . Let ,1(O; R) be the circumcircle and ,2 (T; r) be the incircle. The tangent to ,1 at A and the polar line of A with respect to ,2 intersect at S . The distances from S to AC and AB are denoted by d1 and d2 respectively. Show that (a) ST kBC , (b) jd1 , d2 j = r. [For the bene t of readers who are not familiar with the term \polar line", we give the following de nition as in, for example, Modern Geometries, 4th Edition, by James R. Smart, Brooks/Cole, 1994: The line through an inverse point and perpendicular to the line joining the original point to the centre of the circle of inversion is called the polar of the original point, whereas the point itself is called the pole of the line.]

169

2236.

Proposed by Victor Oxman, University of Haifa, Haifa, Israel. Let ABC be an arbitrary triangle and let P be an arbitrary point in the interior of the circumcircle of 4ABC . Let K , L, M , denote the feet of the perpendiculars from P to the lines AB , BC , CA, respectively.

[ABC ]

Prove that [KLM ]  4 . Note: [XY Z ] denotes the area of 4XY Z . 2237. Proposed by Meletis D. Vasiliou, Elefsis, Greece. ABCD is a square with incircle ,. Let ` be a tangent to ,. Let A0 , B 0 , 0 0 C , D be points on ` such that AA0, BB0, CC 0, DD0 are all perpendicular to `. Prove that AA0  CC 0 = BB 0  DD0 .

Correction

2173. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let n  2 and x1 ; : : : ; xn > 0 with x1 + : : : + xn = 1. Consider the terms ln = and where

s

n X

(1 + xk ) 1 ,x xk k k=1

rn = Cn

n 1 + xk Y p1 , x

k=1

k

Cn = (pn , 1)n+1(pn)n=(n + 1)n,1:

[Ed: there is no x in the line above.] 1. Show l2  r2 . 2. Prove or disprove: ln  rn for n  3.

170

SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 1940. [1994: 108; 1995: 107; 1995: 205; 1996: 321] Proposed by Ji Chen, Ningbo University, China. Show that if x; y;z > 0,

  (xy + yz + zx) (x +1 y )2 + (y +1 z )2 + (z +1 x)2  94 :

Solution by Marcin E. Kuczma, Warszawa, Poland. Let F be the expression on the left side of the proposed inequality. Assume without loss of generality x  y  z  0, with y > 0 (not excluding z = 0), and de ne:

A B C D E

= = = = =

(2x + 2y , z)(x , z )(y , z ) + z (x + y )2; (1=4)z(x + y , 2z)(11x + 11y + 2z ); (x + y)(x + z)(y + z ); (x + y + z)(x + y , 2z ) + x(y , z ) + y (x , z ) + (x , y )2; (1=4)(x + y)z(x + y + 2z )2(x + y , 2z )2:

It can be veri ed that

C 2(4F , 9) = (x , y)2,(x + y)(A + B + C ) + (x + z)(y + z)D=2 + E:

This proves the inequality and shows that it becomes an equality only for x = y = z and for x = y > 0, z = 0. Comment. The problem is memorable for me! It was my \solution" [1995: 107] that appeared rst. According to someone's polite opinion it was elegant, but according to the impolite truth, it was wrong. I noticed the fatal error when it was too late to do anything; the issue was in print already. In [1995: 205] a (correct) solution by Kee-Wai Lau appeared. Meanwhile I found two other proofs, hopefully correct, and sent them to the editor. Like Kee-Wai Lau's, they required the use of calculus and were lacking \lightness", so to say, so the editor asked [1995:206] for a \nice" solution. I became rather sceptical about the possibility of proving the result by those techniques usually considered as \nice", such as convexity/majorization arguments | just because the inequality turns into equality not only for x = y = z, but also for certain boundary con gurations. In response to the editor's prompt, Vedula Murty [1996: 321] proposed a short proof avoiding hard calculations. But I must frankly confess that I do

171 not understand its nal argument: I do not see why the sum of the rst two terms in [1996: 321(3)] must be non-negative. While trying to clarify that, I arrived at the proof which I present here. This proof can be called anything but \nice"! Decomposition into sums and products of several expressions, obviously nonnegative, and equally ugly, has the advantage that it provides a proof immediately understood and veri ed if one uses some symbolic calculation software (with some e ort, the formula can be checked even by hand). But the striking disadvantage of such formulas is that they carefully hide from the reader all the ideas that must have led to them; they take the \background mathematics" of the reasoning away. In the case at hand I only wish to say that the equality I propose here has been inspired by Murty's brilliant idea to isolate the polynomial that appears as the third term in [1996: 321(3)] and to deal with the expression that remains. I once overheard a mathematician problemist claiming lack of sympathy to inequality problems. In the ultimate end, he said, they all reduce to the only one fundamental inequality, which is x2  0!

2124. [1996: 77] Proposed by Catherine Shevlin, Wallsend, England.

Suppose that ABCD is a quadrilateral with \CDB = \CBD = 50 and \CAB = \ABD = \BCD. Prove that AD ? BC .

D

C A

B

I. Solutionby Florian Herzig, student, Perchtoldsdorf, Austria. (Essentially identical solutions were submitted by Jordi Dou, Barcelona, Spain and Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany. The solution by Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USA was very similar.) Let F1 and F2 be the feet of the perpendiculars from D and A to BC respectively. Let p = BC = CD and q = AC . Then, applying the Sine Rule to 4ABC , we have

CF1 = p cos 80;

sin 30 = p cos 70 : CF2 = q cos 70 = psin80  2 sin 80

172 Thus we have

CF1 = cos 80 = 2 sin 80 cos 80 = sin 160 = 1: 70 CF2 2cos cos 70 sin20 sin 80 Thus, F1 = F2 , and this point is the intersection of AD and BC , whence AD ? BC .

II. Solution by Federico Ardila, student, Massachusetts Institute of Technology, Cambridge, Massachusetts, USA. Consider a regular 18{gon P1 P2 : : : P18. P5 P6 We will rst show that P1 P10, P4 P7 P2P12 and P4P15 concur. P3 P8 By symmetry, P1 P10 , P4 P15 and P2 P9 P5P16 are concurrent. Thus it sucient to prove that P1 P10, Q P1 P10 is P2P12 and P5P16 are concurrent. Using the angles version of Ceva's P18 P11 theorem in triangle 4P1 P5P12 , it P17 P12 if sucient to prove that P16 P13 P15 P14 sin(\P1P12P2)  sin (\P12P5 P16)  sin (\P5P1P10) = 1; sin(\P2P12P5) sin(\P16 P5 P1 ) sin(\P10P1P12) or    q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

q

sin(10 )  sin(40 )  sin(50 ) = 1: sin(30 ) sin(30 ) sin(20)

But this is true since

sin 10 sin40 sin 50 = sin10 sin 40 cos  sin80  40 sin10 cos 10  = sin10 = 2 2  sin20 2  = 4 = (sin30 ) sin20 :

So, P1 P10 , P2 P12 and P4 P15 concur at, say, Q. Using this, it is easy to check that \P2 P4 Q = \P4 QP2 = 50 ; and \P2 P1Q = \P4 QP1 = \QP2 P4 (= 80 ): This information clearly determines the quadrilateral P1 P2 P4 Q up to similarity, so P1 P2 P4 Q  ACDB .

173 Since P1 P4 ? P2 Q, it follows that AD ? BC .  Also solved by CLAUDIO ARCONCHER, Jundia, Brazil; SEFKET  ARSLANAGIC, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; SAM BAETHGE, Science Academy, Austin, Texas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; TIM CROSS, King Edward's School, Birmingham, England; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS,   Y,  Ferris State Ursulinengymnasium, Innsbruck, Austria; VACLAV KONECN University, Big Rapids, Michigan, USA; MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; KEE-WAI LAU, Hong Kong; P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; MELETIS VASILIOU, Elefsis, Greece (two solutions); and the proposer. The proposer writes: The genesis of this problem lies in a question asked by Junji Inaba, student, William Hulme's Grammar School, Manchester, England, in Mathematical Spectrum, vol. 28 (1995/6), p. 18. He gives the diagram in my question, with the given information: \CDA = 20; \DBC = 50;

\DAB = 60; \CBA = 30;

and asks the question: \can any reader nd \CDB without trigonometry?" In fact, such a solution was given in the next issue of Mathematics Spectrum by Brian Stonebridge, Department of Computer Science, University of Bristol, Bristol, England. The genesis of the diagram is much older, if one produces BD and AC to meet at E . See Mathematical Spectrum, vol. 27 (1994/5), pp. 7 and 65{ 66. In one reference, the question of nding \CDA is called \Mahatma's Puzzle", but no reference was available. Can any reader enlighten me on the origin of this puzzle?

2125. [1996: 122] Proposed by Bill Sands, University of Calgary, Calgary, Alberta. At Lake West Collegiate, the lockers are in a long rectangular array, with three rows of N lockers each. The lockers in the top row are numbered 1 to N , the middle row N + 1 to 2N , and the bottom row 2N + 1 to 3N , all from left to right. Ann, Beth, and Carol are three friends whose lockers are located as follows:

174

:::

@, @ ,

@,

:::

By the way, the three girls are not only friends, but also next-door neighbours, with Ann's, Beth's, and Carol's houses next to each other (in that order) on the same street. So the girls are intrigued when they notice that Beth's house number divides into all three of their locker numbers. What is Beth's house number? Solution by Han Ping Davin Chor, student, Cambridge, MA, USA. From the diagram, it can be observed that the lockers have numbers

x + 3; N + x + 5 and 2N + x; where 1  x  N , x a positive integer. Here locker x + 3 is in the rst row, locker N + x + 5 is in the second row, and locker 2N + x is in the third row. Let y be Beth's house number, where y is a positive integer. Since y divides into x + 3, N + x + 5 and 2N + x, y must divide into 2(N + x + 5) , (2N + x) , (x + 3) = 7: Therefore y = 1 or 7. However, Beth's house is in between Ann's and

Carol's. Assuming that 0 is not assigned as a house number, it means that Beth's house number cannot be 1 (else either Ann or Carol would have a house number of 0). Therefore Beth's house number is 7. Also solved by SAM BAETHGE, Science Academy, Austin, Texas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; TIM CROSS, King Edward's School, Birmingham, England; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; J. K. FLOYD, Newnan, Georgia, USA; IAN JUNE L. GARCES, Ateneo de Manila University, Manila, the Philippines, and GIOVANNI MAZZARELLO, Ferrovie dello Stato, Firenze, Italy; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, New York, USA; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; JOHN GRANT MCLOUGHLIN, Okanagan University College, Kelowna, B. C.; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CORY PYE, student, Memorial University of Newfoundland, St. John's, Newfoundland; JOEL SCHLOSBERG, student, Hunter College High School, New York NY, USA; ROBERT P. SEALY, Mount Allison University, Sackville,  New Brunswick; HEINZ-JURGEN SEIFFERT, Berlin, Germany; DAVID STONE, Georgia Southern University, Statesboro, Georgia, USA; EDWARD

175 T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. Two solvers eliminated 1 as a possible answer, because the problem said that the girls were \intrigued" that Beth's house number divided all their locker numbers, which would hardly be likely if Beth's house number were just 1! Thus they didn't need the information about the location of Beth's house at all. Another solver, to whom the editor has therefore given the bene t of the doubt, merely stated that \the location of Ann's and Carol's houses doesn't enter into the problem".

2126. [1996: 123] Proposed by Bill Sands, University of Calgary, Calgary, Alberta. At Lake West Collegiate, the lockers are in a long rectangular array, with three rows of N lockers each, where N is some positive integer between 400 and 450. The lockers in the top row were originally numbered 1 to N , the middle row N + 1 to 2N , and the bottom row 2N + 1 to 3N , all from left to right. However, one evening the school administration changed around the locker numbers so that the rst column on the left is now numbered 1 to 3, the next column 4 to 6, and so forth, all from top to bottom. Three friends, whose lockers are located one in each row, come in the next morning to discover that each of them now has the locker number that used to belong to one of the others! What are (were) their locker numbers, assuming that all are three-digit numbers? Solution by Ian June L. Garces, Ateneo de Manila University, Manila, the Philippines, and Giovanni Mazzarello, Ferrovie dello Stato, Firenze, Italy. The friends' locker numbers are 246, 736 and 932. To show this, rst consider any particular locker. Then the original (before the change) number of this locker can be written as iN + j , where 0  i  2 (the row) and 1  j  N (the column). With respect to this original locker number, this particular locker has a new (after the change) number 3(j , 1) + (i + 1) = 3j + i , 2. Consider now the three friends' lockers. Since the three lockers are located one in each row, we can let them be j1 , N + j2 and 2N + j3 where 1  j1; j2; j3  N . For each of these lockers, the corresponding new locker numbers will be 3j1 , 2, 3j2 , 1 and 3j3. Then there will be two possibilities for how their original locker numbers and their new locker numbers were \properly" interchanged: Possibility 1. The rst possibility is when j1 = 3j3;

(1)

N + j2 = 3j1 , 2;

(2)

176

2N + j3 = 3j2 , 1:

(3) Substituting (1) into (2) and solving for j2, we have j2 = 9j3 , 2 , N . Substituting this last equality into (3) and solving for j3 , we have

j3 = 5N26+ 7

which implies that N  9 mod 26. Choosing N between 400 and 450, we have the unique N = 425 and thus j3 = 82, j2 = 311 and j1 = 246. Hence the original locker numbers are 246, 736 and 932 which, after the change, will respectively be 736, 932 and 246 which satisfy what we want. Possibility 2. The other possibility is when

j1 = 3j2 , 1;

2N + j3 = 3j1 , 2: Similar computation as in Possibility 1 yields N = 425, j2 = 115, j3 = 180 and j1 = 344. But this means that one of the lockers will have number 1030

N + j2 = 3j3;

which is contrary to the assumption. Therefore, the only possible locker numbers of the three friends are 246, 736 and 932. Also solved by SAM BAETHGE, Science Academy, Austin, Texas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JOSEPH CALLAGHAN, student, University of Waterloo, Waterloo, Ontario; HAN PING DAVIN CHOR, student, Cambridge, MA, USA; TIM CROSS, King Edward's School, Birmingham, England; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, New York, USA; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; DAVID STONE, Georgia Southern University, Statesboro, Georgia, USA; and the proposer. Many solvers mentioned that the other set of locker numbers arising from the problem is 344, 540 and 1030. Some remarked that the value of N was 425 in both cases. However, apparently nobody noticed that these two triples of numbers enjoy a curious relationship:

246 + 1030 = 736 + 540 = 932 + 344 !

So now readers are challenged to gure out why this relationship is true. When N = 425, the problem says that the numbers 246; 736; 932 are interchanged when the lockers are renumbered. So let's call this set of numbers a \swapset" for N = 425; that is, for a particular N , a swapset is

177 any set of numbers which get swapped among each other by the renumbering. We want true swapping; so we don't allow the sets f1g or f3N g (or the \middle" locker f(3N + 1)=2g when N is odd), which are obviously unchanged by the renumbering, to be in swapsets. Lots of problems concerning swapsets could be looked at. For example, one of the solvers (Stone) points out that there are no swapsets of two numbers when N = 425, but there are when N = 427: lockers 161 and 481 get swapped. Which values of N have swapsets of size two? Here's another problem. It's clear that the set of all numbers from 1 to 3N , minus the two or three numbers that stay the same, will be a swapset for every N . But are there any numbers N which have no other swapsets? If so, can you describe all such N ?

2127. [1996: 123] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is an acute triangle with circumcentre O, and D is a point on the minor arc AC of the circumcircle (D 6= A; C ). Let P be a point on the side AB such that \ADP = \OBC , and let Q be a point on the side BC such that \CDQ = \OBA. Prove that \DPQ = \DOC and \DQP = \DOA. Solution by Florian Herzig, student, Perchtoldsdorf, Austria. First I prove that B is an excentre of 4PDQ. \ABC = 180 , \ADC = 180 , (\ADP + \CDQ + \PDQ) = 180 , (\CBO + \ABO + \PDQ) = 180 , \ABC , \PDQ; (1) \ PDQ ; ) \ABC = 90 , and

2 \PDB = \ADB , \ADP = \ACB , \OCB = \ACO;

\QDB = \CDB , \CDQ = \CAB , \OAB = \CAO: Since 4OAC is isosceles, we have that \PDB = \QDB and thus BD is the internal angle bisector of \PDQ. (2) What is more, we know that, in any 4XY Z , the excentre, M , (whose excircle touches Y Z ), is exactly the point on the internal angle bisector of \Y XZ outside the triangle for which \Y MZ = 180 , \MZY , \MY Z  = \Y + \Z = 180 , \X = 90 , \X :

2

2

2

2

Therefore B is an excentre of 4PDQ because of (1) and (2). Then BP and BQ are the external angle bisectors of \DPQ and \DQP , respectively, whence \APD = \BPQ and \CQD = \BQP: (3)

178 Starting with we obtain

\BOC = 2\BDC

180 , 2\OBC 90 , \OBC 180 , \BDC \BCD + \DBC  (180 , \DAP ) + \DBC

= = = = = \DBC = \DOC =

2\BDC;

\BDC; 90 + \OBC; 90 + \ADP; 90 + (180 , \DAP , \APD); 90 , \APD; 180 , (\APD + \BPQ) [ because of (3) ] = \DPQ;

and analogously \DOA = \DQP: Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; HAN PING DAVIN CHOR, student, Cambridge, MA, USA; P. PENNING, Delft, the Netherlands; WALDEMAR POMPE, student, University of Warsaw, Poland; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.

2128. [1996: 123] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABCD is a square. Let P and Q be interior points on the sides BC and CD respectively, and let E and F be the intersections of PQ with AB and AD respectively. Prove that   \PAQ + \ECF < 5 : 4

Solution by Heinz-Jurgen Sei ert, Berlin, Germany. In cartesian coordinates, let A = (0; 0), B = (1; 0), C = (1; 1), D = (0; 1), P = (1;p) and Q = (q; 1), where 0 < p; q < 1:     Then E = 11,,pqp ; 0 and F = 0; 11,,pqq , tan \PAB = p, tan \DAQ = q , tan \DCF = FD = q(11,,qp) , tan \BCE = BE = p(11,,pq) . Since   \PAQ = , \PAB , \DAQ and \ECF = + \DCF + \BCE; 2 2 it follows that \PAQ + \ECF , p) p(1 , q) =  + arctan q(1 1 , q , arctan q + arctan 1 , p , arctan p   q)2 =  + arctan (1 , p)(1 , q)(1 , pq(1)2,+pq(p)((1p , , q)2 + q(1 , p)2 )(p + q)

179 by the addition formula for arctangents. Since 0 < p; q < 1, it suces to show that

0  (1 , pq)(p , q)2 < (p(1 , q )2 + q (1 , p)2)(p + q ):

The left inequality is obviously true, while the right follows from the identity

(p(1 , q)2 + q(1 , p)2)(p + q) = (1 , pq )(p , q )2 +2pq ((1 , p)2 +(1 , q )2):   University of Sarajevo, SaraAlso solved by SEFKET ARSLANAGIC, jevo, Bosnia and Herzegovina; NIELS BEJLEGAARD, Stavanger, Norway; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JOSEPH CALLAGHAN, student, University of Waterloo; RICHARD I. HESS, Rancho Palos Verdes, California, USA; VICTOR OXMAN, University of Haifa, Haifa, Israel; and the proposer.

2130. [1996: 123] Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. A and B are xed points, and ` is a xed line passing through A. C is a variable point on `, staying on one side of A. The incircle of ABC touches BC at D and AC at E. Show that line DE passes through a xed point. Solutionby Mitko Kunchev, Baba Tonka School of Mathematics, Rousse, Bulgaria. We choose the point P on ` with AP = AB . Let C be an arbitrary point of `, di erent from P but on the same side of A. The incircle of 4ABC touches the sides BC , AC , AB in the points D, E , F respectively. Let ED meet PB in the point Q. According to Menelaus' Theorem applied to 4CBP and the collinear points E, D, Q, we get

PE  CD  BQ = 1: (1) EC DB QP We have EC = CD (because they are tangents from C ). Similarly, AF = AE, so that FB = EP (since AB = AP ). But also, FB = DB, so that DB = PE . Setting EC = CD and DB = PE in (1), we conclude that BQ = QP ; therefore Q is the mid-point of BP . Hence the line DE passes through the xed point Q.

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, UK; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA (two solutions); and the proposer.

180 Seimiya and Yiu used the same argument as Kunchev. Seimiya mentions that the result is easily shown to hold also when C coincides with P (even though the featured argument breaks down). Yiu extends the result to include excircles: The line joining the points where an excircle touches the segment BC and the line ` also passes through Q.

2131. [1996: 124] Proposed by Hoe Teck Wee, Singapore.

Find all positive integers n > 1 such that there exists a cyclic permutation of (1; 1; 2; 2; : : : ; n; n) satisfying: (i) no two adjacent terms of the permutation (including the last and rst term) are equal; and (ii) no block of n consecutive terms consists of n distinct integers. Solution by the proposer. It is clear that 2 does not have the desired property. Suppose 3 has the speci ed property. So there exists a permutation of (1; 1; 2; 2; 3; 3) satisfying the two conditions. WLOG assume that the rst term is 1. From (ii) we know that the second term is not 1, say it is 2. From (i) the third term must be 1. From (i) and (ii) the fourth term must be 2. This leaves the two 3s as the last two terms, contradicting (i). Suppose 4 has the speci ed property. So there exists a permutation of (1; 1; 2; 2; 3; 3; 4; 4) satisfying the two conditions. Arrange these eight (permuted) numbers in a circle in that order so that they are equally spaced. Then the two conditions still hold. Now consider any four consecutive numbers on the circle. If they consist of only two distinct integers, we may assume by (i) that WLOG these four numbers are 1; 2; 1; 2 in that order, and that the other four numbers are 3; 4; 3; 4. Then (ii) does not hold. If they consist of three distinct integers, by (i) and (ii) we may assume WLOG that these four numbers are (a) 1; 2; 3; 1 or (b) 1; 2; 1; 3 or (c) 1; 2; 3; 2, in these orders. By reversing the order, (c) reduces to (b). Next consider (a). If the next number is 2, then by (ii) we have 1; 2; 3; 1; 2; 3, and the two 4s are adjacent, contradicting (i). If the next number is 3, reversing the order to obtain 3; 1; 3; 2; 1 reduces it to (b). Finally consider (b). By (i) and (ii) the next number must be 2, followed by 3, so the two 4s are adjacent, contradicting (ii). Next consider the following permutation for n > 4:

(4; 5; : : : ; n; 1; 2; 3; 2; 3; 4; 5; : : : ; n): Clearly, (i) is satis ed. (ii) follows from the fact that there does not exist a set of four consecutive terms which is a permutation of (1; 2; 3; 4). In conclusion, the answer is: n > 4.

181 Also solved by HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; and DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA. There was one incomplete solution.  2132. [1996: 124] Proposed by Sefket Arslanagi c , Berlin, Germany.

Let n be an even number and z a complex number. Prove that the polynomial P (z ) = (z + 1)n , z n , n is not divisible by z2 + z + n. I. Solution by Richard I. Hess, Rancho Palos Verdes, California, USA. Let Q(z ) = z 2 + z + n. For n = 0 or 1, we have that P (z ) = 0, which is clearly divisible by Q(z ). For any n > 1, suppose that P (z ) is divisible by Q(z ). Then Q(n) divides P (n). But Q(n) = n(n +2)  0 (mod n), while P (n) = (n +1)n , n2 , n  1 (mod n). Thus P (z) is not divisible by Q(z ). II. Composite solution by F.J. Flanigan, San Jose State University, San Jose, California, USA and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Let D(z ) = z 2 + z + n. If n = 0; 1, then P (z ) = 0, which is divisible by D(z ). If n = 2, then P (z ) = 2z , 1, which is clearly not divisible by z2 + z + 2. For n > 2, suppose that D(z ) divides P (z ). Then, since D(z ) is monic, P (z) = Q(z)D(z), where Q(z) is a polynomial of degree n , 3 with integer coecients. Thus P (0) = Q(0)D(0), or 1 , n = nQ(0), which is clearly impossible. III. Solution and generalization by Heinz-Jurgen Sei ert, Berlin, Germany. Let n  2 be an even integer. We shall prove that if a, b, c, are complex numbers such that a 6= 0, then the polynomial

P (z) = (z + b)n , zn , a is not divisible by z 2 + bz + c.

The proposer's result, which does not hold for n = 0, is obtained when a = c = n and b = 1. Let z1 and z2 denote the (not necessarily distinct) roots of z 2 + bz + c. The z1 + z2 = ,b, so that P (z1 ) = z2n , z1n , a, and P (z2) = z1n , z2n , a. Since P (z1) + P (z2) = ,2a = 6 0, our result follows. 6 The example (z + 1) , z 6 = (z 2 + z + 1)(6z 3 + 9z 2 + 5z + 1) shows that the condition a = 6 0 cannot be dropped.

182 Also solved by: CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, Hong Kong; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; NORVALD MIDTTUN (two solutions), Royal Norwegian Naval Academy, Norway; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; and the proposer. Besides the solvers listed in Solutions I and II above, only Janous observed and showed that the assertion holds for all n  2.

2134?. [1996: 124] Proposed by Waldemar Pompe, student, Univer-

sity of Warsaw, Poland. Let fxn g be an increasing sequence of positive integers such that the sequence fxn+1 , xn g is bounded. Prove or disprove that, for each integer m  3, there exist positive integers k1 < k2 < : : : < km , such that xk1 ; xk2 ; : : : ; xkm are in arithmetic progression. Solution by David R. Stone, Georgia Southern University, Statesboro, Georgia, USA, and Carl Pomerance, University of Georgia, Athens, Georgia, USA. An old and well-known result of van der Waerden [4] is that if the natural numbers are partitioned into two subsets, then one of the subsets has arbitrarily long arithmetic progressions. It is not very dicult to show [1] that van der Waerden's theorem has the following equivalent formulation: for every number B and positive integer m, there is a number W (m; B) such that if n  W (m;B) and 0 < a1 < a2 < : : : < an are integers with each ai+1 , ai  B, then m of the ai's form an arithmetic progression. Thus, for the problem as stated, if we let B be the bound on the di erences xn+1 , xn , then for any given m  3, there exists a W (m; B ) with the property stated above. Then, for any n  W (m; B ), any nite subsequence of length n will have an arithmetic progression of length m as a sub-subsequence. That is, the original sequence contains in nitely many arithmetic progressions of length m. In 1975, Szemeredi [3] proved a conjecture of Erdos } and Turan which improves on van der Waerden's Theorem, relaxing the condition that the sequence's di erences have a uniform upper bound, requiring only that the sequence have a positive upper density. Hence the problem posed here also follows from the theorem of Szemeredi, who, we believe, received (for this result) the highest cash prize ever awarded by Pal Erdos } | $1,000.

183 Comment by the solvers. Do we know how Pompe became interested in this problem?

References [1] T.C. Brown, Variations on van der Waerden's and Ramsey's theorems, Amer. Math. Monthly 82 (1975), 993{995. [2] Carl Pomerance, Collinear subsets of lattice point sequences | an analog of Szemeredi's Theorem, J. Combin. Theory 28 (1980), 140{149. [3] E. Szemeredi, On sets of integers containing no k elements in arithmetic progression, Acta Arith. 27 (1975), 199{245. [4] B.L. van der Waerden, Beweis einer Baudetschen Vermutung, Nieuw. Arch. Wisk. 15 (1928), 212{216. Also solved by THOMAS LEONG, Staten Island, NY, USA; and JOEL SCHLOSBERG, student, Hunter College High School, New York NY, USA; both using van der Waerden's theorem or its variation. Leong gave the reference: Ramsey Theory by R.L. Graham, B.L. Rothschild and J.H. Spencer. Schlosberg remarked that van der Waerden's theorem was discussed in the July 1990 issue of Scienti c American. The proposer showed that van der Waerden's theorem follows easily from the statement of his problem. His intention (and hope) was to nd a proof independent of van der Waerden's theorem. This would establish a new \proof" of the latter. In view of his comment and the solution above, it should be obvious that the two statements are equivalent, and hence such a proof is unlikely.

2135. [1996: 124] Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Let n be a positive integer. Find the value of the sum bn= X2c k=1

(,1)k(2n , 2k)! : (k + 1)!(n , k)!(n , 2k)!

Solution by Florian Herzig, student, Perchtoldsdorf, Austria. [Modi ed slightly by the editor.] Let Sn denote the given summation. Note that S1 is an \empty" sum,  2 n which we shall de ne to be zero. We prove that Sn = , .

n+2 Since n  2. Follow3 = 0, this is true for n = 1. Assume that  ing standard convention, for k = 0, 1, 2, : : : , let xk (f (x)) denote the

2

184 coecient of xk in the series expansion of the function f (x). Let

P (x) = ,1 , x2n+1 (1 , x),(n+1): Then, by the binomial expansion, itsgeneralization   (by Newton), and the , n n + k , 1 k well-known fact that , we have: k = (,1) k x2k+2  ,1 , x2n+1 = h,x2k+1i ,1 , x2n+1 n + 1 k +1 = (,1) k+1 ; for k = ,1; 0; 1; 2; : : : , and xn,2k  (1 , x),(n+1) = (,1)n,2k,n , 1 2n , 2k n , 2k = n , 2k for k  n=2. Hence xn+2  (P (x)) = x2k+2  xn,2k  (P (x)) bn= X2c k+1n + 12n , 2k = (,1) k + 1 n , 2k : k=,1   On the other hand, since P (x) = (1 + x)n+1 , we have xn+2 (P (x)) = 0. Therefore bn= X2c k+1n + 12n , 2k (,1) k + 1 n , 2k = 0: k=,1 Since bn= X2c k+1n + 12n , 2k 1 Sn = , n + 1 (,1) k + 1 n , 2k ; k=1 we get

      Sn = , n +1 1 n +1 1 2nn , n +0 1 2nn++22 n + 2)! , (2n)! = (n +(21)( n + 2)!n! n!n! n + 1)g  (2n)! = f2(2n + 1)(n,+(n2)!+n2)( ! 1) (2n)! = ,(2n)! = ,n((nn +, 2)! (n + 2)!(n , 2)!  2n  n! = , n+2 :

185 Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck,  Austria; HEINZ-JURGEN SEIFFERT, Berlin, Germany; and the proposer. The correct answer, without a proof, was sent in by RICHARD I. HESS, Rancho Palos Verdes, California, USA. If, in the given summation, one lets k start from zero (this was, in fact, the proposer's original idea), then it is easy to see that the answer becomes

1 2n + 2; n+2 n+1

the (n + 1){th Catalan number.

2136. [1996: 124] Proposed by G.P. Henderson, Campbellcroft, On-

tario.

Let a; b; c be the lengths of the sides of a triangle. Given the values of p = P a and q = P ab, prove that r = abc can be estimated with an error of at most r=26. Solution by P. Penning, Delft, the Netherlands. Scale the triangle down by a factor (a + b + c). The value of p then beq r comes 1, the value of q becomes Q = 2 , and R = 3.

a + b and v = a , b : Introduce s = 2

(a + b + c)

(a + b + c)

2

a = s + v; b = s , v; c = 1 , 2s; Q = 2s , 3s2 , v2; R = (1 , 2s)(s2 , v2):

Since a, b, c, represent the sides of a triangle, we must require 0 < c < a + b and , c < a , b < c: [Ed: in other words, the triangle is not degenerate | a case which must be discarded as inappropriate.] This translates to

1 y if and only if 1997 , x  1996 , y . 7. Show that xp log x ! 0 as x ! 0 for all p > 0. Use induction. 8. Expanding out, we see that the elementary symmetric functions appear as the coecients of the polynomial. Suppose the ai 's are distinct. Show that if any ai is negative, then x > 0 can be chosen so that the polynomial is negative. 9. Symmetry suggests trying the substitution a = c to nd a counterexample. [This does not work!] To investigate this question further, prove the inequality is true when strict inequality is relaxed to . The inequality is equivalent to [h(a; b; c)]2  0 for an appropriate function h. 10. Having found two pairs of adjacent Fibonacci numbers with the same last four digits, now work backwards. Compare fm,1 and fn,1 etc. Keep going. 11. Note that

1 1 2k = , 2 2 2 k +k +1 k ,k+1 k +k+1 4

Using partial fractions, show that the sum can be written as a telescoping sum.

259 12. Apply Fermat's Great Theorem. 13. There is more than one way to skin a cat: use integration by parts and the transformation t = tan(x=2). 14. Descartes Rule of Signs tells us that if z is the number of positive zeros of a polynomial, and c is the number of changes of sign of the sequence of coecients, then c , z is a nonnegative even number. 15. Show that

[f (x + a)]2 , f (x + a) = ,[f (x) , 1=2]2: Then write out a formula for f (x +2a) using the original equation given in the question.

ANSWERS 1. True 2. True 3. False 4. True 5. False 6. False 7. False 8. True 9. False 10. False 11. True 12. True 13. False 14. True 15. True

Comment For the information of readers, to the knowledge of the editor, other versions of the Bernoulli Trials have been held at least in his home institution and at the Canadian IMO Residential Training Program. In the latter instance, local high school students were added to the list of competitors. On both occasions, all the competitors reported that it was an enjoyable experience, and a pleasant social event as well.

260

THE OLYMPIAD CORNER No. 183

R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. We begin with the 1997 Canadian Mathematical Olympiad which we reproduce with the permission of the Canadian Mathematical Olympiad Committee of the Canadian Mathematical Society. Thanks go to Daryl Tingley, University of New Brunswick, and Chair of the Canadian Mathematical Olympiad Committee of the Canadian Mathematical Society, for forwarding the questions to me.

PROBLEMS FROM THE 1997 CANADIAN MATHEMATICAL OLYMPIAD 1. How many pairs of positive integers x; y are there, with x  y, and such that gcd (x; y ) = 5! and lcm (x; y ) = 50! ? Note: gcd (x; y ) denotes the greatest common divisor of x and y , lcm (x; y ) denotes the least common multiple of x and y , and n! = n  (n , 1)      2  1. 2. The closed interval A = [0; 50] is the union of a nite number of closed intervals, each of length 1. Prove that some of the intervals can be removed so that those remaining are mutually disjoint and have total length  25. Note: For a  b, the closed interval [a; b] := fx 2 R : a  x  bg has length b , a; disjoint intervals have empty intersection. 3. Prove that 1 < 1  3  5    1997 < 1 : 1999 2 4 6 1998 44

4. The point O is situated inside the parallelogram ABCD so that \AOB + \COD = 180 :

Prove that \OBC = \ODC .

261

5. Write the sum in the form

n X

(,1)k,nk 3 2 k=0 k + 9k + 26k + 24

p(n) , where p and q are polynomials with integer coecients. q(n)

The next set of problems are from the twenty-sixth annual United States of America Mathematical Olympiad written May 1, 1997. These problems are copyrighted by the Committee on the American Mathematical Competitions of the Mathematical Association of America and may not be reproduced without permission. Solutions, and additional copies of the problems may be obtained from Professor Walter E. Mientka, AMC Executive Director, 917 Oldfather Hall, University of Nebraska, Lincoln, NE, USA 68588{0322. As always, we welcome your original, \nice" solutions and generalizations which di er from the published solutions.

26th UNITED STATES OF AMERICA MATHEMATICAL OLYMPIAD Part I

9 a.m. - 12 noon

May 1, 1997

1. Let p1; p2; p3; : : : be the prime numbers listed in increasing order, and let x0 be a real number between 0 and 1. For positive integer k, de ne 8 < 

0  if xk,1 = 0; p xk = : k if xk,1 6= 0; xk,1 where fxg denotes the fractional part of x. (The fractional part of x is given by x , bxc where bxc is the greatest integer less than or equal to x.) Find, with proof, all x0 satisfying 0 < x0 < 1 for which the sequence x0 ; x1 ; x2 ; : : : eventually becomes 0. 2. Let ABC be a triangle, and draw isosceles triangles BCD, CAE, ABF externally to ABC , with BC , CA, AB as their respective bases. ! !Prove ! that the lines through A, B , C perpendicular to the lines EF; FD; DE , respectively, are concurrent. 3. Prove that for any integer n, there exists a unique polynomial Q with coecients in f0; 1; : : : ; 9g such that Q(,2) = Q(,5) = n.

262

Part II

1 p.m. - 4 p.m.

4. To clip a convex n-gon means to choose a pair of consecutive sides AB, BC and to replace them by the three segments AM , MN , and NC , where M is the midpoint of AB and N is the midpoint of BC . In other words, one cuts o the triangle MBN to obtain a convex (n + 1)-gon. A regular hexagon P6 of area 1 is clipped to obtain a heptagon P7. Then P7 is clipped (in one of the seven possible ways) to obtain an octagon P8, and so on. Prove that no matter how the clippings are done, the area of Pn is greater than 1=3, for all n  6. 5. Prove that, for all positive real numbers a; b; c, (a3 + b3 + abc),1 + (b3 + c3 + abc),1 + (c3 + a3 + abc),1  (abc),1: is es

6. Suppose the sequence of nonnegative integers a ; a ; : : : ; a 1

2

1997

sat-

ai + aj  ai+j  ai + aj + 1 for all i; j  1 with i + j  1997. Show that there exists a real number x such that an = bnxc (the greatest integer  nx) for all 1  n  1997. As a third Olympiad for this issue we give selected problems of the 3rd Ukrainian Mathematical Olympiad, written March 26 and 27, 1994. My thanks go to Richard Nowakowski, Canadian Team leader to the IMO in Hong Kong for collecting the set.

3rd UKRAINIAN MATHEMATICAL OLYMPIAD Selected Problems March 26{27, 1994

1. (class 9) A convex quadrangle ABCD is given. Bisectors of external angles of ABCD form a new quadrangle PQRS . Prove that sum of diagonals of PQRS is more than the perimeter of ABCD. 2. (9{10) A convex polygon and point O inside it are given. Prove that for any n ,,! > 1 there ! exist points A1 ; A2 ;, : : : ; An on the sides of the polygon such that OA1 + ,, OA2 + : : : + ,,! OAn = ! O. 3. (10) A sequence of natural numbers ak , k  1, such that for each k, ak < ak+1 < ak + 1993 is given. Let all prime divisors of ak be written for every k. Prove that we receive an in nite number of di erent prime numbers. 4. (10) Inside the triangle ABC the point D is given such that angles \ABC and \DBC are equal. Let K and L be the projections of D on lines (AC ) and (BD) respectively. Prove that the midpoints of AB , CD and KL are collinear.

263

5.

(11) 19942 squares 1  1 of k di erent colours are given. Find all natural numbers k for which it is always possible to construct a 1994  1994 square coloured symmetrically with respect to the diagonal from any collection of 19942 such squares. 6. (11) A mathematical olympiad consists of two days. The Jury intends to place the participants in rooms so that any two students will be in rooms with di erent numbers of participants in one of the days. Is it possible for (a) 9 participants? (b) 14 participants? (The Jury has the necessary number of rooms.) 7. (11) Three points P , Q and S in space are given. Two rays from P and two rays from Q are taken such that each ray from P intersects two rays from Q. Let A, B , C , D be the intersection points of these four rays. It is known that pyramid SABCD has a section KLMN of rectangular form (K 2 SA, L 2 SB , M 2 SC , N 2 SD). The area of ABCD equals 1. Prove that the volume of SABCD is not larger than PQ=6. 8. (11) Two circles with radii R and r (R > r) in the plane are tangent at point M with one circle inside the other. Chord AB of the bigger circle is tangent to the smaller circle. Find the largest possible value of the perimeter of ABM . Eagle-eyed readers Dan Velleman, Stan Wagon and Andy Liu spotted trouble with problem 4 of the Swedish Mathematics Contest 1993 as we gave it in the last number of the Corner. Indeed, as they point out, setting ab = 1 for all a, b satis es the conditions of the problem, but does not give a unique solution! I must not have been seeing stars when I proof-read that one. Here is the correct version to try.

4. Swedish Mathematics Contest 1993 [1997: 196]

To each pair of real numbers a and b, where a 6= 0 and b 6= 0, there is a real number a  b such that

a  (b  c) = (a  b)c a  a = 1:

Solve the equation x  36 = 216.

264 In somewhat of a departure from past practice we now give the solutions given to the Canadian Mathematical Olympiad given earlier this issue! (I hope you have solved them by now.) The \ocial solutions" provided for me by Daryl Tingley, University of New Brunswick, and Chair of the Canadian Mathematical Olympiad Committee of the Canadian Mathematical Society, feature solutions by contestants in the CMO itself. Each solution is credited to one of the contestants.

PROBLEM SOLUTIONS FOR THE 1997 CANADIAN MATHEMATICAL OLYMPIAD 1. How many pairs of positive integers x; y are there, with x  y, and

such that gcd (x; y ) = 5! and lcm (x; y ) = 50! ? Note: gcd (x; y ) denotes the greatest common divisor of x and y , lcm (x; y ) denotes the least common multiple of x and y , and n! = n  (n , 1)      2  1. Solution by Deepee Khosla, Lisgar Collegiate Institute, Ottawa, ON Let p1 ; : : : ; p12 denote, in increasing order, the primes from 7 to 47. Then and

5! = 23  31  51  p01  p02     p012

50! = 2a1  3a2  5a3  pb11  pb22     pb1212 : Note that 24 ; 32 ; 52 ; p1 ; : : : ; p12 all divide 50!, so all its prime powers di er from those of 5! Since x; y j50!, they are of the form

x = 2n1  3n2      pn1215 y = 2m1  3m2      pm1215 : Then max(ni; mi) is the ith prime power in 50! and min(ni; mi) is the ith prime power in 5! Since, by the above note, the prime powers for p12 and under di er in 5! and 50!, there are 215 choices for x, only half of which will be less than y . (Since for each choice of x, we have that y is forced and either x < y or y < x.) So the number of pairs is 215=2 = 214. 2. The closed interval A = [0; 50] is the union of a nite number of closed intervals, each of length 1. Prove that some of the intervals can be removed so that those remaining are mutually disjoint and have total length greater than or equal to 25. Note: For a  b, the closed interval [a; b] := fx 2 R : a  x  bg has length b , a; disjoint intervals have empty intersection.

265 Solution by Byung-Kuy Chun, Harry Ainlay Composite High School, Edmonton, AB. Look at the rst point of each given unit interval. This point uniquely de nes the given unit interval. Lemma. In any interval [x; x + 1) there must be at least one of these rst points (0  x  49). Proof. Suppose the opposite. The last rst point before x must be x , " for some " > 0. The corresponding unit interval ends at x , " + 1 < x + 1. However, the next given unit interval cannot begin until at least x + 1. This implies that points (x , " + 1; x + 1) are not in set A, a contradiction. Therefore there must be a rst point in [x; x + 1). Note that for two rst points in intervals [x; x + 1) and [x + 2; x + 3) respectively, the corresponding unit intervals are disjoint, since the intervals are in the range [x; x + 2) and [x + 2; x + 4) respectively. Thus, we can choose a given unit interval that begins in each of

[0; 1) [2; 3) : : : [2k; 2k + 1) : : : [48; 49): Since there are 25 of these intervals, we can nd 25 points which correspond to 25 disjoint unit intervals.

Solution by Colin Percival, Burnaby Central Secondary School, Burnaby, BC. S I prove the more general result, that if [0; 2n] = i Ai ; jAi j = 1; Ai are intervals then 9 a1 : : : an , such that Aa \ Aa = ;. Let 0 < "  n,2 1 and let bi = (i , 1)(2 + "); i = 1 : : : n. Then i

j

 2 minfbig = 0; maxfbig = (n , 1)(2 + ")  (n , 1) 2 + n , 1   2 n = (n , 1) n , 1 = 2n: 

So all the bi are in [0; 2n]. S Let ai be such that bi 2 Aa . Since Ai = [0; 2n], this is possible. Then since (bi , bj ) = (i , j )(2+ ")  2+ " > 2 (i > j ), and the Ai are intervals of length 1, min Aa , max Aa > 2 , 1 , 1 = 0, so Aa \ Aa = ;. Substituting n = 25, we get the required result. 3. Prove that i

i

j

1 < 1  3  5    1997 < 1 : 1999 2 4 6 1998 44

i

j

266 Solution by Mihaela Enachescu, Dawson College, Montreal, PQ. 1 3 1997 . Then 1 > 1 [because 2 < 3], 3 > 3 Let P =   : : : 

2 4

1998 2 3 1997 1997 [because 4 < 5], : : : ; : : : 1998 > 1999 [because 1998 < 1999].

4

5

So

1997 = 1 : P > 13  35  : : :  1999 1999 Also 12 < 32 [because 1  3 < 2  2], 34 < 45 [because 3  5 < 4  4], : : : , 1997 < 1998 [because 1997  1999 = 19982 , 1 < 19982]. 1998 1999

(1)

  2 4 1998 2 4 6 1998 1 . So P <   : : :  =    : : :  3 5 1999 | 1 3 5 {z 1997 } 1999

=P

1

1 < 1 = 1 and P < 1 : Hence P < 1999 1936 442 44 2

(2)

1 Then (1) and (2) give 1999 < P < 441 . 4. The point O is situated inside the parallelogram ABCD so that \AOB + \COD = 180 : Prove that \OBC = \ODC . Solution by Joel Kamnitzer, Earl Haig Secondary School, North York, ON.

O0

A

B O

D

C a translation which maps D to A. It will map O ! O0 with ,OO ,!0 =Consider ,! DA, and C will be mapped to B because ,! CB = ,! DA.

267 This translation keeps angles invariant, so \AO0 B = \DOC =

180 , \AOB.

Therefore AOBO0 is a cyclic quadrilateral. And further, \ODC = \O0 AB = \O0 OB . But, since O0 O is parallel to BC , \O0 OB = \OBC: Therefore

\ODC = \OBC:

Solution by Adrian Chan, Upper Canada College, Toronto, ON.

A

B

 , 

180

O



 ,

180

D

C Let \AOB =  and \BOC = . Then \COD = 180 ,  and \AOD = 180 , . Since AB = CD (parallelogram) and sin  = sin(180 , ), the sine law on 4OCD and 4OAB gives sin \CDO = sin(180 , ) = sin  = sin \ABO OC CD AB OA

so

so

OA = sin \ABO : OC sin \CDO Similarly, the Sine Law on 4OBC and 4OAD gives sin \CBO = sin = sin(180 , ) = sin \ADO OC BC AD OA OA = sin \ADO : OC sin \CBO

Equations (1) and (2) show that sin \ABO  sin \CBO = sin \ADO  sin \CDO:

(1)

(2)

268 Hence

1 [cos(\ABO + \CBO) , cos(\ABO , \CBO)] 2 = 12 [cos(\ADO + \CDO) , cos(\ADO , \CDO)]:

Since \ADC = \ABC (parallelogram) and \ADO + \CDO = \ADC and \ABO + \CBO = \ABC it follows that

cos(\ABO , \CBO) = cos(\ADO , \CDO):

There are two cases to consider. Case (i): \ABO , \CBO = \ADO , \CDO. Since \ABO+\CBO = \ADO+\CDO, subtracting gives 2 \CBO = 2 \CDO so \CBO = \CDO, and we are done. Case (ii): \ABO , \CBO = \CDO , \ADO. Since we know that \ABO + \CBO = \CDO + \ADO, adding gives 2 \ABO = 2 \CDO so \ABO = \CDO and \CBO = \ADO. Substituting this into (1), it follows that OA = OC . Also, \ADO + \ABO = \CBO + \ABO = \ABC . Now, \ABC = 180 , \BAD since ABCD is a parallelogram. Hence \BAD + \ADO + \ABO = 180 so \DOB = 180 and D; O; B are collinear. We now have the diagram

A

B



O

,

180

D

C

Then \COD + \BOC = 180 , so \BOC =  = \AOB . 4AOB is congruent to 4COB (SAS, OB is common, \AOB = \COB and AO = CO), so \ABO = \CBO. Since also \ABO = \CDO we conclude that \CBO = \CDO.

269 Since it is true in both cases, then \CBO = \CDO.

5. Write the sum

n X

(,1)k nk ; 3 2 k=0 k + 9k + 26k + 24 , 

in the form pq((nn)) , where p and q are polynomials with integer coecients. Solution by Sabin Cautis, Earl Haig Secondary School, North York, ON. We rst note that

k3 + 9k2 + 26k + 24 = (k + 2)(k + 3)(k + 4): n X

kn ( , 1) k Let S (n) = . 2 + 9k 2 + 26k + 24 k k=0 , 

Then

n X

(,1)k n! k=0 k!(n , k)!(k + 2)(!k + 3)(k + 4)   n k (n + 4)! X ( , 1) k + 1 =  (n + 1)(n + 2)(n + 3)(n + 4) : k=0 (k + 4)!(n , k)!

S(n) =

Let

T (n) = (n +1)(n +2)(n +3)(n +4)S(n) = Now, for n  1,

n X i=0

since

(1 , 1)n =

(,1)i

n  X k=0

(,1)k

n + 4(k + 1) : k+4



n =0 i

 

n , n + n + : : : + (,1)nn = 0: 0 1 2 n

 

()

270 Also

n X

  n X (,1)i ni i = (,1)i i!  i(n n,! i)! + (,1)0  0!0  nn!!

i=0

i=1 n X

n! (,1)i (i , 1)!( n , i)!

=

i=1 n X

 n , 1 = i,1 i=1   n X = n (,1)i ni ,, 11 i=1   n X i , 1 n,1 = ,n (,1) i , 1 : i=1 

(,1)in

Substituting j = i , 1, (*) shows that n X

    nX ,1 (,1)i ni i = ,n (,1)j n ,j 1 = 0.

i=0

Hence

j =0

n X

()

 n + 4 T (n) = k + 4 (k + 1) k=0  n X + 4(k + 1) = (,1)k+4 nk + 4 k=0   n X k +4 n + 4 = (,1) k + 4 (k + 1) k=,4    n + 4 , ,3 + 2(n + 4) , 2 :

(,1)k



Substituting j = k + 4, gives nX +4

    (,1)j n +j 4 (j , 3) , 2n + 8 , 3 , (n + 4)(2 n + 3) j =0   nX +4 n + 4 j = (,1) j j

T (n) =

j =0

nX +4

 n + 4 1 (4n + 10 , n2 , 7n , 12) ,3 , j 2 j =0 The rst two terms are zero because of results () and () so

(,1)j



271 2 T (n) = n + 32n + 2 :

Then

Therefore

(n) S(n) = (n + 1)(n + T2)( n + 3)(n + 4) 2 n + 3 = 2(n + 1)(n + 2)(nn++23)(n + 4) n + 2) = 2(n + 1)((nn++1)( 2)(n + 3)(n + 4) 1 = 2(n + 3)( n + 4) : n X

(,1)k nk 1 = : 3 2 k=0 k + 9k + 26k + 24 2(n + 3)(n + 4) , 

That completes the Corner for this issue. Send me Olympiad contests and your nice solutions and generalizations for use in the Corner.

Congratulations! The following contributors to CRUX with MAYHEM distinguishedthemselves at the 38th IMO held in Argentina: Name Florian Herzig Sabin Cautis Adrian Chan Byung Kyu Chun Mansur Boase Toby Gee Carl Bosley

Country Austria Canada Canada Canada United Kingdom United Kingdom USA

Award Gold Medal Bronze Medal Silver Medal Silver Medal Silver Medal Bronze Medal Gold Medal

CRUX with MAYHEM o ers hearty congratulations to these students. We trust that their participation in solving problems in CRUX with MAYHEM was a contributory factor in their successes.

272

BOOK REVIEWS Edited by ANDY LIU The Puzzle Arcade, by Jerry Slocum, published in 1996 by Klutz, 2121 Staunton Court, Palo Alto, CA 94306, USA., ISBN# 0-888075-851-4, coil bound, 48 pages, US$20 plus handling. Quantum Quandaries, edited by Timothy Weber, published in 1996 by the National Science Teachers' Association, 1840 Wilson Boulevard, Arlington, VA 22201-3000, USA., ISBN# 0-87355-136-2, softcover, 208 pages, US$7.95 plus handling. Reviewed by Andy Liu, University of Alberta. These two books are the most delightful additions to the literature of popular mathematics recently, and should appeal to everyone. Jerry Slocum, a retired executive in the aircraft industry, has a collection of mechanical puzzles numbering over 22000. The Puzzle Arcade features a very small portion of this amazing trove. The book is profusely illustrated with striking colours, and comes with several pockets containing either complete puzzles or equipment needed for others. Apart from the standard tangram-like puzzles, matchstick puzzles, topological entanglement puzzles, optical illusions, mazes and word puzzles, there are many unlikely to be familiar to the readers. The readers are asked to put together two identical polyhedral pieces to form a familiar object. This is perhaps the most annoying of the puzzles with only two pieces. A three-piece puzzle features Sam Loyd's tantalizing ponies. There is also a puzzle with four or ve pieces. With four, the readers can form the back of a playing card, but it takes all ve to come up with the King of Hearts. The list goes on and on, and the readers simply have to see the book to relish it. Hints and answers are provided. The publisher has done a fantastic job packing so much into a relatively compact o ering. Purchased separately, the puzzles will cost many times more than this bargain of a book! Timothy Weber of the NSTA is the managing editor of the magazine Quantum. It is the sister publication of the Russian magazine Kvant, a Russian word which means quantum. About two-thirds of the material in Quantum are translated from Kvant, the rest being original contributions in English. About two-thirds of the material are mathematical, the rest being in physics. It is published six times a year by Springer-Verlag New York Inc., P.O.Box 2485, Secaucas, NJ 07096-2485, USA.. Student subscription at US$18 per volume is a steal. One of the most popular columns in Quantum is the Brainteasers. There are ve in each issue, usually four in mathematics and one in physics. Many are presented in whimsical style, and there are always the most delightful

273 cartoon illustrations. Quantum Quandaries, a pocket-sized book, contains the rst 100 Brainteasers and their solutions. Each Brainteaser is on one page, with its solution overleaf for easy reference. Additional cartoon illustrations are provided by Sergey Ivanov. The only regret is that the book is in black-and-white, whereas the original presentation in the magazine was in full colour. We conclude this review with a sample Brainteaser. Solve the alphametic equation USA+USSR=PEACE

where letters stand for di erent digits if and only if they are di erent.

What are the parametric equations, and the range of the parameter, of the following curve? 6

-

274

A Fermat-Fibonacci Collaboration K.R.S. Sastry

Fermat begins Fermat the mischievous proposed the impossible: Find three squares in arithmetic progression (AP ), the common di erence (cd) being a square, [Dickson [1] pp. 435]. Throughout this paper we deal with integer squares but, we nd that, what is impossible for squares is possible for triangles: 6, 21, 36 are triangular numbers in AP with cd of 15 also being triangular! Let us see how we can generate triples of triangular numbers in AP whose cd is also triangular. Curiously, the supplier is Fibonacci! 1 We recall that the nth triangular number is given by Tn = n(n + 1). 2 We also note that 8Tn + 1 = (2n + 1)2 is a perfect square, an odd square to be exact. Simply retrace the preceding steps to see that the converse, every odd square leads to a triangular number, holds. Furthermore, multiplying each term of an AP by 8 and then adding 1 to them results in a new AP , but the new cd is just eight times the old. Take an example: AP : 2, 7, 12 results in the AP : 17, 57, 97 with cd's of 5 and 40: (1) The upshot of all this is that the triangular numbers in AP can be made to depend on the square numbers in AP .

An expression for squares in AP

Suppose the squares x2 , y 2, z 2 are in AP . Then x2 + z 2 = 2y 2. We may assume x, y , z are relatively prime in pairs and that x < y < z holds. Hence x and z are both odd. Put x = p , q , z = p + q where p and q are relatively prime integers having opposite parity. Then we have the famous Pythagorean relation p2 + q 2 = y 2 whose primitive solutions are well known: p = m2 , n2, q = 2mn, y = m2 + n2. Here m and n are relatively prime integers with opposite parity. This yields x = jm2 , n2 , 2mnj; y = m2 + n2; z = m2 , n2 + 2mn: (2) The cd: y 2 , x2 , equals 4mn(m2 , n2). This cannot be a square in view of Fermat's proof of the fact mn(m2 , n2 ), the area of the (Pythagorean) right-angled triangle with sides m2 , n2, 2mn, m2 + n2 cannot be a square [Dickson [1] pp. 615-616].

275

Triangular numbers in AP whose cd is triangular Suppose the triangular numbers

Ta = 12 a(a + 1); Tb = 12 b(b + 1); Tc = 12 c(c + 1);

are in AP . Then

8Ta + 1 = (2a + 1)2 = A2; 8Tb + 1 = (2b + 1)2 = B 2; 8Tc + 1 = (2c + 1)2 = C 2 ;

are three odd squares in AP . Hence from the expressions in (2)

A = jm2 , n2 , 2mnj; B = m2 + n2; C = m2 , n2 + 2mn: (3) The cd of the AP squares A2 , B 2, C 2 is 4mn(m2 , n2 ). Therefore, from the remark in (1), the triangular numbers Ta , Tb , Tc in AP have cd: 1 (4mn(m2 , n2)) = 1 mn(m2 , n2 ): 8 2

The very form of this cd suggests that we choose m2 , n2 = mn  1. In that case the cd:

1 mn(m2 , n2) = 1 mn(mn  1) 2 2

would be triangular. Thus we seek, in natural numbers, the solution of the equation

m2 , n2 = mn  1:

Fibonacci enters Before solving the equation m2 , n2 = mn  1, let us recall the Fibonacci sequence

1; 1; 2; 3; 5; 8; 13; 21; 34;    given recursively by F1 = F2 = 1, Fk = Fk,1 + Fk,2 , k > 2. It is also given by Binet's formula

p

p

k , 5)k ; k = 1; 2; 3;    : Fk = (1 + 5)2k+p(1 5

(4)

276 Also, the Lucas sequence

1; 3; 4; 7; 11; 18; 29;    is likewise de ned: L1 = 1, L2 = 3, Lk = Lk,1 + Lk,2 , k > 2. We leave it to the reader to deduce the following relations:

p5 !k 1 , p5 !k 1 + Lk = Fk,1 + Fk+1 = + : (5) 2 2 Now consider the solution of the equation m2 , n2 = mn  1, put in the form (2m , n)2 , 5n2 = 4. The rst few trial solutions of the equation (2m , n)2 , 5n2 = ,4 are 2m , n 1 4 11    n m

1 2 1 3

5  8 

It is already apparent that 2m , n = L2k,1 , m = F2k , n = F2k,1 , k = 1, 2, 3,    . Direct calculations using (4) and (5) show that

L22k,1 , 5F22k,1 = ,4; k = 1; 2; 3;    :

We leave the veri cation to the reader. Likewise the rst few trial solutions of the equation are

(2m , n)2 , 5n2 = 4

2m , n 3 7 18   

n 1 3 8  m 2 5 13    This time we observe that 2m , n = L2k , m = F2k+1 , n = F2k and again we leave the veri cation that to the reader.

L22k , 5F22k = 4; k = 1; 2;   

A numerical example Choose the solution m = 8, n = 5. Then from (3), we have

2a + 1 = A = jm2 , n2 , 2mnj = 41; 2b + 1 = B = m2 + n2 = 89; 2c + 1 = C = m2 , n2 + 2mn = 119:

277 This yields a = 20, b = 44, c = 59, Ta = 210, Tb = 990, Tc = 1770 in AP 1 whose cd, 780 = (39)(40), is triangular too. The reader may have spotted 2 a like parity solution m = 5, n = 3. This makes a, b, c non-integral, since A, B, C are even. This happens when m = F6i,1, n = F6i,2, i = 1; 2;    , a fact easily shown by induction.

Fermat concludes The next question to settle would be: Are there three pentagonal numbers in AP whose cd is pentagonal? The cases that have been settled are n = 3 and n = 4, instances of the general problem: For what values of n are there three n-gonal numbers in AP whose cd is n-gonal too? It may be recalled that the rth n-gonal number is 2 p(n; r) = (n , 2) r2 , (n , 4) r2 ; n  3; r = 1; 2; 3;    :

For n > 4, is this Fermat's newest last problem?

Reference [1] L.E. Dickson, History of the Theory of Numbers, Vol. II, Chelsea, NY, (1971), pp. 1-39, 435, 615-616. 2943 Yelepet Dodballapur 561 203 Karnataka, India

278

THE SKOLIAD CORNER No. 23 R.E. Woodrow

This number we give the problems of the 1995 Prince Edward Island Mathematics Competition. Thanks go to Gordon MacDonald, University of Prince Edward Island, organizer of the contest, for supplying us with the problems, and the solutions which we will give next issue.

1995 P.E.I. Mathematics Competition

1. Find the area of the shaded region inside the circle in the following

gure.

61

,1

1-

P

?,1

2. \I will be n years old in the year n ", said Bob in the year 1995. How old is Bob? 3. Draw the set of points (x; y) in the plane which satisfy the equation jxj + jx , yj = 4. 4. An autobiographical number is a natural number with ten digits 2

or less in which the rst digit of the number (reading from left to right) tells you how many zeros are in the number, the second digit tells you how many 1's, the third digit tells you how many 2's, and so on. For example, 6; 210; 001; 000 is an autobiographical number. Find the smallest autobiographical number and prove that it is the smallest.

279

5. A solid cube of radium is oating in deep space. Each edge of the cube is exactly 1 kilometre in length. An astronaut is protected from its radiation if she remains at least 1 kilometre from the nearest speck of radium. Including the interior of the cube, what is the volume (in cubic kilometres) of space that is forbidden to the astronaut? (You may assume that the volume of a sphere of radius r is 43 r3 and the volume of a right circular cylinder of radius r and height h is r2 h.) 6. Which is greater, 999! or 500999? (Where 999! denotes 999 factorial, the product of all the natural numbers from 1 to 999 inclusive.) Explain your reasoning.

Where do those errors come from? Several readers wrote about the error in the solution of Problem 2(a) of the Manitoba Contest 1995 [1995: 219]. 2. (a) Find two numbers which di er by 3 and whose squares di er by 63. Corrections by Jamie Batuwantudawa, student, Fort Richmond Collegiate, Winnipeg, Manitoba; by Paul{Olivier Dehaye, Brussels, Belgium; and by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA. The solution of the equations 6x + 9 = 63 or 6x + 9 = ,63 is obvious: x = 9 or x = ,12, and not x = 7 or x = ,8 as given.

Next we give solutions for the problems of the Twelfth W.J. Blundon Contest, written February 22, 1995, [1997: 218{219].

THE TWELFTH W.J. BLUNDON CONTEST February 22, 1995

1. (a) From a group of boys and girls, 15 girls leave. There are then left two boys for each girl. After this, 45 boys leave. There are then 5 girls for each boy. How many boys and how many girls were in the original group? Solution. Call the number of boys b and the number of girls g . When 15 girls leave b = 2(g , 15). After 45 boys leave 5(b , 45) = g , 15. So b = 2(5(b , 45)), b = 10b , 450, 9b = 450, b = 50; g , 15 = 50=2 = 25 and g = 40. Thus the total number of boys and girls was originally 90.

280 (b) A certain number of students can be accommodated in a hostel. If two students share each room then two students will be left without a room. If three students share each room then two rooms will be left over. How many rooms are there? Solution. Let s be the number of students and r the number of rooms. At two students per room 2r = s , 2. Consider three per room, then 3(r , 2) = s. Now 2r = 3(r , 2) , 2 so r = 8. There are 8 rooms (and 18 students).

2. How many pairs of positive integers (x; y) satisfy the equation

x y 19 + 95 = 1? Solution. Note that 95 = 19  5. For x = 1; 2; : : : ; 18 we have y = 1 , x = 19 , x = 95 , 5x ; 95 19 19 95 giving a solution. There are then 18 such pairs. 3. A book is to have 250 pages. How many times will the digit 2 be used in numbering the book? Solution. Two is used once in each decade in the one's place for a total of 25 such occurrences. It is used 10 times in each hundred in the ten's place for another 30 occurrences. And it occurs 51 times in the hundred's place, for a grand total of 25 + 30 + 51 = 106 times.

4. Without using a calculator p p p

p

(a) Show that 7 + 48 + 7 , 48 is a rational number. p p p p Solution. Let x = 7 + 48 + 7 , 48. Then q p p q p p x2 = 7 + 48 + 2 7 + 48 7 , 48 + 7 , 48 = 14 + 2p49 , 48 = 16

Thus x = 4, as x > 0. (b) Determine the largest prime factor of 9919. Solution. Considering the units digit we are led to try 7 as a factor, yielding 9919 = 7  1417. We are now led to try for a factor of 1417 ending in 3 or 9 less than 38, (since 382 = 1444), and 1417 = 13  109. Since 109 is prime, the answer is 109.

281

5. A circle is inscribed in a circular sector which is one sixth of a circle of radius 1, and is tangent to the three sides of the sector as shown. Calculate the radius of the inscribed circle. Solution. r

r

r

Call the radius of the inscribed circle r. The central angle is 16  360 =  60 , so each small triangle is a r30, 60, 90 right triangle. The hypotenuse, h, has length h = r sin 30 = . So r + h = 1 gives 3 r = 1, and r = 2 . 2

6. Determine the units digit of the sum

2

2626 + 3333 + 4545:

Solution. The units digit is the same as for

626 + 333 + 545 : Now 6n  6 mod 10 for n  1 and 8 3 n  1 mod 4 > > < 2 mod 4 ; so 333  3 mod 10: 3n  > 97 nn   3 mod 4 > : 1 n  4 mod 4 Finally 5n  5 mod 10 for n  1. Hence 2626 + 3333 + 4545  6 + 3 + 5  4 mod 10 and the last digit is 4. 7. Find all solutions (x; y) to the system of equations x + y + xy = 19

x(x + y) = 60: y

3

282 Solution. Now

x = 60 so y x+y

x + y + x 60 + y = 19;

(x + y )2 , 19(x + y ) + 60 = 0

and x + y = 4 or x + y = 15. If x + y = 4, xy = 60, we get x = 60y , and 61y = 4, so y = 614 and x = 240 . 61 If x + y = 15, xy = 5 giving y = 3, x = 12. ,  4 The solutions for (x; y ) are 240 61 ; 61 and (12; 3). 8. Find the number of di erent divisors of 10800. Solution. Now 10800 = 9  12  100 = 24  33  52 . The factors must be of the form 2a 3b 5c where 0  a  4, 0  b  3 and 0  c  2, so the number of factors is 5  4  3 = 60. 9. Show that n4 , n2 is divisible by 12 for any positive integer n > 1. Solution. Now n4 , n2 = (n , 1)n2(n + 1). In any three consecutive numbers, at least one is divisible by 3. Also if n is even, 4 divides n2 , and if n is odd, both n 2, 1 and n + 1 are even so 4 divides n2 , 1. In either case 4 4 divides n , n and we are done. 10. Two clocks now indicate the correct time. One gains a second every hour, and the other gains 3 seconds every 2 hours. In how many days will both clocks again indicate the correct time? Solution. There is a bit of a problem about whether the clock is a 12 hour or 24 hour clock. Let us assume a 12 hour clock in each case. The rst clock will gain 12 hours in 12  60  60 hours, or 1800 days, which is the rst time it will again,tell the correct time. The second clock will gain 12 hours in 12  60  60  32 hours, or 1200 days. At the end of 3600 days they will both tell the correct time. That completes the Corner for this number. I need suitable exam materials, and would welcome your suggestions for the evolution of this feature of Crux with Mayhem.

283

MATHEMATICAL MAYHEM Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. All material intended for inclusion in this section should be sent to the Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University, PO Box 208283 Yale Station, New Have, CT 06520{8283 USA. The electronic address is still [email protected] The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto). The rest of the sta consists of Richard Hoshino (University of Waterloo), Wai Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices Playful Palindromes What are palindromes? They are any sort of objects which are the same when read backwards or forwards. An example of a palindromic word is NOON. An example of a palindromic sentence is ABLE WAS I ERE I SAW ELBA, accredited to Napoleon. Here we concentrate on palindromic numbers such as 12321, and henceforth the word palindrome will refer to palindromic numbers. Quickly work out the following problems. 1. How many two digit palindromes are there? 2. How many three digit palindromes are there? 3. How many palindromes are there from 0 to 1998? 4. How many n-digit palindromes are there? (Consider the cases when n is even and odd. Palindromic Sums Ask a friend to pick a positive integer. Tell this friend to reverse the digits of the integer and add it to the original. If the sum is not a palindrome then take the sum's reverse and add it to itself. Continue doing this until the sum is a palindrome.

284 Examples

123 + 321

139 + 931

444

1070 + 0701 1771 It is clear that if all the digits are less than 5 then the process terminates after the rst step. The second example shows that a number with a digit greater than 5 can also have a palindromic sum.

Now try these examples: 1. 1990 2. 89 3. 196 Answers 1. 68486 2. 8813200023188 3. No one has yet found the answer! Unsolved Problem Is it true that every integer has a palindromic sum (that is, the process will eventually terminate)? This is an example of where better algorithms and faster computers will not help. Even if a super-fast computer with terabytes of memory were to nd the solution of 196, this leaves the problem open still as we do not know that ALL integers have a palindromic sum. Perhaps for the eager reader, an attempt should be made to classify as much as possible those numbers which DO have palindromic sums. We have already made one such class of numbers, namely those whose digits are all less than 5. Are there any other classes of numbers that guarantee to give palindromic sums?

Best of the Web We present here some web sites which may be of interest to the mathematical web surfer: http://www.math.princeton.edu/~kkedlaya/competitions.html

Kiran Kedlaya's Math Competition Archive, with copies of old USAMO, IMO, and Putnam papers, and many links to other related sites. http://www.math.toronto.edu/mathnet/

285 U of T Mathnet. A collection of material for high school students and teachers. http://www.research.att.com/~njas/sequences/

Sloane's On-line Sequence Dictionary. Thanks to Ben Chia, ex Cedarbrae C.I.

Decimal Expansion of Fractions Cyrus Hsia While programming on the computer, a couple of my friends, Peter Plachta and Jacob Mouka noticed some interesting patterns. One such pattern came about as follows: Stumped on a hard programming problem, they tried examining their output. Through the complex mess of numbers, they noticed that the decimal expansion of fractions, of the form p1 , where p is a prime number (other than 2 or 5) always had a repeating decimal pattern. Furthermore, this repeating pattern appeared immediately after the decimal point. For example, 13 = 0:3. With many such examples generated on the computer, they conjectured that this must be the case for all primes p, p 6= 2; 5. In fact, their conjecture is correct. We will show this result and some interesting generalizations. But rst let us give some de nitions so that there will be no ambiguity in their meaning. The decimal expansion of a fraction is of the form

0:a1a2 : : : an b1 : : : bm : : : b1 : : : bm : : : ;

where the sequence b1 : : : bm repeats inde nitely. We write a bar over it to indicate that it repeats, that is,

0:a1 a2 : : : an b1 : : : bm:

De nition 1.1. The repeating part in a decimal expansion is the number (that is, the sequence of digits b1 : : : bm ) under the bar. The astute reader would notice that the bar could have been placed one decimal place over, as 0:a1 a2 : : : an b1b2 : : : bm b1, or that a bar could have been placed over twice the above sequence and still give the same value. So to make this de nition unambiguous, we will always mean the earliest and smallest such pattern after the decimal point. Finally, we will say that fractions, like 21 , that terminate do not have a repeating part. De nition 1.2. With the above de nition of the repeating part in mind, the leading part is the number (that is, the sequence of digits a1 : : : an ) immediately following the decimal point before the repeating part. Note that fractions may not have a leading part, as in 13 .

286 Finally, let us call a number good if it is relatively prime to 10, that is it is not divisible by 2 or 5. With these de nitions, let us show that the conjecture above is true by rst proving some simple results that we need. We will assume from now on that we are looking at the decimal expansions of 1p , where p is a good prime. Lemma 1.3. The decimal expansion of 1p , where p is a good prime, must have a repeating part, that is, p1 = 0:a1 a2 : : : an b1 : : : bm , m  1. Proof. Left to the reader as an exercise. Lemma 1.4. If 0:a1 a2 : : : an b1 : : : bm is the decimal expansion of a fraction, then an 6= bm . Proof. Left to the reader as an exercise. Lemma 1.5. For a decimal expansion,

0:a1 : : : anb1 : : : bm n,1 n,2 m,1 + 10m,2 b2 +    + bm = 10 a1 + 1010n a2 +    + an + 10 b1 10 ; n (10m , 1)

which we write

a1 : : : an + b1 : : : bm 10n 10n(10m , 1)

in short form. Proof. Left to the reader as an exercise. Now with these lemmas we can prove the conjecture. Conjecture 1.6. (Peter Plachta and Jacob Mouka) The fraction 1p in decimal notation consists only of a repeating part (that is, it has no leading part) if p is a good prime. Proof. Assume on the contrary that p1 = 0:a1 a2 : : : an b1 : : : bm , where n  1, so that there is at least one digit in the leading part. Case I: m = 0, that is, the decimal expansion terminates. Then

1 = 0:a : : : a = a1 : : : an ) 10n = p(a : : : a ) ) p j 10n ; 1 n 1 n p 10n

but p is good, so this is not possible. Case II: m  1. Then

1 = a1 : : : an + b1 : : : bm p 10n 10n (10m , 1) (by Lemma 1.5). Multiplying both sides by 10n (10m , 1)p, we get 10n (10m , 1) = p[(10m , 1)(a1 : : : an ) + b1 : : : bm] = p[10m a1 : : : an + b1 : : : bm , a1 : : : an ]:

287 Since n  1, 10 j 10n(10m , 1), so 10 divides the right hand side of the equation. In particular, since m  1,

10 j (b1 : : : bm , a1 : : : an ) ) 10 j (bm , an ): Now 0  an ; bm  9 ) an = bm , which by Lemma 1.4 is impossible.

Why does the fraction have to have a prime as its denominator? The prime is what makes the particular types of fractions of Conjecture 1.6 have no leading part. Then a natural question to ask is whether the converse holds, that is, given a decimal expansion with only a repeating part, is the denominator in its fractional representation a prime. To see that the converse does not hold, look at the decimal expansion of 19 , which is 0:1. It has only a repeating part but the denominator is not prime. It is not the fact that the denominator is a prime that makes Conjecture 1.6 true as the following theorem will show. Theorem 1.7. For all positive integers N > 1, N1 in decimal notation has a repeating part but no leading part if and only if N is good. First, we need a result that most of the readers will already be familiar with. In fact, it has popped up a number of times in olympiad and preolympiad type problems. Consider the numbers Mi consisting of i digits made up entirely of 1's, for example, M1 = 1, M2 = 11, M3 = 111. Then we have: Lemma 1.8. For any good positive integer N , N divides some Mi . Proof. Consider the N + 1 numbers M1 , M2 , : : : , MN +1 modulo N . Then by the Pigeonhole Principle, at least two of the Mi 's are congruent, say Mj  Mk (mod N ), where j > k. Then we have

Mj  Mk (mod N ) ) 11 : : : 1}  11 : : : 1} (mod N ) | {z | {z j

k

) 11 : : : 1} 00 : : : 0}  0 (mod N ) | {z | {z j ,k

k

) 11 : : : 1}  0 (mod N ): | {z j ,k

This can be done since neither 2 nor 5 divides N . This means that Mj ,k is divisible by N . Proof of Theorem 1.7. (Only if part) If N1 has a repeating part but no leading part, then we have

1 = 0:b : : : b = 10m,1 b1 +    + bm 1 m N 10m , 1

by Lemma 1.5. Cross-multiplying, we get 10m ,1 = N (10m,1 b1 +   +bm ). Thus N is good.

288 (If part) If N is good, then by Lemma 1.8 there is an M = Mi divisible by N . Let M = kN , where k is an integer, k  1. We have then

1 = k = k = 9k = 9k : N M 11 : : : 1} 99 : : : 9} 10i , 1 | {z | {z

Hence,

i

i

1 < 1 =) k < M =) 9k < 9M = 10i , 1 < 10i : N

This means that 9k is at most an i-digit number; that is, 9k can be written as b1 : : : bi, where some of the bi 's can be 0. Thus

1 = b1 : : : bi = 0:b : : : b 1 i N 10i , 1

as required. Now it is time for the readers to roll up their sleeves and tackle some problems. Exercises 1.1 Prove Lemmas 1.3, 1.4, and 1.5. 1.2 (a) Show that any fraction of the form ab in lowest terms, where a and b are positive integers, a < b, also has no leading term if and only if b is good. (b) Show that (a) holds even when a  b and with the same conditions. Thus ab = a1 a2 : : : an :b1 : : : bm if and only if b is good. 1.3 (Related exercise) Find the smallest positive integer n such that

n3 + 9n2 + 9n + 7 1996

in decimal notation terminates (that is, it has no repeating part, just a nite leading part). 1.4 (Related exercise) Write a computer program to nd the repeating part in a fraction of the form p1 , where p is a prime. To nish o , fractions with denominators that have factors of 2 and 5 still need to be considered. In fact, it turns out that there is a nice relationship between the number of factors of 2 and 5 in the denominator and the number of terms in the leading part of its decimal representation. Lemma 2.1. If N is of the form 2k M , where M is good, then N1 has k terms in its leading part, and it has a repeating part if M > 1.

289 Proof. If M = 1 then it is obvious that 21 = 105 has k terms in the leading part and no repeating part. If M > 1, then we have k

k

k

!

1 = 1 = 5k = 1 5k : N 2kM 10kM 10k M

5 = a1 : : : an :b1 : : : bm. Dividing through by By Exercise 1.2, we have M 1 10k and noting N is less than 1, we must have n < k. Now once we show that an 6= bm, then we are nished, as then k

!

1 = 1 5k = 0: 0 : : : 0a : : : a b : : : b : 1 n} 1 m | {z N 10k M k terms

If an = bm then

!

5k,1 = 1 5k = a : : : a :a b : : : b ; 1 n,1 n 1 m,1 2M 10 M

contradicting Exercise 1.2 that a fraction has a repeating part if and only if the denominator is good. Lemma 2.2. Similarly, if N is of the form 5k M , where M is good, then N1 has k terms in its leading part, and it has a repeating part if M > 1. Proof. Similar (in fact identical) to the proof of Lemma 2.1. Now for the grand nale... Theorem 2.3. If N is of the form 2k 5l M , where again M is good, then the fraction N1 has max(k; l) terms in the leading part, and if M > 1 then it also has a repeating part. Proof. Once again, this just takes from material shown before. If k or l equals 0, then the result becomes Lemma 2.2 or 2.1. If k = l, then the result is immediate, and in fact the leading part has k = l 0's. If k > l > 1, then we have 2 51 M = 10 2 1, M . By Lemma 2.1, we have k , l terms in the leading part of 2 ,11 M . Dividing by 10l just shifts the decimal point l places to the left giving (k , l) + l = k terms in the leading part (and similarly l terms for l > k > 1). More Exercises 2.1 (Exploration) In our long analysis, we have shown (not on purpose), that the length of the leading part can be determined. However, the length of the repeating part was never discussed. Determine any relationships between the numbers in the fraction and the length of the repeating part in its decimal representation. 2.2 (a) If we had 6 ngers (3 on each hand?), then perhaps our system of counting would be in base 6. In general, if the base was pq , where p k

l

k

l

k

l

290 and q are distinct primes, then show an analogous result to the above theorems. (b) Do the same for any base B . 2.3 (More down to earth problem) Of the numbers 1995, 1996, : : : , 1999, which number has the greatest number of terms in the leading part of its reciprocal? With this, we hope you will be a leading part in this year! Acknowledgements Peter Plachta and Jacob Mouka, 3rd year students, University of Toronto at Scarborough, for the conjecture.

An Ambivalent Sum Naoki Sato A conditionally P convergent series an is a series which converges, but such that the series jan j diverges. One of the most famous examples of such a series, which appears in textbooks everywhere, is the following: P

1 , 12 + 31 , 14 + 15 , 16 +    = ln2:

A basic, but unusual property of conditionally convergent series is that we may rearrange the terms, so that the sum comes out to be any value we wish (see Calculus, 3rd Ed., M. Spivak, pp. 476-8), so we must be careful when manipulating the terms in determining the sum. For example, a question of the 1954 Putnam asks to verify that

1 , 1 +    = 3 ln2: 1 + 13 , 21 + 15 + 71 , 14 + 19 + 11 6 2

Same terms, di erent sum. These results suggest a generalization, and the form of the alternating harmonic series makes for an elegant analysis. For positive integers r and s, let

S(r; s) = 1 + 13 +    + 2r 1, 1 , 12 , 14 ,    , 21s {z } | {z } | s terms

r terms 1 1 + 2r + 1 + 2r + 3 +    + 4r 1, 1 , 2s 1+ 2 , 2s 1+ 4 ,    , 41s | {z } | {z } r terms s terms + :

291 A rst step in summing S (r;s) will be looking at its partial sums, so let Sn (r; s) be the sum of the rst n(r + s) terms of S (r; s). Assume for the moment that r > s. Then,    1 + 13 +    + 2r 1, 1 , 21 + 14 +    + 21s   1 1 1 + 2r + 1 + 2r + 3 +    + 4r , 1   1 1 1 , 2s + 2 + 2s + 4 +    + 4s +      1 1 1 + 2(n , 1)r + 1 + 2(n , 1)r + 3 +    + 2nr , 1  1  , 2(n , 11)s + 2 + 2(n , 11)s + 4 +    + 2ns   = 1 + 13 +    + 2r 1, 1 , 12 , 14 ,    , 21r + 12 + 14 +    + 21r , 21 , 14 ,    , 21s  + 2r 1+ 1 + 2r 1+ 3 +    + 4r 1, 1  , 2r 1+ 2 , 2r 1+ 4 ,    , 41r + 2r 1+ 2 + 2r 1+ 4 +    + 41r , 2s 1+ 2 , 2s 1+ 4 ,    , 41s +     + 2(n , 11)r + 1 + 2(n , 11)r + 3 +    + 2nr1, 1  1 1 1 , 2(n , 1)r + 2 , 2(n , 1)r + 4 ,    , 2nr 1 + 2(n , 11)r + 2 + 2(n , 11)r + 4 +    + 2nr 1 , 2(n , 11)s + 2 , 2(n , 11)s + 4 ,    , 2ns 1 = A2nr + 2ns1+ 2 + 2ns1+ 4 + 2ns1+ 6 +    + 2nr  1 ; = A2nr + 12 ns 1+ 1 + ns 1+ 2 + ns 1+ 3 +    + nr

Sn(r; s) =

where



An = 1 , 12 + 31 , 14 +    + (,1)n+1 n1 :

292 We may estimate a more general sum by integrals. Since y = x1 is decreasing on (0; 1), for 0 < d < a, a+md 1 x dx < a

Z

so that

Z a+(m,1)d d + d +  + d 1 dx < a a+d a + (m , 1)d x a,d

1 ln  a + md  < 1 + 1 +    + 1 d a a a + d a+ (m , 1)d < 1 ln a + (m , 1)d

d

a,d

(if you are not sure why these inequalities are true, draw a graph). So, plugging in a = ns + 1, d = 1, m = n(r , s), we obtain

   nr + 1 1 1 1 ln ns + 1 < ns + 1 + ns + 2 +    + nr < ln rs  1 + 1 +    + 1  = ln  r  : =) nlim !1 ns + 1 ns + 2 nr s 

Note that we could have also used the well-known result

 1 + 1 +    + 1 , ln n = ; lim 1 + n!1 2 3 n

where is Euler's constant. Finally, we have that

1 ln  r  = 1 ln  4r  : S(r; s) = nlim S ( r; s ) = ln2 + !1 n 2 s 2 s

The case r < s is argued similarly (the case r = s is obvious), and we obtain the same value. We make some observations: (1) The value S (r;s) depends only on the ratio r=s, as can be predicted. (2) The case r = 1, s = 4, produces the intriguing-looking formula

1 , 1 , 1 , 1 +    = 0: 1 , 12 , 14 , 16 , 81 + 13 , 10 12 14 16

(3) The set fS (r;s)g = f 12 ln( 4sr ) : r;s 2 Z+g is dense in R.

293 (4) Going back to a result mentioned above, to rearrange the terms of a conditionally convergent series so that it sums to a given value, say S , we use the following algorithm: Assuming S  0, we successively add the rst positive terms of the series until we go over S ; then we add the rst negative terms of the series until we go under S , and so on (if S < 0, we merely start with the negative terms). In fact, this is the idea behind the proof. The nature of the conditionally convergent series guarantees this algorithm produces a series with sum S . It has been observed that if we try this with the alternating harmonic series, the number of positive terms and negative terms taken each time becomes constant. We can see this is because of (3); that is, any real number can be approximated by a sum S (r;s).

J.I.R. McKnight Problems Contest 1978 The J.I.R. McKnight Problems Contest is a problem solving contest in Scarborough, Ontario. It began in 1975 in honour of one of Scarborough's nest Mathematics teachers and continues today as a scholarship paper for senior students. This paper, although tailored to students taking OAC (Ontario Academic Credit) Mathematics courses, contains many problems that are accessible to all high school students. With some ingenuity, most problems reveal some very beautiful results and we hope our readers will enjoy this as well. 1. Verify that 13! = 1122962 , 798962. 2. Show that, for any real numbers x, y , and z , 2 2 (x + y)z  (x +2 y ) + z 2:

3. Sum to k terms the series whose nth term is

n4 + 2n3 + n2 , 1 : n2 + n

4. Show that in order that the quadratic function

3x2 + 2pxy + 2y2 + 2ax , 4y + 1

may be resolved into factors linear in x and y , p must be a root of the equation p2 + 4ap + 2a2 + 6 = 0. 5. Find the sum of all 3{digit odd numbers that can be made using the digits 1, 3, 5, 7, 8, 9, where no repeated digits are used in any number.

294 6. Find a cubic function f (x) with the following properties: (a) It is a curve passing through (1; 1). (b) The tangent to the curve at (1; 1) has slope 1. (c) The sum of the roots of its corresponding equation is 51 . (d) The sum of the squares of the roots of its corresponding equation . is 121 25

1984 Swedish Mathematics Olympiad Qualifying Round 1. Solve the following system of equations (in 5 unknowns):

ab bc cd de ea

= = = = =

1; 2; 3; 4; 6:

2. Show that if n is an odd natural number, then

n12 , n8 , n4 + 1

is divisible by 29. 3. Find all positive numbers x such that

x8,3x > x7 : 4. ABC is an isosceles triangle with AB = AC . Choose a point D on the side AB , and a point E on AC produced through C , such that AD + AE = AB + AC . Show that DE > BC . 5. The numbers 1, 2, : : : , 9, are placed in a 3  3 grid, so that no row, no

column and neither of the two major diagonals contains a sequence of numbers ordered by size (either increasing or decreasing). Show that the number in the central square must be odd. 6. A woman, who is not yet 100 years old, and one of her grandchildren have the same birthday. For six years in a row, the woman's age was a multiple of her grandchild's age. How old was the woman on the sixth of these birthdays?

295

Final Round 1. Let A and B be two arbitrary points inside a circle C . Show that there always exists a circle through A and B which lies completely inside C . 2. The squares in a 3  7 grid are coloured either blue or yellow. Consider all rectangles of m  n squares in this grid, where 2  m  3, 2  n  7. Show that at least one of these rectangles has all four of its corner squares the same colour. 3. Show that if a and b are positive numbers, then

a + 1 b+1   a b : b+1 b 4. Find positive integers p and q such that all the roots of the polynomial (x2 , px + q )(x2 , qx + p) 

are positive integers. 5. Solve the system of equations

a3 , b3 , c3 = 3abc; a2 = 2(a + b + c);

where a, b, and c are natural numbers. 6. The numbers a1 , a2 , : : : , a14 are positive integers. It is known that 14 X

i=1

ai = 6558:

Show that the numbers a1 , a2 , : : : , a14 consist of the numbers 1, 2, : : : , 7 each taken twice.

296

Swedish Mathematics Olympiad Solutions 1983 Final Round 1. The positive integers are added in groups as follows: 1, 2+3, 4+5+6, 7 + 8 + 9 + 10, and so on. What is the sum of the nth group? Solution by Bob Prielipp, University of Wisconsin-Oshkosh, Oshkosh, WI, USA. Let Tk = k(k +1)=2 be the kth triangular number. Then the n numbers in the nth group are Tn,1 + 1, Tn,1 + 2, : : : , Tn,1 + n = Tn . Using the formula

S = n(a2+ l) ;

where S is the sum of a nite arithmetic progression, a is the rst term, l is the nal term, and n is the number of terms in the progression, the desired sum is

n[(Tn,1 + 1 + Tn)] = n 2



n,1)n + 1 + n(n+1) 

(

2

2

2

3 = n 2+ n :

Mayhem Problems The Mayhem Problems editors are: Richard Hoshino Mayhem High School Problems Editor, Cyrus Hsia Mayhem Advanced Problems Editor, Ravi Vakil Mayhem Challenge Board Problems Editor. Note that all correspondence should be sent to the appropriate editor | see the relevant section. In this issue, you will nd only problems | the next issue will feature only solutions. We warmly welcome proposals for problems and solutions. With the new schedule of eight issues per year, we request that solutions from the previous issue be submitted by 1 October 1997, for publication in the issue 5 months ahead; that is, issue 2 of 1998. We also request that only students submit solutions (see editorial [1997: 30]), but we will consider particularly elegant or insightful solutions from others. Since this rule is only being implemented now, you will see solutions from many people in the next few months, as we clear out the old problems from Mayhem.

297

High School Problems Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario, Canada. L3S 3E8 H225. In Cruxmayhemland, stamps can only be bought in two denominations, p and q cents, both of which are at least 31 cents. It is known that if p and q are relatively prime, the largest value that cannot be created by these two stamps is pq , p , q . For example, when p = 5 and q = 3, one can ax any postage that is higher than 15 , 5 , 3, or 7 cents, but not 7 cents itself. The governor of Cruxmayhemland tells you that 1997 is the largest value that cannot be created by these stamps. Find all possible pairs of positive integers (p; q ) with p > q . H226. In right-angled triangle ABC , with BC as hypotenuse, AB = x and AC = y, where x and y are positive integers. Squares APQB, BRSC , and CTUA are drawn externally on sides AB, BC , and CA respectively. When QR, ST , and UP are joined, a hexagon is formed. Let K be the area of hexagon PQRSTU . (a) Prove that K cannot equal 1997. (HINT: Try to nd a general formula for K .) (b) Prove that there is only one solution (x; y ) with x > y so that K = 1998.

H227. The numbers 2, 4, 8, 16, : : : , 2n are written on a chalkboard. A student selects any two numbers a and b, erases them, and replaces them by their average, namely (a+b)=2. She performs this operation n,1 times until only one number is left. Let Sn and Tn denote the maximum and minimum possible value of this nal number, respectively. Determine a formula for Sn and Tn in terms of n. H228. Verify that the following three inequalities hold for positive reals x, y , and z : (i) x(x , y )(x , z ) + y (y , x)(y , z ) + z (z , x)(z , y )  0 (this is known as Schur's Inequality),

(ii) x4 + y 4 + z 4 + xyz (x + y + z )  2(x2 y 2 + y 2z 2 + z 2x2 ), (iii) 9xyz + 1  4(xy + yz + zx), where x + y + z = 1. (Can you derive an ingenious method that allows you to solve the problem without having to prove all three inequalities directly?)

298

Advanced Problems Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada. M1G 1C3 A201. Consider an in nite sequence of integers a1, a2, : : : , ak, : : : with the property that every m consecutive numbers sum to x and every n consecutive numbers sum to y . If x and y are relatively prime then show that all numbers are equal. A202. Let ABC be an equilateral triangle and , its incircle. If D and E are points on AB and AC , respectively, such that DE is tangent to ,, show that

AD + AE = 1: DB EC

(8th Iberoamerican Mathematical Olympiad, Mexico '93) A203. Let Sn = 1+ a + aa +    + aa , where the last term is a,1tower of n , 1 a's. Find all positive integers a and n such that Sn = na . p A204. Given a quadrilateral ABCD as shown, with AD = 3, AB + CD = 2AD, \A = 60 and \D = 120, nd the length of the :::a

Sn n

line segment from D to the midpoint of BC .

D

120

p3 A

C

60

B

Challenge Board Problems Editor: Ravi Vakil, Department of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544{1000 USA C73. Proposed by Matt Szczesny, 4th year, University of Toronto. 1 X The sequence fan g consists of positive reals, such that an diverges. 1 X

n=1

an diverges, where s is the nth partial sum, that is, Show that n n=1 sn sn = a1 + a2 +    + an.

299

PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later than 1 March 1998. They may also be sent by email to [email protected]. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication. Correction Problem 2235 [1997: 168] was incorrectly attributed to Walther Janous, Ursulinengymnasium, Innsbruck, Austria, when it was in fact proposed by D.J. Smeenk, Zaltbommel, the Netherlands. The editor apologizes for this error.

Solutions submitted by FAX There has been an increase in the number of solutions sent in by FAX, either to the Editor-in-chief's departmental FAX machine in St. John's, Newfoundland, or to the Canadian Mathematical Society's FAX machine in Ottawa, Ontario. While we understand the reasons for solvers wishing to use this method, we have found many problems with it. The major one is that hand-written material is frequently transmitted very badly, and at times is almost impossible to read clearly. We have therefore adopted the policy that we will no longer accept submissions sent by FAX. We will, however, continue to accept submissions sent by email or regular mail. We do encourage email. Thank you for your cooperation.

300

2251.

Proposed by Victor Oxman, University of Haifa, Haifa, Israel. In the plane, you are given a circle (but not its centre), and points A, K , B , D, C on it, so that arc AK = arc KB and arc BD = arc DC . Construct, using only an unmarked straightedge, the mid-point of arc AC . 2252. Proposed by K.R.S. Sastry, Dodballapur, India. Prove that the nine-point circle of a triangle trisects a median if and only if the side lengths of the triangle are proportional to its median lengths in some order. 2253. Proposed by Toshio Seimiya, Kawasaki, Japan. ABC is a triangle and Ib, Ic are the excentres of 4ABC relative to sides CA, AB respectively. Suppose that Ib A2 + Ib C 2 = BA2 + BC 2 and that Ic A2 + Ic B 2 = CA2 + CB2. Prove that 4ABC is equilateral. 2254. Proposed by Toshio Seimiya, Kawasaki, Japan. ABC is an isosceles triangle with AB = AC . Let D be the point on side AC such that CD = 2AD. Let P be the point on the segment BD such that \APC = 90 . Prove that \ABP = \PCB . 2255. Proposed by Toshio Seimiya, Kawasaki, Japan. Let P be an arbitrary interior point of an equilateral triangle ABC . Prove that j\PAB , \PAC j  j\PBC , \PCB j. 2256. Proposed by Russell Euler and Jawad Sadek, Department of Mathematics and Statistics, Northwest Missouri State University, Maryville, Missouri, USA. If 0 < y < x  1, prove that

ln(x) , ln(y ) > ln  1 .

x,y

y

2257. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. The diagonals AC and BD of a convex quadrilateral ABCD intersect at the point O. Let OK , OL, OM , ON , be the altitudes of triangles 4ABO, 4BCO, 4CDO, 4DAO, respectively. Prove that if OK = OM and OL = ON , then ABCD is a parallelogram.

301

2258. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. In a right-angled trianglepABC (with \C = 90 ), D lies on the segment BC so that BD = AC 3. E lies on the segment AC and satis es AE = CDp3. Find the angle between AD and BE. 2259. Proposed by Paul Yiu, Florida Atlantic University, Boca Raton, Florida, USA. Let X , Y , Z , be the projections of the incentre of 4ABC onto the sides BC , CA, AB respectively. Let X 0 , Y 0 , Z 0 , be the points on the incircle diametrically opposite to X , Y , Z , respectively. Show that the lines AX 0 , BY 0, CZ 0, are concurrent. 2260. Proposed by Vedula N. Murty, Andhra University, Visakhapatnam, India. Let n be a positive integer and x > 0. Prove that n+1

(1 + x)n+1  (n +n1) n

x:

2261.

Proposed by Angel Dorito, Geld, Ontario. Assuming that the limit exists, nd

2 + N1++:::::: lim 1 + ; 1+::: N !1 N + 2+ ::: !

where every fraction in this expression has the form

a + bc++:::::: b + ac++:::::: for some cyclic permutation a, b, c of 1, 2, N .

[Proposer's comment: this problem was suggested by Problem 4 of Round 21 of the International Mathematical Talent Search, Mathematics and Informatics Quarterly, Vol. 6, No. 2, p. 113.] 2262. Proposed by Juan-Bosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Consider two triangles 4ABC and 4A0 B 0 C 0 such that \A  90 and 0 \A  90 and whose sides satisfy a > b  c and a0 > b0  c0 . Denote the altitudes to sides a and a0 by ha and h0a . Prove that (a)

1 1 1 1 1 1 hah0a  bb0 + cc0 , (b) hah0a  bc0 + b0c .

302

SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 2145. [1996: 170] Proposed by Robert Geretschlager, Bundesrealgymnasium, Graz,nAustria. n , Y, Y  k , 1 Prove that ak + b  ak + bn,k for all a, b > 1. k=1

k=1

Editor's composite solutionbased on the nearly identical solutionssubmitted by almost all the solvers. Note that if n is odd, then both sides have the positive factor 

n + 1 a + b 2

,2 1

n

:

Hence the given inequality is equivalent to bn= Y2c , k=1



ka + bk,1 ,(n , k + 1)a + bn,k

bn= Y2c , k=1

ka + bn,k ,(n , k + 1)a + bk,1  :

Therefore, it suces to show that, for all k = 1, 2, : : : , b n2 c,

ka +,bk,1 ,(n , k, + 1)a + bn,k    ka + bn,k (n , k + 1)a + bk,1 :

,

(1)

After expanding and cancelling equal terms, (1) becomes

(n , 2k + 1)abk,1  (n , 2k + 1)abn,k: (2) ,  We have n , 2k + 1  n , 2 n2 + 1 = 1 > 0, and bn,2k+1  0, which

implies bk,1  bn,k . Thus, (2) follows immediately. Solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER   Y,  FerJANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV KONECN ris State University, Big Rapids, Michigan, USA; MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; KEE-WAI LAU, Hong Kong; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; L. RICE, University of Toronto Schools, Toronto, Ontario; KRISTIAN SABO, student,

303 Osijek, Croatia; JOEL SCHLOSBERG, student, Hunter College High School,  New York NY, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; and the proposer. Several solvers noted that the given conditions can be relaxed to a > 0 (or even a  1) and b  1. This is obvious from the proof above.

2146. [1996: 171] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle with AB > AC , and the bisector of \A meets BC at D. Let P be an interior point on the segment AD, and let Q and R be the points of intersection of BP and CP with sides AC and AB respectively. Prove that PB , PC > RB , QC > 0. Solution by the proposer. Let E be a point on AB such that AE = AC . Then we get 4AEP  4ACP , so that \AEP = \ACP and PE = PC . Since \BEP = 180 , \QCP > 180 , \ACB > \ABC > \EBP , we have PB > PE = PC . Because AE = AC < AB, we obtain \ABP < \AEP = \ACP . Since \BAP = \PAC , we have \APR = \ACP + \PAC > \ABP + \BAP = \APQ. Let F be a point on AB such that \APF = \APQ, then \APF < \APR, so F is between A and R, and 4APF  4APQ, so that AF = AQ, PF = PQ and \AFP = \AQP . Therefore, AR > AF = AQ. In addition, \ARP = \RPB + \RBP = \RPB + \ABQ and \PFR = \PQC = \BQC = \BAC + \ABQ: If \RPB = \BAC; \ARP = \PFR; so PR = PF = PQ; if \RPB < \BAC; \ARP < \PFR; so PR > PF = PQ; if \RPB > \BAC; \ARP > \PFR; so PR < PF = PQ: We consider 2 cases. Case I: \BAC  90: Because \RPB < \ARC < 90  \BAC , we have PR > PQ. Let G be a point on PR such that PG = PQ and H a point on PB such that PH = PC . Then 4 PGH  4PQC ; thus we get GH = QC . As \BRG > \RAC  90 , we get GB > RB , so BH > GB , GH > RB , GH . Hence PB , PC = BH > RB , QC:

304 Case II: \BAC < 90: Then there exists a point O on AD such that \BOC = 180 , \BAC . (i) If O coincides with P , we have \RPB = \BAC and so PR = PQ. We take a point J on PB such that PJ = PC . Then 4PRJ  4PQC , thus RJ = QC: Because BJ > RB , RJ = RB , QC , we have PB , PC > RB , QC . (ii) If P is between A and O, then \BPC < \BOC = 180 , \BAC , so \RPB = 180 , \BPC > \BAC and from above, PR < PQ. We take a point K on PQ such that PK = PR and a point M on PC produced beyond C such that PM = PB (> PC ). Let T be the intersection of KM and QC . As 4PKM  4PRB , we get KM = RB and \PKM = \PRB , so \QKM = \ARP = \RPB + \RBP = \RPB + \ABQ > \BAC + \ABQ = \BQC:

Therefore, \QKT > \KQT , thus KT < QT . Hence

RB , QC = KM , QC = (KT + TM ) , (QT + TC ) = (KT , QT ) + (TM , TC ) < TM , TC: As TM , TC < CM = PB , PC , we obtain RB , QC < PB , PC . (iii) If P is between O and D, then \BPC > \BOC = 180 , \BAC , so \RPB = 180 , \BPC < \BAC . Thus, from the above, PR > PQ. We take points X and Y on PR and PB , respectively, such that PX = PQ and PY = PC . Then 4PXY  4PQC , so XY = QC . As \BPC > \BOC = 180 , \BAC , we have \QPR + \RAQ > 180, thus \ARP +\AQP < 180 : Since \AQP > \ARP , we get \ARP < 90 , so \BRP > 90 ; hence RB < XB . Therefore, Y B > XB , XY > RB , XY = RB , QC . Thus PB , PC = Y B > RB , QC . No other solutions were submitted.

2147. [1996: 171] Proposed by Hoe Teck Wee, Singapore.

Let S be the set of all positive integers x such that there exist positive integers y and m satisfying x2 + 2m = y 2. (a) Characterize which positive integers are in S . (b) Find all positive integers x so that both x and x + 1 are in S .

305 Solution by Thomas Leong, Staten Island, NY, USA. (a) We show that S = fx : x = 2r , 2s ; r > s  0; r; s 2 Zg. First suppose that x2 + 2m = y 2 with x positive. Then 2m = y 2 , x2 = (y + x)(y , x), and so y + x = 2r and y , x = 2s where r > s  1 (s 6= 0 since x and y are of the same parity). Thus 2x = 2r , 2s , or x = 2r,1 , 2s,1 where r , 1 > s , 1  0. Conversely, if x = 2r , 2s with r > s  0, then with m = r + s + 2 and y = 2r + 2s , we have x2 + 2m = y 2. (b) The desired set is fx : x = 2r , 2; r  2 or x = 2r , 1; r  1g. Now every odd number in S is of the form 2r , 1, r  1. If x + 1 2 S is odd, then x + 1 = 2r , 1 and x = 2r , 2 are both in S provided r  2. If x 2 S is odd, then x = 2r , 1 and x + 1 = 2r = 2r+1 , 2r are both in S provided r  1. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,  OCHOA, Logro~no, Spain; THEODORE UK; MIGUEL ANGEL CABEZON CHRONIS, student, Aristotle University of Thessaloniki, Greece; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; KEITH EKBLAW, Walla Walla, Washington, USA; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; DAVID HANKIN, Hunter College Campus Schools, New York, NY, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;  WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV   KONECNY, Ferris State University, Big Rapids, Michigan, USA; KEE-WAI LAU, Hong Kong; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; GIOVANNI MAZZARELLO, Florence, Italy and IAN JUNE GARCES, Ateneo de Manila University, Manila, The Philippines; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; HEINZ JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; LAWRENCE SOMER, The Catholic University of America, Washington, DC; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There were 6 incorrect or incomplete solutions.  Y remarks the similarity between this problem and problem 13, KONECN page 162 of Introduction to Number Theory by W.W. Adams and L.J. Goldstein, which claims that the Diophantine equation x2 , y 2 = mk is solvable for x; y for any given m; k  3. (In the present problem m = 2 and x and y are interchanged.) Several solvers described S in terms of the binary expansions of its members: namely any positive integer in S has a block of 1s followed by a (possibly empty) block of 0s.

306

2148. [1996: 171] Proposed by Aram A. Yagubyants, Rostov na Donu, Russia. Suppose that AD, BE and CF are the altitudes of triangle ABC . Suppose that L, M , N are points on BC , CA, AB , respectively, such that BL = DC , CM = EA, AF = NB. Prove that: 1. the perpendiculars to BC , CA, AB at L, M , N , respectively are concurrent; 2. the point of concurrency lies on the Euler line of triangle ABC . Solution by Florian Herzig, student, Perchtoldsdorf, Austria. Let H be the orthocentre and O the circumcentre. Re ect H in O to get a point H 0 on OH , and let L0 , M 0 , N 0 be the feet of the perpendiculars from H 0 onto BC , CA and AB , respectively. If X , Y , Z denote the midpoints of BC , CA, AB , then HD k OX k H 0 L0 (and similarly for the other sides). As HO = OH 0 we have DX = XL0 and, together with XC = XD + DC = BL0 + L0 X 0= BX , yields that DC = BL0 . Hence from the de nition of L and since L is also an interior point of BC , we have L = L0 and analogously0 M = 0M 0 and N 0 = N 0 . The perpendiculars to the sides of the triangle LH ; MH , and NH intersect at H 0 , which lies on the Euler line HO of 4ABC , as we wanted to prove. Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; L. RICE, Woburn CI, Scarborough Ontario; TOSHIO SEIMIYA, Kawasaki, Japan; SHAILESH SHIRALI, Rishi Valley School, India; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer. One solution was incomplete. Several readers reduced the result to the theorem, attributed sometimes to Steiner and sometimes to Carnot, that appeared recently in CRUX with MAYHEM [1997: 122] in an alternative solution to problem 2120. Although this approach leads to a nice solution here, it takes more work to set up the machinery than to solve the problem without it.

2149. [1996: 171] Proposed by Juan-Bosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Let ABCD be a convex quadrilateral and O is the point of the intersection of the diagonals AC and BD. Let A0 B 0 C 0 D0 be the quadrilateral whose vertices, A0 , B 0 , C 0 , D0 , are the feet of the perpendiculars drawn from the point O to the sides BC , CD, DA, AB , respectively. Prove that ABCD is an inscribed (cyclic) quadrilateral if and only if A0 B0C 0D0 is a circumscribing quadrilateral (A0 B0, B0C 0, C 0D0, D0 A0 are tangents to a circle).

307 Solution by D.J. Smeenk, Zaltbommel, the Netherlands (with notation modi ed by the editor) Assume ABCD is cyclic. We denote: \DBA = \DCA = ; \ACB = \ADB = ; \BDC = \BAC = ; \CAD = \CBD = : [Because of the right angles at C 0 and D0 ,] AD0 OC 0 is cyclic, so that \OD0 C 0 = \OAC 0 =  . Likewise BD0 OA0 is cyclic, so that \A0 D0 O = \A0 BO =  . Thus D0 O is the bisector of \A0 D0 C 0 . Arguing in this manner we conclude that the angle bisectors of A0 B 0 C 0 D0 all pass through O and, therefore, that this quadrilateral has an incircle. For the converse assume that A0 B 0 C 0 D0 has an incircle and obtain A, B, C , D as the appropriate intersection points of the lines through A0 0 perpendicular to OA , through B 0 perpendicular to OB 0 , etc. We are to show that ABCD is cyclic and O is the point where AC and BD intersect. We denote: \OA0 D0 = \B 0 A0 O = ; \OB 0 A0 = \C 0 B 0 O = ; \OC 0 B 0 = \D0 C 0 O = ; \OD0 C 0 = \A0 D0 O = : with + + +  = : (1) 0 OC 0 cyclic implies \OAC 0 = \OD0 C 0 =  , and \D0 AO = AD 0 0 \D C O = . In a similar way, \C 0 DO = and \ODB 0 = ; \B 0 CO = and \OCA0 = ; \A0 BO =  and \OBD0 = : Consider the quadrangle AOCD. We have \OAD + \ADC + \DCO + \COA = 2 ; that is,  + ( + ) + + \COA = 2: (2) Equations (1) and (2) imply \COA =  , so that O lies on AC . Arguing analogously, O lies also on BD, so that O is the desired intersection point. Finally, opposite angles have the sum \BAD + \DCB = ( +  ) + ( + ) =  ; thus ABCD is cyclic.

308 Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and  LOPEZ  MARI A ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain (with two proofs of the \only if" portion); CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK (\only if" portion); TOSHIO SEIMIYA, Kawasaki, Japan; SHAILESH SHIRALI, Rishi Valley School, India; and the proposer. FLORIAN HERZIG, student, Perchtoldsdorf, Austria commented that the \only if" portion follows immediately from problem 3 in the 1990 Canadian Mathematical Olympiad, CRUX [1990; 161, 198]. Seimiya added a warning that the parenthetical comment appended to the statement of the problem actually leads to a di erent result which turns out to be not quite correct; if the lines A0 B 0 ; B 0 C 0 ; C 0 D0 ; D0 A0 are tangents to a circle then it is not always true that ABCD is an inscribed quadrilateral. For a counterexample, let \BAC + \BDC =  , and \BAC 6= =2. If T is the point where AB intersects CD, then T , A, O, D are concyclic, so that by Simson's theorem, B 0 , C 0 , D0 are collinear. The incircle of 4A0 B 0 D0 is tangent to the lines A0 B 0 , B 0 C 0 , C 0 D0 , D0 A0 ; yet A, B , C , D are not concyclic.  2150. [1996: 171] Proposed by Sefket Arslanagi c , Berlin, Germany. Find all real solutions of the equation

p1 , x = 2x2 , 1 + 2xp1 , x2:

I. Solution by Kee-Wai Lau, Hong Kong. By squaring both sides of the equation and simplifying we obtain p

,x = 4x(2x2 , 1) 1 , x2:

(1)

Clearly x = 0 is not a solution. So by (1) we obtain p

p

,1 = 4(2x2 , 1) 1 , x2:

(2)

Let y = 1 , x2 . Then (2) becomes

8y 3 , 4y , 1 = 0 or (2y + 1)(4y 2 , 2y , 1) = 0:

p

s

p

1 + 5 and x =  5 , 5 . For the negative Since y is positive, y = 4 p 8 value of x, both 2x2 , 1 and 2x 1 , x2 are negative and so cannot be a solution. For 0  x  1 the continuous function

p1 , x , (2x2 , 1) + 2xp1 , x2

309 is positive when x = 0 and negative when x = 1. Thus the function vanishes at least once in the s interval [0; 1]. We conclude that the only solution of the

p

5 , 5. equation is x = 8

II. Solution by David Doster, Choate Rosemary Hall, Wallingford, Connecticut, USA. Squaring both sides of the given equation and simplifying the resulting equation yields p

4(1 , 2x2) 1 , x2 = 1: Since x lies between ,1 and 1, we may setp x = sin , where ,=2    =2. Then we get 4(1 , 2 sin2 ) 1 , sin2  = 1, or equivalently, 4(2 cos2  , 1) cos  = 1. Now the equivalent cubic equation 8 cos3  , 4 cos  , 1 = 0 can be rewritten as (2 cos  + 1)(4 cos2  , 2 cos  , 1) = 0:

p

p

p

1 + 5 . Thus sin  =  10 , 2 5 . Since cos   0, we must have cos  = 4 4 Sinceponly thep positive solution satis es the original equation, we have x = 10 ,4 2 5 . (, incidentally, is =5.)

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL  OCHOA, Logro~no, Spain; THEODORE CHRONIS, student, ArisCABEZON totle University of Thessaloniki, Greece; TIM CROSS, King Edward's School, Birmingham, England; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; RUSSELL EULER, NW Missouri State University, Maryville, Missouri, USA; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; DAVID HANKIN, Hunter College Campus Schools, New York, NY, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; AMIT KHETAN, Troy,   Y,  Ferris State University, Big Rapids, Michigan, MI, USA; VACLAV KONECN USA; MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; BEATRIZ MARGOLIS, Paris, France; J.A. MCCALLUM, Medicine Hat, Alberta; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; L. RICE, University of Toronto Schools, Toronto, Ontario; KRISTIAN SABO, student, Os  ijek, Croatia; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon,   Spain; HEINZ-JURGEN SEIFFERT, Berlin, Germany; SKIDMORE COLLEGE PROBLEM GROUP, Saratoga Springs, New York, USA; D.J. SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH, Mount Royal College, Calgary, Alberta; DAVID R. STONE, Georgia Southern University, Statesboro, Geor-

310 gia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; NISHKA VIJAY, student, Mount Allison University, Sackville, New Brunswick; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; and the proposer. There were 2 incorrect solutions. Most solvers who used a trigonometric approach began immediately with the statement that ,1  x  1, and then used either a sine or cosine substitution. This assumes that we are restricting our attention to an equation with real values, but we are told only that the variable x must be real-valued. Only HANKIN explicitly showed that there were no real-valued solutions to be found outside the above range.

2151. [1996: 217] Proposed by Toshio Seimiya, Kawasaki, Japan.

4ABC is a triangle with \B = 2\C . Let H be the foot of the perpendicular from A to BC , and let D be the point on the side BC where the excircle touches BC . Prove that AC = 2(HD). Solution by Cristobal  S a nchez{Rubio, I.B. Penyagolosa, Castellon,  Spain. Let s = (a + b + c)=2 be the semiperimeter of 4ABC ; we have b + CD = c + BD = s, so that DC = s , b = (a , b + c)=2: Let C 0 be the image of A under the rotation about B through the angle 180 , \B. Because 4ABC 0 is isosceles, from the condition \B = 2\C it follows that \CAC 0 is also isosceles so that HC = (a + c)=2. Finally, HD = HC , DC = a + c , a , b + c = b = AC : 2

2

2

2

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,  OCHOA, Logro~no, Spain; GEORGI Spain; MIGUEL ANGEL CABEZON DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON,  LOPEZ  Beaverton, OR, USA; MARI A ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; LUIZ A. PONCE, Santos, Brazil; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer. Many solvers used the fact that \B = 2\C is equivalent to the condition b2 = c(c + a), which has appeared before in CRUX: [1976: 74], [1984: 278], and [1996: 265{267]. For a recent problem concerning integer{sided triangles with \B = 2\C see the solution to problem 578 in The College Math J. 28:3 (May, 1997) 233{235. Further references are given there.

311

2152. [1996: 217] Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. n X Let n  2 and 0  x1  : : :  xn  2 be such that sin xk = 1. k=1 Consider the set Sn of all sums x1 + : : : + xn . 1. Show Sn is an interval. 2. Let ln be the length of Sn. What is nlim !1 ln ? I. Solution by Florian Herzig, student, Perchtoldsdorf, Austria. 1. First I prove that

n arcsin n1  x1 + x2 +    + xn  2 :

Since the graph of sin x is concave down for x 2 [0; =2], the chord joining the points (0; sin 0) and (=2; sin =2) lies below the graph. Hence [since this chord has slope 2= ]

2x  sin x



for all x 2 [0; =2]

and we can deduce the right-hand side of the claim:

2 (x + x +    + x )  sin x + sin x +    + sin x = 1: n 1 2 n  1 2

The left-hand side follows immediately from Jensen's inequality, since sin x is concave down for x 2 [0; =2] and 0  (x1 +    + xn )=n < =2:

1 = sin x1 + sin x2 +    + sin xn  sin  x1 + x2 +    + xn  : n n n

Next I will show that the set Sn of all sums x1 + x2 +    + xn is the interval  

n arcsin n1 ; 2 : First let 0  x  y  arcsin A be such that sin x + sin y = A, where 0  A  1 is xed. I prove that x + y is a continuous function f depending on x only: for each value of x there is exactly one value of y such that all conditions are ful lled, since

sin y = A , sin x 2 [0; A] if and only if y = arcsin(A , sin x): Thus f (x) = x + arcsin(A , sin x). It is clear that f is continuous for x 2 [0; arcsin A].

312 Now we consider the following process. Initially let x1 =    = xn = arcsin(1=n) [so that P sin xi = 1]. In the rst step keep x3 ; : : : ; xn constant and vary x1 ; x2 such that sin x1 + sin x2 remains constant and x1 becomes zero. Thereby sin x1 + sin x2 +    + sin xn remains constant (equal to 1) throughout and x1 + x2 +    + xn varies continuously because of the previous result. In the second step keep x1 = 0, x4 ; : : : ; xn constant and vary x2 and x3 as before, with x2 ! 0. Again x1 + x2 +    + xn varies continuously. It is clear now how to vary the xi step by step such that in the end x1 = x2 =    = xn,1 = 0 and as a consequence xn = =2. As we see, x1 +    + xn has varied continuously from n arcsin(1=n) to =2. Hence   1  Sn = n arcsin n ; 2 : 2. By l'H^opital's rule,

  , n arcsin 1  lim l = lim n!1 n n!1 2 n  arcsin x = 2 , xlim !0 x  1 = 2 , xlim !0 p1 , x2 = 2 , 1:

II. Solution by Thomas C. Leong, The City College of City University of New York, New York, NY, USA. Equivalently, we consider the set Y = fy = (y1; : : : ; yn) j 1  y1      yn  0; y1 +    + yn = 1g  Rn and the image f (Y ) of Y under f (y) = arcsin y1 +    + arcsin yn . Note that Sn = f (Y ). 1. Since Y is a connected subspace of Rn and f is a continuous function, the image f (Y )  R is also connected, and we know that the only connected subspaces of R are intervals. Thus Sn is an interval. 2. Since arcsin x is convex in [0; 1], we can use the majorization inequality. Since 

1 ; 1 ; : : : ; 1   (y ; y ; : : : ; y )  (1; 0; : : : ; 0) 1 2 n n n n

[Editor's note: here x  y means that x is majorized by y; see for example page 45 and then Theorem 108, page 89 of Hardy, Littlewood and Polya's  Inequalities],

n arcsin n1  arcsin y1 +    + arcsin yn  arcsin 1 = 2 ;

313 with equality when (y1; : : : ; yn ) is equal respectively to (1=n;: : : ; 1=n) and (1; 0; : : : ; 0). Thus ln = 2 , n arcsin(1=n) and      1  arcsin(1 =n )  , 1: = nlim !1 ln = 2 , nlim !1 n arcsin n = 2 , nlim !1 1=n 2

Also solved by KEITH EKBLAW, Walla Walla, Washington, USA; and RICHARD I. HESS, Rancho Palos Verdes, California, USA.  2153. [1996: 217] Proposed by Sefket Arslanagi c , Berlin, Germany.

Suppose that a, b, c 2 R. If, for all x 2 [,1; 1], jax2 + bx + cj  1, prove that

jcx2 + bx + aj  2:

Solution (virtually identical) by Florian Herzig, student, Perchtoldsdorf, Austria, and Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let f (x) = ax2 + bx + c and g (x) = cx2 + bx + a. Then by assumption ja + b + cj = jf (1)j  1, ja , b + cj = jf (,1)j < 1 and jf (0)j  1. Hence jg(x)j = c(x2 , 1) + (a + b + c) 1 +2 x + (a , b + c) 1 ,2 x  jcjjx2 , 1j + ja + b + cj j1 +2 xj + ja , b + cj j1 ,2 xj  jx2 , 1j + j1 +2 xj + j1 ,2 xj = 1 , x2 + 1 + x + 1 , x

= 2 , x2  2 for all x 2 [,1; 1].

2

2

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; DIGBY SMITH, Mount Royal College, Calgary, Alberta; and the proposer. There was one incorrect solution. Several solvers pointed out that this is not a new problem. Herzig cited the book International Mathematical Olympiad 1978{1985 (published by the Mathematical Association of America, 1986). Chronis gave the reference Selected Problems and Theorems in Elementary Mathematics | Arithmetic and Algebra by D.O. Shklyarsky, N.N. Chentov, and I.M. Yaglom (exercise 304 on page 60). Demizev and Kunchev remarked that it can be found as problem #86 on page 35 of the book Mathematics Competitions by L. Davidov,

314 V.Petnob, I. Tonov, and V. Chukanov (So a, 1977). Bellot Rosado mentioned that the same problem was proposed during the 2nd round of the 1996 Spanish Mathematical Olympiad. Janous remarked that the problem can be restated as \Prove that if p(x) is a real polynomial of2 degree 2 satisfying the condition that jp(2x)j  1 for all x 2 [,1; 1], then jx p(1=x)j  2 for all x 2 [,1; 1]." (Ed: x p(1=x) is to be written as a polynomial before substituting values of x.) He then ventured the following conjecture: Conjecture: If p(x) is a real polynomial with degree n such that jp(x)j  1 for all x 2 [,1; 1] then jxn p(1=x)j  2n for all x 2 [,1; 1]. Can any reader prove or disprove this?

2154. [1996: 217] Proposed by K.R.S. Sastry, Dodballapur, India.

In a convex pentagon, the medians are concurrent. If the concurrence point sections each median in the same ratio, nd its numerical value. (A median of a pentagon is the line segment between a vertex and the midpoint of the third side from the vertex.)

A

D0

B

C0 E

P

E0 C

B0 A0

D

I Solution by Toshio Seimiya, Kawasaki, Japan. Let the convex pentagon be ABCDE , and let A0 , B 0 , C 0 , D0 , E 0 be the midpoints of CD, DE , EA, AB , BC , respectively. We assume that AA0 , BB0 , CC 0, DD0 , EE0 are concurrent at P , and that

AP = BP = CP = DP = EP = : PA0 PB0 PC 0 PD0 PE0 Then set D0 E 0 = 1. Since D0 , E 0 are midpoints of AB , BC , respectively, we have AC = 2D0 E 0 = 2, and D0 E 0 k AC . Since DP = EP = ; PD0 PE0 we have D0 E 0 k DE and DE=D0 E 0 = , so that DE = D0E0 = : (1)

315 Since AP=PA0 = CP=PC 0 = , we have AC=A0 C 0 = AP=PA0 = , so that

2: A0 C 0 = AC = (2)   Since AC k D0 E 0 and D0 E 0 k DE , we get AC k DE , and since A0 , C 0 are midpoints of CD, AE , respectively, we have AC + DE = 2A0C 0: Therefore we have from (1) and (2)

2 +  = 4 ;

p

from which we have 2 + 2 , 4 = 0. Thus we obtain  = 5 , 1. II Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. (We will use the same notation as used by Seimiya above. Each boldface letter represents the vector to that point from the origin.) Then we have D0 = 12 (A + B), E0 = 12 (B + C), etc. There must exist real numbers ; > 0, with + = 1, such that P = A + A0 = B + B0 = C + C0 = D + D0 = E + E0 : Now D + D0 = A + A0 () 2 D + (A + B) = 2 A + (C + D); (3) A + A0 = B + B0 () 2 A + (C + D) = 2 B + (D + E): (4) Equation (4) can be simpli ed to 2 A + C = 2 B + E () 2 (B , A) = (C , E); which implies that AB k EC . Similarly, BC k AD; CD k BE; DE k CA; EA k DB: (5) Now equation (3) can be written as: 2 D + A , 2 A , D = C , B; (2 , )(D , A) = (C , B): On the other hand from (5) we have 2 (C , B) = (D , A): Hence, (C , B) = 2 (C , B); D,A =

2 ,



316 whence we obtain

2 , 2 4 , 2 , 2 4 2 , 2 (1 , ) , (1 , )2 5 2

= = = = =

2



0 0 1p

5 5:

Thus the ratio we are considering has the numerical value

p : = : (1 , ) = 1 : ( 5 , 1)

Also solved by SAM BAETHGE, Nordheim, Texas, USA; MIHAI CIPU,  OCHOA, Logro~no, Romanian Academy, Bucharest; MIGUEL ANGEL CABEZON Spain; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka Schoolof Mathematics, Rousse, Bulgaria; HANS ENGELHAUPT, Franz{ Ludwig{Gymnasium, Bamberg, Germany; RICHARD I. HESS, Rancho Palos Verdes, California, USA; D. KIPP JOHNSON, Valley Catholic High School,  Beaverton, Oregon; HEINZ-JURGEN SEIFFERT, Berlin, Germany; and the proposer. Seimiya also observes the following: 1. If four medians are concurrent at a point P , then the fth median also passes through P . 2. We assume that ve medians are concurrent at a point P . If any three medians are divided at P into the same ratio  : 1, then the other two medians are divided at P into the same ratio  : 1. 3. This gure is an image of a regular pentagon by an ane transformation.

2155. [1996: 218] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. Prove there is no solution of the equation 1 + 1 = 1 x2 y8 z2

in which y is odd and x; y; z are positive integers with highest common factor 1. Find a solution in which y = 15, and x and z are also positive integers. Solution by D. Kipp Johnson (modi ed slightly by the editor), Valley Catholic High School, Beaverton, Oregon, USA.

317 First, we nd the general solution in N to the auxiliary equation

1 + 1 = 1 where t is odd. (1) x2 t2 z2 (1) can be written as z 2 (x2 + t2 ) = x2 t2 , which shows that x2 + t2 must be a square. From well{known results, we then have x = k(2uv ), t = k(u2 , v 2), where k 2 N is odd, and u, v are relatively prime integers of opposite parity with u > v . Substituting into (1) and simplifying gives

z(u2 + v2) = 2kuv(u2 , v2): (2) Since u2 + v 2 is odd, z = 2w for some w 2 N and (2) becomes w(u2 + v2) = kuv(u2 , v2): (3) Since (u; v ) = 1, (u2 + v 2; uv ) = 1. Furthermore, (u2 + v 2; u2 , v 2) = 1 since if p is a prime such that pju2 + v 2, u2 , v 2, then pj2u2 , 2v 2 implies pj(2u2; 2v2). But (2u2; 2v2) = 2(u; v2)2 =22 and clearly p = 6 2 as u2 + v2 is odd. This is a contradiction. Hence u + v jk. Letting k = j (u2 + v 2), then (3) becomes w = juv(u2 , v2) where j 2 N is odd. Therefore, we may choose u, v subject to the stated conditions and arbitrarily odd j 2 N to nd the general solution of the original equation, which is given by:

x = 2juv(u2 + v2); y4 = t = j (u2 , v2)(u2 + v2); z = 2juv(u2 , v2): Since (x; y;z ) = 1, we must have j = 1 and so y 4 = u4 , v 4, which is impossible by Fermat's Last Theorem. To nd a solution in which y = 15, let

y4 = 154 = 34  54 = j (u , v)(u + v)(u2 + v2): (4) Since u , v , u + v , and u2 + v 2 are pairwise relatively prime, and the left side of (4) contains only two distinct prime factors, u , v = 1 and (4) becomes 154 = j (2v + 1)(2v2 + 2v + 1): (5) From (5) we see that 154 > 4v 3 and so v < 24 resulting in 2v + 1 < 49. Now, 2v +1 is odd and the only odd factors of 154 which are less than 49 are 3, 5, 9, 15, 25 and 45 with corresponding values of v = 1, 2, 4, 7, 12, and 22. A quick check shows that only in the case v = 1 is 2v 2 + 2v + 1 also a factor of 154. Thus u = 2 and (5) gives j = 33  53 = 3375 resulting in the unique solution:

x = 67500; y = 15; z = 40500:

318 Also solved (both parts) by WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria, and the proposer. The second part of the problem was also solved by GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria (jointly); HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; and DIGBY SMITH, Mount Royal College, Calgary, Alberta. All these solvers gave the same solution (67500; 15; 40500). However, most of their \proofs" erred in assuming that gcd(x; y; z ) = 1 implies that x, y , z are pairwise relatively prime. Furthermore, Johnson was the only one who gave a completely valid proof for the uniqueness of the solution when y = 15, though Janous found this to be the case by using DERIVE. Both Hess and Klamkin considered the given equation without the restriction that y be odd and (x; y;z ) = 1. Hess observed that an in nite family of solutionsis given by x = 2mn(m2 + n2)4 (m2 , n2)3 , y = m4 , n4, and z = 2mn(m2 + n2)3(m22 , n32)4 where m, n 2 N with m > n while Klamkin gave the family: x = 2  3  54  n4, y = 15n, z = 22  34  53  n4 . Both formulas yield the solution given above when m = 2 and n = 1.

2156. [1996: 218] Proposed by Hoe Teck Wee, Singapore.

ABCD is a convex quadrilateral with perpendicular diagonals AC and BD. X and Y are points in the interior of sides BC and AD respectively such that BX = BD = DY : CX AC AY Evaluate BC  XY : BX  AC Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria. Let Z be the point in the interior of side AB such that BZ : AZ = BX : CX . Then it follows from

BZ = BX = DY = BD  AZ CX AY AC that XZ and AC as well as ZY and BD are parallel, which, considering AC ? BD, implies that 4XY Z is a right triangle with legs XZ and Y Z . Thus XZ = XZ  AC  BD = BZ  AZ  AB = 1 ZY AC BD ZY AB BZ AZ [where XZ=AC = BZ=AB and BD=ZY = AB=AZ by similar triangles while AC=BD = AZ=BZ by the de nition of Z ]. This means that 4XY Z

319 is an equilateral right triangle, whence

p XY : ZX = 2 : 1

and, consequently,

BC  XY = AC  XY = XY = p2: BX  AC ZX  AC ZX

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val OCHOA, Logro~no, Spain; GEORGI ladolid, Spain; MIGUEL ANGEL CABEZON DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; TOSHIO SEIMIYA, Ka wasaki, Japan; HEINZ-JURGEN SEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer. Bellot comments that facts about orthodiagonal quadrilaterals can be found in Jordan Tabov, Simple properties of the orthodiagonal quadrilaterals, Matematyka & Informatyka, 1:1 (1991) 1{5. Janous, using rectangular coordinates, shows that if the angle from AC to BD were , then

BC  XY = 2 sin  : BX  AC 2

 2157. [1996: 218] Proposed by Sefket Arslanagi c , Berlin, Germany. 

Prove that 21997 1996 , 1 is exactly divisible by 19972 . I. Solution by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. From the Fermat-Euler theorem we have a'(m)  1(mod m), for every a; m such that gcd(a; m) = 1. We also have '(pk) = pk,1 (p , 1), when p is a prime number. So '(19972) = 1997  1996 and 2'(19972)  1(mod 1)9972. II. Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. We have '(1997) = 1996 = 22  499. Furthermore, 2449  1585(mod 1)997 and 22449  ,1(mod 1)997. Hence 2 is of order 1996 modulo 1997; that is

:= 21996 , 1  0 (mod 1997):

On the other hand,

 672989 (mod 19972); that is, 6 0 (mod 19972):

320 But

:= 219971996 , 1= ((21996 , 1) + 1)1997 , 1 = ( + 1)1997 , 1  1997 3 +    + 1997 , 1 2 = 1 + 1997 + 1997 + 2 3 3 = 1997 + 1997  f; where f is an integer. Thus  0(mod 1)9972, but 6 0(mod 1)9973.

 OCHOA, Logro~no, Spain; Also solved by MIGUEL ANGEL CABEZON MIHAI CIPU, Romanian Academy, Bucharest, Romania; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; D. KIPP JOHNSON, Valley Catholic High School, Beaverton, Oregon; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; THOMAS LEONG, Staten Island, NY, USA; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; CAN ANH MINH, University of California, Berkeley, California; SOLEDAD ORTEGA and JAVIER  students, Universidad de La Rioja, Logro~no, Spain; YOLANDA GUTIERREZ, PELLEJERO, student, Universidad de La Rioja, Logro~no, Spain; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; HEINZ JURGEN SEIFFERT, Berlin, Germany; and the proposer. It is interesting to note that only Janous and Manes addressed the question of \exact" divisibility; that is, they showed that no higher power of 1997 than 2 would work.

Crux Mathematicorum

Founding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell Editors emeriti / Redacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem

Founding Editors / Redacteurs-fondateurs: Patrick Surry & Ravi Vakil Editors emeriti / Redacteurs-emeriti: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

321

THE ACADEMY CORNER No. 13

Bruce Shawyer All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Mathematics Competitions There are many mathematical competitions at many di erent levels. There is in fact a journal devoted to this. It is called Mathematics Competitions, and is the Journal of the World Federation of National Mathematics Competitions. It is published by the Australian Mathematics Trust, and information on it can be obtained by contacting them at PO Box 1, Belconnen, ACT 2616, Australia. The journal covers all sorts of mathematical competitions, and articles not only give problems and solutions, but also cover the philosophy of competition and alternative ways of stimulating talented students. The World Federation also makes two awards, the David Hilbert International Award (which recognises contributions of mathematicians who have played a signi cant role over a number of years in the development of mathematical challenges at the international level and which have been a stimulus for mathematical learning), and the Paul Erdos National Award (which recognises contributions of mathematicians who have played a signi cant role over a number of years in the development of mathematical challenges at the national level and which have been a stimulus for the enrichment of mathematics learning).

322

THE OLYMPIAD CORNER No. 184

R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. We begin this number with the problems of the Mock Test for the International Mathematical Olympiad team of Hong Kong. My thanks go to Richard Nowakowski, Canadian Team Leader, for collecting them at the 35 IMO in Hong Kong. th

INTERNATIONAL MATHEMATICAL OLYMPIAD 1994 Hong Kong Committee | Mock Test, Part I Time: 4.5 hours

1. In 4ABC , we have \C = 2\B. P is a point in the interior of 4ABC satisfying AP = AC and PB = PC . Show that AP trisects the angle \A.

2. In a table-tennis tournament of 10 contestants, any two contestants meet only once. We say that there is a winning triangle if the following situation occurs: i contestant defeated j contestant, j contestant defeated k contestant, and k contestant defeated i contestant. Let Wi and Li be respectively the number of games won and lost by the i contestant. Suppose Li + Wj  8 whenever the i contestant beats the j contestant. Prove that there are exactly 40 winning triangles in this tournament. 3. Find all the non-negative integers x, y, and z satisfying that th

th

th

th

th

th

th

th

th

7x + 1 = 3y + 5z :

Mock Test, Part II Time: 4.5 hours

1. Suppose that yz + zx + xy = 1 and x, y, and z  0. Prove that x(1 , y

2)(1

,z

2) +

y(1 , z

2 )(1

,x

2) +

z(1 , x

2 )(1

p

, y  4 9 3: 2)

323

2. A function f (n), de ned on the natural numbers, satis es:

f (n) = n , 12 if n > 2000; and f (n) = f (f (n + 16)) if n  2000: (a) Find f (n). (b) Find all solutions to f (n) = n. 3. Let m and n be positive integers where m has d digits in base ten and d  n. Find the sum of all the digits (in base ten) of the product (10n , 1)m. As a second Olympiad set we give the problems of the Final Round of the 45 Mathematical Olympiad written in April, 1994. My thanks go to Marcin E. Kuczma, Warszawa, Poland; and Richard Nowakowski, Canadian Team leader to the 35 IMO in Hong Kong, for collecting them. th

th

45th MATHEMATICAL OLYMPIAD IN POLAND Problems of the Final Round | April 10{11, 1994 First Day | Time: 5 hours

1. Determine all triples of positive rational numbers (x; y; z) such that 2. In the plane there are given two parallel lines k and l, and a cir-

x + y + z, x,1 + y,1 + z,1 and xyz are integers.

cle disjoint from k. From a point A on k draw the two tangents to the given circle; they cut l at points B and C . Let m be the line through A and the midpoint of BC . Show that all the resultant lines m (corresponding to various points A on k) have a point in common. 3. Let c  1 be a xed integer. To each subset A of the set f1, 2, : : : : , ng we assign a number w(A) from the set f1, 2, : : : , cg in such a way that w(A \ B) = min(w(A);w(B)) for A; B  f1; 2; : : : ; ng: p Suppose there are a(n) such assignments. Compute limn!1 n a(n). Second Day | Time: 5 hours

4. We have three bowls at our disposal, of capacities m litres, n litres and m + n litres, respectively; m and n are mutually coprime natural numbers. The two smaller bowls are empty, the largest bowl is lled with water. Let k be any integer with 1  k  m + n , 1. Show that by pouring water (from any one of those bowls into any other one, repeatedly, in an unrestricted manner) we are able to measure out exactly k litres in the third bowl.

324

5. Let A1; A2; : : : ; A8 be the vertices of a parallelepiped and let O be its centre. Show that 4(OA21 + OA22 +    + OA28)  (OA1 + OA2 +    + OA8)2:

6. Suppose that n distinct real numbers x1; x2; : : : ; xn (n  4) satisfy the conditions x1 + x2 +    + xn = 0 and x21 + x22 +    + x2n = 1. Prove that one can choose four distinct numbers a, b, c, d from among the xi 's in such a way that a + b + c + nabc  x31 + x32 +    + x3n  a + b + d + nabd:

We now give three solutions to problems given in the March 1996 Corner as the Telecom 1993 Australian Mathematical Olympiad [1996: 58].

TELECOM 1993 AUSTRALIAN MATHEMATICAL OLYMPIAD Paper 1 Tuesday, 9th February, 1993 (Time: 4 hours)

6. In the acute-angled triangle ABC , let D, E, F be the feet of altitudes through A, B , C , respectively, and H the orthocentre. Prove that AH + BH + CH = 2: AD BE CF

land.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

AH + BH + CH = 3 ,  HD + HE + HF  AD BE CF AD BE CF   [ BHC ] [ CHA ] [ AHB ] = 3 , [ABC ] + [ABC ] + [ABC ] ] = 2: = 3 , [[ABC ABC ]

7. Let n be a positive integer, a1; a2; : : : ; an positive real numbers and s = a1 + a2 +    + an . Prove that n X

ai  n i=1 s , ai n , 1

and

n X

s , ai  n(n , 1): i=1 ai

325 Solutions by Mansur Boase, student, St. Paul's School, London, England and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give the solution by Boase. n X

n X

n s s , 1 = X ,n i=1 ai i=1 ai n s X n ai X X and  ( 1)2 = n2; a s i i=1 i=1

s , ai = i=1 ai

ai = 1 i=1 s

n X



by the Cauchy{Schwarz inequality. Thus n

s  n2: i=1 ai X

n X

s,a

i  n2 , n = n(n , 1). Hence a i i=1 To prove the rst inequality, rst note that n X

1

n X

i=1 i=1

n X

a2i 

n X i=1

ai

!2

= s2 :

s2

Hence a2i  . n i=1 By the Cauchy{Schwarz inequality, n X

n X

ai(s , ai) s ,aia  i i=1 i=1

n X i=1

ai

!2

= s2 :

Therefore

n X

ai  P s2 s2 P = P n ai (s , ai ) s n ai , n a2 i=1 i=1 i=1 2 i=1 s , ai 2  s2 s, s2 = 1 ,1 1 = n ,n 1 : n n

So, both inequalities are proved. 8. [1996: 58] Telecom 1993 Australian Mathematical Olympiad. The vertices of triangle ABC in the xy {plane have integer coordinates, and its sides do not contain any other points having integer coordinates. The interior of ABC contains only one point, G, that has integer coordinates. Prove that G is the centroid of ABC .

326 land.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

A C0 B

By Pick's Theorem

r

A0

G

B0 C

  [ABC ] = 1 + 3 12 , 1 = 32 ;   [ABG] = 0 + 3 12 , 1 = 12 ; [BCG] = 12 and [CAG] = 12 :

Therefore Hence

[ABG] = [BCG] = [CAG] = 1 : [ABC ] [ABC ] [ABC ] 3

GA0 = GB0 = GC 0 = 1 : AA0 BB0 CC 0 3

The unique point satisfying this above is well-known to be the centroid. Next we give one solution from the Japan Mathematical Olympiad 1993 given in the March 1996 Corner. 2. [1996: 58] Japan Mathematical Olympiad 1993. Let d(n) be the largest odd number which divides a given number n. Suppose that D(n) and T (n) are de ned by

D(n) = d(1) + d(2) +    + d(n); T (n) = 1 + 2 +    + n:

Prove that there exist in nitely many positive numbers n such that 3D(n) = 2T (n). Solutions by Mansur Boase, student, St. Paul's School, London, England, and Zun Shan and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give the solution by Boase.

327

T (n) = n(n2+ 1) :

Thus we need to prove that there are in nitely many n for which

D(n) = n(n3+ 1)

so that

3D(n) = 2T (n)

holds:

Consider

D(2n) = d(1) + d(3) +    + d(2n , 1) + d(2) + d(4) +    + d(2n) = 1 + 3 +    + (2n , 1) + d(1) + d(2) +    + d(2n,1 ) = 1 + 3 +    + (2n , 1) + D(2n,1 ):

Now

n,1 n,1 n n 1 + 3 +    + (2n , 1) = 2 (2 2 + 1) , 2 2 (2 2 + 1) = 2n,1 (2n , 2n,1 ) = 22n,2: Thus D(2n ) = D(2n,1 ) + 22n,2. 22n + 2 for Now, D(21 ) = 2 and we shall prove by induction that D(2n ) = 3 n  0. This holds for n = 0 and for n = 1. Suppose it holds for n = k. 22k + 2 . Thus D(2k ) = 3

Then

2k D(2k+1) = D(2k) + 22k = 2 3+ 2 + 22k 2k = 4(2 3) + 2 2k+2 = 2 3+2

and the result follows by induction. Now consider D(2n , 2).

D(2n , 2) = D(2n) , d(2n , 1) , d(2n) 2n = 2 3+ 2 , (2n , 1) , 1

2n = 2 3+ 2 , 2n 2n n) + 2 (2n , 1)(2n , 2) = 2 , 3(2 = : 3 3

328

x(x + 1) for x = 2n , 2, and there are in nitely many such x.

Thus D(x) =

3

Next we turn to comments and solutions from the readers to problems from the April 1996 number of the Corner where we gave the selection test for the Romanian Team to the 34 IMO as well as three contests for the Romanian IMO Team [1996: 107{109]. th

SELECTION TESTS FOR THE ROMANIAN TEAM, 34th IMO. Part II | First Contest for IMO Team 1st June, 1993

1. Find the greatest real number a such that p

y

x

+ pz 2y+ x2 + p 2z 2 > a z x +y

2+ 2

is true for all positive real numbers x, y , z . Solutions by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We claim that a = 2. Let

+ pz 2y+ x2 + p 2z 2 : y z x +y We show that f (x;y;z ) > 2. Since f (x; y;z ) ! 2 as x ! y and z ! 0, the lower bound 2 is sharp. Without loss of generality, assume that x  y  z .

f (x;y;z) =

p

x

2+ 2

Since by the arithmetic{harmonic{mean inequality, we have it suces to show that or equivalently,

pz2 + x2 py2 + z2 p y2 + z2 + pz2 + x2  2;

pz2 + x2 py2 + z2 f (x;y;z) > py2 + z2 + pz2 + x2

pz2 + x2 , x py2 + z2 , y p p x2 + y2 > y2 + z2 + pz2 + x2 : z

By simple algebra, this is easily seen to be equivalent to

z p y2 + z2( z2 + x2 + x) + p 2 2 pz 2 2 z + x ( y + z + y)
x and 2 x  x2 + y 2, we have p

y

z z = p1  p 21 2 : < p p 2 2 2 z ( x + x ) 2 2x z z + x + x) 2 x +y

2 + 2(

Thus to establish (1), it remains to show that

pz2 + x2(pzy2 + z2 + y) < 2px21+ y2

or equivalently Since

2z p y2 + z2 + y
2y

2z y2 + z2 + y ;

p

or equivalently

p p y2 + z2 + y y2 + z2 > 2 2 yz: Since y 2 + z 2  2yz , we have p p y2 + z2 + y y2 + z2  2yz +py 2z2 p = (2 + 2)yz > 2 2 yz

(2)

and thus (2) holds. This completes the proof. 2. Show that if x, y, z are positive integers such that x2 + y2 + z2 = 1993, then x + y + z is not a perfect square. Solutions by Mansur Boase, student, St. Paul's School, London, England; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's solution. We show that the result holds for nonnegative integers x, y , and z . Without loss of generality, we may assume that 0  x  y  z . Then

3z2  x2 + y 2 + z 2 = 1993

330 implies that

z2  665; z  26: On the other hand z 2  1993 implies that z  44 and thus 26  z  44. Suppose that x + y + z = k2 for some nonnegative integer k. By the Cauchy{Schwarz Inequality we have

k4 = (x + y + z)2  (12 + 12 + 12)(x2 + y2 + z2) = 5979 p and so k  b 4 5979c = 8. Since k2  z  26, k  6. Furthermore, since x2 + y2 + z2 is odd, it is easily seen that x + y + z must be odd, which implies that k is odd. Thus k = 7 and we have x + y + z = 49. Let z = 26 + d, where 0  d  18. Then x + y = 23 , d ) y  23 , d ) x2 + y2  2y2 = 2(23 , d)2 = 1058 , 92d + 2d2 : (1) On the other hand, from x2 + y 2 + z 2 = 1993 we get x2 + y2 = 1993 , z2 = 1993 , (26 + d)2 = 1317 , 52d , d2: (2) From (1) and (2), we get

1317 , 52d , d2  1058 , 92d + 2d2

or

3d2 , 40d  259 which is clearly impossible since 3d2 , 40d = d(3d , 40)  18  14 = 252. This completes the proof. Remark: It is a well-known (though by no means easy) result in classical number theory that a natural number n is the sum of three squares (of nonnegative integers) if and only if n 6= 4l (8k + 7) where l and k are nonnegative integers. Since 1993  1(mod 8) it can be so expressed and thus the condition given in the problem is not \vacuously" true. In fact, 1993 can be so expressed in more than one way; for example,

1993 = = = =

02 + 122 + 432 22 + 152+  422 22 + 302 + 332 112 + 242 + 362:

These representations also show that the conclusion of the problem is false if we allow x, y , and z to be negative integers; e.g. if x = ,2, y = ,30, z = 33 then x2 + y2 +2 z2 2= 1993 and x + y + z = 12 ; and if x = ,11, 2 y = 24, z = 36 then x + y + z = 1993 and x + y + z = 49 = 72.

331

4. Show that for any function f : P (f1; 2; : : : ; ng) ! f1; 2; : : : ; ng there exist two subsets, A and B , of the set f1; 2; : : : ; ng, such that A 6= B and f (A) = f (B ) = maxfi j i 2 A \ B g. Comment by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. The problem, as stated, is clearly incorrect since for maxfi : i 2 A \ B g to make sense, we must have A \ B 6= ;. For n = 1 clearly there are no subsets A and B with A 6= B and A \ B 6= ;. A counterexample when n = 2 is provided by setting f (;) = f (f1g) = f (f2g) = 1 and f (f1; 2g) = 2. This counterexample stands if max is changed to min. The conclusion is still incorrect if A \ B is changed to A [ B . A counterexample would be f (;) = 2 and f (f1g) = f (f2g) = f f(1; 2)g = 1. Part III | Second Contest for IMO Team 2nd June, 1993

3. Prove that for all integer numbers n, with n  6, there exists an n{point set M in the plane such that every point P in M has at least three other points in M at unit distance to P . Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. The three diagrams displayed below illustrate the existence of such a set. M1 is for all n = 3k, M2 is for all n = 3k+1 and M3 is for all n = 3k+2, where k = 2; 3; 4; : : : . In each diagram, the solid lines connecting two points all have unit length and the dotted lines, also of unit length, indicate how to construct an (n + 3){point set with the described property from one with n points. c

c c c

s

s

c s

s

s

s

s M1

(n = 6; 9; 12; : : : )

c

c

s

s

s s

s s

c c

s

s

s

s

s s M2

(n = 7; 10; 13; : : : )

s

s M3

(n = 8; 11; 14; : : : )

332

Part IV | Third Contest for IMO Team 3rd June, 1993

1. The sequence of positive integers fxngn1 is de ned as follows: x1 = 1, the next two terms are the even numbers 2 and 4, the next three terms are the three odd numbers 5, 7, 9, the next four terms are the even numbers 10, 12, 14, 16 and so on. Find a formula for xn . Solutions by Mansur Boase, student, St. Paul's School, London, England; and by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give the solution of Shan and Wang. We arrange the terms of the sequence fxn g in a triangular array according to the given rule so that the 1 level consists of a single 1 and for all k  2, the k level consists of the k consecutive even (odd) integers that follow the last odd (even) integer in the (k , 1) level: st

th

st

17

10

5 19

2 12

1 7 21

4 14

9 23

16

25

For any given n 2 N , let ln denote the unique integer such that

ln  < n  ln + 1;



2

that is,

2

(ln , 1)ln < n  ln(ln + 1) : 2 2

(1)

Note that ln is simply the number of the level which xn is on. We claim that

xn = 2n , ln : (2) When n = 1, clearly l1 = 1 and thus 2n , ln = 1 = x1 . Suppose (2) holds for some n  1. There are two possible cases: ,  Case (i): If n + 1  ln2+1 ; that is, if xn+1 and xn are on the same level, then ln+1 = ln and hence xn+1 = xn + 2 = 2n , ln + 2 = 2(n + 1) , ln+1 :

333 Case (ii): If n + 1 > ln2+1 then xn is the last number on the ln level and xn+1 is the rst number on the ln+1 level. Thus ,



th

th

ln+1 = ln + 1

and

xn+1 = xn + 1 = 2n , ln + 1 = 2(n + 1) , ln+1 :

Hence, by induction, (2) is established. From (1) we get

l2n , ln < 2n  (ln + 1)2 , (ln + 1): Solving the inequalities, we easily obtain Hence

p1 + 8n 1 + ln <  ln + 1: 2 p ln = 1 + 21 + 8n , 1 &

'

(3)

where dxe denotes the least integer greater than or equal to x (that is, the ceiling function). From (2) and (3) we conclude that

p1 + 8n 1 + xn = 2n + 1 , : 2 &

'

That completes this number of the Corner. Send me your nice solutions as well as Olympiad contests.

334

BOOK REVIEWS Edited by ANDY LIU

A Mathematical Mosaic, by Ravi Vakil, published by Brendan Kelly Publishing Inc., 1996, 2122 Highview Drive, Burlington, ON L7R3X4, ISBN# 1-895997-04-6, softcover, 253+ pages, US$16.95 plus handling. Reviewed by Jim Totten, University College of the Cariboo. So, you have a group of students who have decided they want the extra challenge of doing some mathematics competitions. You want a source of problems which will pique the students' interest, and which also lead to further exploration. The problem source should lend itself well to independent work. The question is: where do you nd the appropriate level enrichment material? Many of us have already tried to answer this question and have a collection of such problem books. Well, here is a book to be added to your collection! It is certainly a problem book, but it is much more than that. The author at one moment guides the reader through some very nice mathematical developments, and throws out problems as they crop up in the development, and in the next moment uses a problem as a starting point for some interesting mathematical development. With a few exceptions the problems in this book are not new, nor are the solutions. They are, however, well organized, both by topic and by level of mathematical maturity needed. Answers are NOT always provided; instead there is often simply a solution strategy or hint given, and occasionally there is simply a reference to some other source for a full-blown treatment. Even when answers are provided, they are not tucked away at the end of the book, but rather they are worked into another topic (usually later in the book, but not always), where they become part of the development of another topic or problem. The author is a PhD candidate in pure mathematics at Harvard University (at the time the book was written). Being still very young, he knows how to speak to today's teenagers. His sense of humour and general puckishness is present throughout: just when you are lulled into some serious computation in probability, he deviously throws a trick question at you, that has a totally non-obvious answer (non-obvious, that is, until you CAREFULLY re-read the question). Many mathematics books published today include short biographies on famous mathematicians through history, especially those whose names come up in the theory developed in the book. This book is no exception. But what is unique about this book is the inclusion of Personal Pro les of young mathematicians from several countries that he has met at International Mathematical Olympiads (IMOs) in the past. The pro les are quite diverse, which

335 means that most bright students could nd one to identify with and to use as a role model. The author and those he pro les have taken a risk in doing this: they have tried to predict some of the important mathematicians in the early part of the next century. It should be interesting to follow their careers and see if those predictions can come true, or if by placing them in the spotlight, they nd too much pressure to deal with. The problems range from puzzles that elementary school children can do to problems that provide training for Putnam candidates (toward the end of the book). There are many cross-references and connections between seemingly unrelated problems from di erent areas of mathematics, connections that most students would be unable to make. Many of these connections are new to this reviewer. However, once made these connections are quite clear. As for his credentials, Ravi Vakil placed among the top ve in the Putnam competition in all four of his undergraduate years at the University of Toronto. Before that he won two gold medals and a silver medal in IMOs and coached the Canadian team to the IMO from 1989 to 1995.

Tests for Divisibility

Some tests for divisibility are well known, such as the test for divisibility by 3: nd the sum of the digits of the whole number n | if that sum is divisible by 3, then the original whole number n is divisible by 3. Many people know this test, but do not know why it works. So, why does it work? The basic principle is that the remainder obtained when 10k is divided by 3, is 1. This is easy to check, since 10k = 3  3  10k,1 + 10k,1 . This enables one to step down one power at a time. So, if d is a digit, the \remainder" obtained when d  10k is divided by 3 is d. (NB: this is not the true remainder one gets when dividing by 3!) If n = abc : : : def is a whole number, then we write it as

n = a  10k + b  10k,1 + c  10k,2 + : : : + d  100 + e  10 + f: When we divide by 3, we get a \remainder" of a + b + c + : : : + d + e + f . If this is divisible by 3, the the true remainder is 0, and so n is divisible by 3.

Now, do you know, or can you construct, a test for divisibility by 7?

336

Packing Boxes with N {tetracubes Andris Cibulis Riga, Latvia

Introduction With the popularity of the video game Tetris, most people are aware of the ve connected shapes formed by four unit squares joined edge to edge. They are called the I {, L{, N {, O{ and T {tetrominoes, after the letter of the alphabet whose shapes they resemble. They form a subclass of the polyominoes, a favourite topic in research and recreational mathematics founded by Solomon Golomb. Here is a problem from his classic treatise, Polyominoes. Is it possible to tile a rectangle with copies of a particular tetromino? Figure 1 shows that the answer is armative for four of the tetrominoes but negative for the N {tetromino, which cannot even ll up one side of a rectangle.

Figure 1 Getting o the plane into space, we can join unit cubes face to face to form polycubes. By adding unit thickness to the tetrominoes, we get ve tetracubes, but there are three others. They are shown in Figure 2, along with the N {tetracube. Is it possible to pack a rectangular block, or box, with copies of a particular tetracube? The answer is obviously armative for the I {, L{, O{ and T {tetracubes, and it is easy to see that two copies of each of the three tetracubes not derived from tetrominoes can pack a 2  2  2 box. Will the N {tetracube be left out once again? Build as many copies of it as possible and experiment with them. If the k  m  n box can be packed with the N {tetracube, we call it an N {box. Are there any such boxes? Certain types may be dismissed immediately.

337

PPP  P PPPP P PP PPP PP PP PPP PP  P PPPP  PPP PP P   P PPPP PPP

PPPPPP PPPPPPPPPPP PPPPPP PPPPPPP PPP PPPP P PPPPP PPPP  P

Figure 2

Observation 1. The k  m  n box cannot be an N {box if it satis es at least one of the following conditions: (a) one of k; m and n is equal to 1; (b) two of k; m and n are equal to 2; (c) kmn is not divisible by 4. It follows that the 2  3  4 box is the smallest box which may be an N {box. Figure 3 shows that this is in fact the case. The box is drawn in two layers, and two dominoes with identical labels form a single N {tetracube.

0 1 3

0 2

1

2 3

Figure 3 So there is life in this universe after all! The main problem is to nd all N {boxes.

N {cubes

If k = m = n, the k  m  n box is called the k{cube, and a cube which can be packed by the N {tetracube is called an N {cube. We can easily assemble the 12{cube with the 2  3  4 N {box, which makes it an N {cube. This is a special case of the following result.

338 Observation 2. Suppose the k  m  n and `  m  n boxes are N {boxes. Let a, b and c be any positive integers. Then the following are also N {boxes: (a) (k + `)  m  n; (b) ak  bm  cn. The 12{cube is not the smallest N {cube. By Observation 1, the rst candidate is k = 4. It turns out that this is indeed an N {cube. It can be assembled from the 2  4  4 N {box, whose construction is shown in Figure 4.

0 1

3

0 1

2

2 3

Figure 4 The next candidate, the 6{cube, is also an N {cube, but a packing is not that easy to nd. In Figure 5, we begin with a packing of a 2  6  6 box, with a 1  2  4 box attached to it. To complete a packing of the 6{cube, add a 2  3  4 N {box on top of the small box, ank it with two 2  4  4 N {boxes and nally add two more 2  3  4 N {boxes.

0

1

2

0 3

4

5

1 6 7

2

3

8 9 4

6 7 8 9

5

Figure 5 Can we pack the 8{cube, the 10{cube, or others? It would appear that as size increases, it is more likely that we would have an N {cube. However, it is time to stop considering one case at a time. We present a recursive construction which expands N {cubes into larger ones by adding certain N {boxes. Theorem 1. The k{cube is an N {cube if and only if k is an even integer greater than 2. Proof: That this condition is necessary follows from Observation 1. We now

339 prove that it is sucient by establishing the fact that if the k{cube is an N { cube, then so is the (k + 4){cube. We can then start from either the 4{cube or the 6{cube and assemble all others. From the 2  3  4 and 2  4  4 N {boxes, we can assemble all 4  m  n boxes for all even m; n  4, via Observation 2. By attaching appropriate N {boxes from this collection, we can enlarge the k{cube rst to the (k + 4)  k  k box, then the (k + 4)  (k + 4)  k box and nally the (k + 4){cube. This completes the proof of Theorem 1.

Further Necessary Conditions Observation 1 contains some trivial necessary conditions for a box to be an N {box. We now prove two stronger results, one of which supersedes (c) in Observation 1. Lemma 1. The k  m  n box is not an N {box if at least two of k; m and n are odd. Proof: We may assume that m and n are odd. Place the box so that the horizontal cross-section is an m  n rectangle. Label the layers L1 to Lk from bottom to top. Colour the unit cubes in checkerboard fashion, so that in any two which share a common face, one is black and the other is white. We may assume that the unit cubes at the bottom corners are black. It follows that Li has one more black unit cube than white if i is odd, and one more white unit cube than black if i is even. Suppose to the contrary that we have a packing of the box. We will call an N {tetracube vertical if it intersects three layers. Note that the intersection of a layer with any N {tetracube which is not vertical consists of two or four unit cubes, with an equal number in black and white. The intersection of a vertical N {tetracube with its middle layer consists of one unit cube of each colour. Since L1 has a surplus of one black unit cube, it must intersect `1 vertical N {tetracubes in white and `1 + 1 vertical N {tetracubes in black, for some non-negative integer `1 . These N {tetracubes intersect L3 in `1 black unit cubes and `1 + 1 white ones. Hence the remaining part of L3 has a surplus of two black. They can only be packed with `3 vertical N {tetracubes intersecting L3 in white, and `3 + 2 vertical N {tetracubes in black, for some non-negative integer `3. However, the surplus in black unit cubes in L5 is now three, and this surplus must continue to grow. Thus the k  m  n box cannot be packed with the N {tetracube. This completes the proof of Lemma 1. Lemma 2. The k  m  n box is not an N {box if kmn is not divisible by 8.

340 Proof: Suppose a k  m  n box is an N {box. In view of Lemma 1, we may assume that at least two of k; m and n are even. Place the packed box so that the horizontal cross-section is an m  n rectangle, and label the layers L1 to Lk from bottom to top. De ne vertical N {tetracubes as in Lemma 1 and denote by ti the total number of those which intersects Li ; Li+1 and Li+2; 1  i  k , 2. Since at least one of m and n is even, each layer has an even number of unit cubes. It follows easily that each ti must be even, so that the total number of vertical N {tetracubes is also even. The same conclusion can be reached if we place the box in either of the other two non-equivalent orientations. Hence the total number of N {tetracubes must be even, and kmn must be divisible by 8. This completes the proof of Lemma 2.

N {Boxes of Height 2 We now consider 2  m  n boxes. By Observation 1, m  3 and n  3. By Lemma 2, mn is divisible by 4. We may assume that m is even. First let m = 4. We already know that the 2  4  3 and 2  4  4 boxes are N {boxes. Figure 6 shows that so is the 2  4  5 box. If the 2  4  n box is an N {box, then so is the 2  4  (n + 3) box by Observation 2. It follows that the 2  4  n box is an N {box for all n  3.

0

0

1 3

2

1

2 3

Figure 6 Now let m = 6. Then n is even. We already know that the 2  6  4 box is an N {box. However, the 2  6  6 box is not. Our proof consists of a long case-analysis, and we omit the details. On the other hand, the 2  6  10 box is an N {box. In Figure 7, we begin with the packing of a 2  3  6 box with a 2  2  3 box attached to it. We then build the mirror image of this solid and complete the packing of the 2  6  10 box by adding a 2  4  3 N {box. If the 2  6  n box is an N {box, then so is the 2  6  (n + 4) box by Observation 2. It follows that the 2  6  n box is an N {box for n = 4 and all even n  8. Theorem 2. The 2  m  n box is an N {box if and only if m  3; n  3 and mn is divisible by 4, except for the 2  6  6 box.

341

0

1

0 1

3 4

5

2

3

2 4

5

Figure 7

Proof: If m is divisible by 4, the result follows immediately from Observation 2. Let m = 4` + 2 for some positive integer `. We already know that the 2  10  6 box is an N {box. If the 2  (4` + 2)  n box is an N {box, then so is the 2  (4` + 6)  n box by Observation 2. This completes the proof of Theorem 2.

The Main Result Theorem 3. The k  m  n box, k  m  n, is an N {box if and only if it satis es all of the following conditions: (a) k  2; (b) m  3; (c) at least two of k, m and n are even; (d) kmn is divisible by 8; (e) (k;m; n) 6= (2; 6; 6). Proof: Necessity has already been established, and we deal with suciency. We may assume that k  3, since we have taken care of N {boxes of height 2. Consider all 3  m  n boxes. By (c), both m and n must be even. By (d), one of them is divisible by 4. All such boxes can be assembled from the 3  2  4 box. Consider now the k  m  n box. We may assume that m and n are even. If k is odd, then one of m and n is divisible by 4. Slice this box into one 3  m  n box and a number of 2  m  n boxes. Since these are all N {boxes, so is the k  m  n box. Suppose k is even. Slice this box into a number of 2  m  n boxes, each of which is an N {box unless m = n = 6. The 4  6  6 box may be assembled from the 4  2  3 box, and we already know that the 6{cube is an N {cube. If the k  6  6 box is an N {box, then so is the (k + 4)  6  6 box. This completes the proof of Theorem 3.

342

Research Projects Problem 1. Try to prove that the 2  6  6 box is not an N {box. It is unlikely that any elegant solution exists. Problem 2. An N {box which cannot be assembled from smaller N {boxes is called a prime N {box. Find all prime N {boxes. Problem 3. Prove or disprove that an N {box cannot be packed if we replace one of the N {tetracubes by an O{tetracube. Problem 4. For each of the other seven tetracubes, nd all boxes which it can pack.

Bibliography Bouwkamp, C. J. & Klarner, David, Packing a Box with Y {pentacubes, Journal of Recreational Mathematics, 3 (1970), 10-26. Gobel, F. & Klarner, David, Packing Boxes with Congruent Figures, Indagationes Mathematicae, 31 (1969), 465-472. Golomb, Solomon G., Polyominoes | Puzzles, Patterns, Problems & Packings, Princeton University Press, Princeton, 1994. Klarner, David, A Search for N{pentacube Prime Boxes, Journal of Recreational Mathematics, 12 (1979), 252-257.

Acknowledgement This article has been published previously in a special edition of delta-k, Mathematics for Gifted Students II, Vol. 33, 3 1996, a publication of the Mathematics Council of the Alberta Teachers' Association and in AGATE, Vol. 10, 1, 1996, the journal of the Gifted and Talented Education Council of the Alberta Teachers' Association. It is reprinted with permission of the author and delta-k.

343

THE SKOLIAD CORNER No. 24 R.E. Woodrow

This number we give the problems of the 1995 Concours Mathematique du Quebec. This contest comes to us from the organizers via the Canadian Mathematical Society which gives partial support to the contest. My thanks go to Therese Ouellet, secretary to the contest. The contest was written by over 2000 students February 2, 1997 over a three hour period. This will also test your French! We will, of course, accept solutions in either English or French.

  CONCOURS MATHEMATIQUE DU QUEBEC 1995 February 2, 1995

Time: 3 hours 1. LA FRACTION SIMPLIFIE E Simpli er la fraction

1 358 024 701 : 1 851 851 865

 2. LA FORMULE MYSTERE

Considerons les e quations suivantes

xy = p; x + y = s; x1993 + y1993 = t; x1994 + y1994 = u: En faisant appel aux lettres p, s, t, u mais pas aux lettres x, y , donnez une formule pour la valeur x1995 + y1995:  3. LA DIFFE RENCE ETONNANTE

Lorsque la circonference d'un cercle est divisee en dix parties e gales, les cordes qui joignent les points de division successifs forment un decagone regulier convexe. En joignant chaque point de division au troisieme suivant, on obtient un decagone regulier e toile. Montrer que la di erence entre les c^otes de ces decagones est e gale au rayon du cercle.

344

4. LE TERRAIN EN FORME DE CERF-VOLANT

Abel Belgrillet est membre du Club des aerocervidophiles (amateurs de cerfs-volants) du Quebec. Il dispose de quatre troncons de cl^oture rectilignes AB, BC , CD, DA pour delimiter un terrain (en forme de cerf-volant, voir gure) sur lequel il s'adonnera a son activite favorite cet e te. Sachant que les troncons AB et DA mesurent 50 m chacun et que les troncons BC et DA mesurent 120 m chacun, determiner la distance entre les points A et C qui maximisera l'aire du terrain.

50

D

120

A

C 50

B

120

 5. L'INEGALIT E MODIFIE E D'AMOTH DIEUFUTUR

(a) (2 points) L'inegalite x2 + 2y 2  3xy est-elle vraie pour tous les entiers? (b) (8 points) Montrer que l'inegalite x2 + 2y 2  145 xy est valide pour tous reels x et y .  6. L'ECHIQUIER COQUET Trouver l'unique facon de colorier les 36 cases d'un e chiquier 6  6 en noir et blanc de sorte que chacune des cases soit voisines d'un nombre impair de cases noires. (Note: deux cases sont voisines si elles se touchent par un c^ote ou par un coin). On ne demande pas de demontrer que la solution est unique. 7. LA FRACTION D'ANNE GRUJOTE  En base 10, 13 = 0:333 : : : . Ecrivons 0:3 pour ces decimles repetees. 1 Comment e crit-on 3 dans une base b, ou b est de la forme (i) b = 3t (trois points), (ii) b = 3t + 1 (trois points), (iii) b = 3t + 2 (quatre points), ou t est un entier positif quelconque?

345 To nish this number of the Skoliad Corner we give the ocial solutions to the 1995 P.E.I. Mathematics Competition given last issue [1997: 278]. My thanks go to Gordon MacDonald, University of Prince Edward Island for forwarding the materials.

1995 P.E.I. Mathematics Competition

1. Find the area of the shaded region inside the circle in the following

gure.

61 ,1

T2

O

T1

P

S

-1

?,1

Solution. First determine the coordinates of the point P . Since P lies on the line y = x, P = (a; a) for some value of a. Since P lies on the circle x2 + y2 = 1, a2 + a2 = 1 and so a = p,12 . Hence P = ( p,12 ; p,12 ). Partition the shaded region as shown. Then the area of the quarter circle S is 4 and the area of the triangle T1 is 12 (base)(height). The base is 1 and the height is p12 . (Turn the page upside down.) So the area of T1 is 2p1 2 . The triangle T2 has the same area so the total area of the shaded region is

 + p1 : 4 2

2. \I will be n years old in the year n2", said Bob in the year 1995.

How old is Bob? Solution. Assuming Bob's age is an integer, Bob must have a reasonable chance to be alive in a year that is a perfect square. Since 442 = 1936, 452 = 2025 and 462 = 2116, Bob must be 45 years old in the year 2025. Hence Bob is now 15 years old (or 14 years old if he hasn't had his birthday yet this year). 3. Draw the set of points (x; y) in the plane which satisfy the equation jxj + jx , yj = 4.

346 Solution. Consider the following four possible cases: Case 1: When x  0 and x  y . Then x + x , y = 4 so 2x , y = 4. Case 2: When x  0 and x < y . Then x , (x , y ) = 4 so y = 4. Case 3: When x < 0 and x  y . Then ,x + x , y = 4 so y = ,4. Case 4: When x < 0 and x < y . Then ,x , (x , y ) = 4 so ,2x + y = 4. Thus the set of points is made up of four line segments. When these line segments are drawn we obtain the following parallelogram: y

4

{axis

6r

r

2

,4

d

,2

2

4

- {axis x

,2 r

,4

r

4. An autobiographical number is a natural number with ten digits or less in which the rst digit of the number (reading from left to right) tells you how many zeros are in the number, the second digit tells you how many 1's, the third digit tells you how many 2's, and so on. For example, 6; 210; 001; 000 is an autobiographical number. Find the smallest autobiographical number and prove that it is the smallest. Solution. When you add the digits of an autobiographical number, you count the total number of digits. (For example the digits of the above 10 digit autobiographical number must sum to 10.) Using this fact and the process of elimination we can nd the smallest autobiographical number. The only possible one-digit number whose digits sum to 1 is 1 and it is not autobiographical so there are no one digit autobiographical numbers. The only possible two digit numbers whose digits sum to two are (in increasing order) 11 and 20 and they are not autobiographical so there are no two digit autobiographical numbers. The only possible three digit numbers whose digits sum to three are (in increasing order) 102; 111; 120; 201; 210 and 300 and they are not autobiographical so there are no three digit autobiographical numbers. The possible four digit numbers whose digits sum to four are (in increasing order) 1003; 1012; 1021; 1030; 1102; 1111; 1120; 1201; : : : .

347 Checking these numbers we nd that the rst autobiographical number in this list, and hence the smallest autobiographical number is 1201.

5. A solid cube of radium is oating in deep space. Each edge of the cube is exactly 1 kilometre in length. An astronaut is protected from its radiation if she remains at least 1 kilometre from the nearest speck of radium. Including the interior of the cube, what is the volume (in cubic kilometres) of space that is forbidden to the astronaut? (You may assume that the volume of a sphere of radius r is 43 r3 and the volume of a right circular cylinder of radius r and height h is r2 h.) Solution. Along with the cube of radium, on each of the six faces of the cube there will be a cube with sides of length 1 km of forbidden space. Along each of the 12 edges of the cube of radium there will be a quarter of a cylinder of radius 1 km and length 1 km of forbidden space, and at each of the eight corners of the cube of radium there will be an eighth of a sphere of radius 1 km of forbidden space. Putting it all together, there are 7 cubes (with sides of length 1 km), 3 cylinders of radius 1 km and length 1 km, and 1 sphere of radius 1 km of forbidden space. So the volume of forbidden space is: 7 + 3 + 34  km3 = 7 + 16  km3 : 3

6. Which is greater, 999! or 500999? (Where 999! denotes 999 factorial,

the product of all the natural numbers from 1 to 999 inclusive.) Explain your reasoning. Solution. Write

999! = 1  2  3  4    995  996  997  998  999

and successively pair o numbers from the left with numbers from the right so

999! = (1  999)(2  998)(3  997)(4  996)    (449  501)(500): (Note that no number is paired with 500 and there are 499 such pairs.) Now use the fact that (500 , k)(500+ k) = 5002 , k2 < 5002 for each value of k from 1 to 499 so 999! < (5002)(5002)(5002)(5002)    (5002)(500) where 5002 is repeated 499 times. Hence 999! < (5002)499(500) = 5002(499)+1 = 500999; so 500999 is greater.

348

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. All material intended for inclusion in this section should be sent to the Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University, PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic address is still [email protected] The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto). The rest of the sta consists of Richard Hoshino (University of Waterloo), Wai Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices

Invariants of Inscribed Regular n-gons There are two unexpected (in their simultaneity) invariants of inscribed regular n-gons. Let P1P2    Pn be a regular n-gon, and let P be a point on its circumcircle. Then both PP12 + PP22 +    + PPn2 and PP14 + PP24 +    + PPn4 are independent of the position of P . Proof. We will show the result using complex numbers. Without loss of generality, assume the radius of the circumcircle is 1, and that Pk is at the complex number ! k,1 in the complex plane, where ! = cis( 2n ). Let z be the complex number corresponding to P , so jz j = 1, and zz = jz j2 = 1. Note

PPk2+1 = jz , !kj2 = (z , !k)(z , !k) = zz , !kz , ! kz + ! k ! k = 2 , ! kz , ! n,k z; k2 since ! k = j!!kj = !1k = !!nk = ! n,k . Hence, PP12 + PP22 +    + PPn2 =

nX ,1 k=0

jz , !kj2 = nX ,1

= 2n , k=0 = 2n;

!k

!

nX ,1 k=0

z,

(2 , ! kz , ! n,k z ) nX ,1 k=0

!n,k

!

z

349 since

nX ,1 k=0

!k = 11,,!! = 0: n

Therefore, the sum is indeed independent of z . Similarly,

so

PPk4+1 = (2 , !kz , !n,kz)2 = 4 + ! 2kz2 + ! 2(n,k)z 2 , 4!kz , 4! n,k z + 2 = 6 + ! 2kz2 + ! 2(n,k)z 2 , 4!kz , 4! n,k z; PP14 + PP24 +    + PPn4 nX ,1 = (6 + ! 2k z 2 + ! 2(n,k)z 2 , 4! kz , 4! n,k z ) k=0

= 6n +

,4

nX ,1 k=0 nX ,1 k=0

!

2k !

!

z

2+

!k z , 4

nX ,1 k=0 nX ,1 k=0

!

2(n,k)

!

z2

!

!n,k z = 6n:

Four Large Spheres and a Small Sphere Wai Ling Yee Problem: Given four identical spheres of radius R which are mutually tangent, nd the radius r of the sphere that will t in the hole at the centre of the tetrahedron that they form, in terms of R. When I rst tried this problem, I looked at several triangles and, after many calculations, produced an answer. I came across this problem again in chemistry when calculating the size of tetrahedral holes in ion lattices. Try this: Construct a cube so that the centres of the spheres are at alternate corners of the cube. The length of a face diagonalpof the cube is 2R. Therefore, the length of a body diagonal of the cube is 6R. Note that we can also express the length of the body diagonal as 2(R + r). Hence,

p

p 2(R + r) = 6R implies that r = 62, 2 R:

Rider. This is the radius of the sphere that is externally tangent to all four. What is the radius of the sphere that is internally tangent to all four?

350

IMO REPORT Adrian Chan Two weeks of training at St. Mary's University in Halifax in early July kicked o the month-long journey the 1997 Canadian IMO team would endure. Four ights later, the team found itself in Mar del Plata, Argentina, ready to participate in the 37th International Mathematical Olympiad. This year's team members included: Adrian \Terrible Taco" Birka, Sabin \Cursed Corner" Cautis, Adrian \Flash Flood" Chan, Jimmy \Fruits and Veggies" Chui, Byung Kyu \Argentinian Polite" Chun, and Mihaela \My Dearest" Enachescu. We gave our team leader Dr. Richard \Richard" Nowakowski and our deputy leader Naoki \Raspberry" Sato all they could handle. Special thanks goes out to our coaches Dr. Bruce Shawyer and leader observer Dr. Chris \Bob's Your Uncle" Small. This year's contest was not as dicult as last year's killer in India, but still challenging. It was the largest IMO ever, with a record 82 countries competing. Canada was well represented, bringing back 2 silver, 2 bronze, and an honourable mention. Our team's scores were as follows: CAN1 Adrian Birka 6 CAN2 Sabin Cautis 16 Bronze Medal CAN3 Adrian Chan 25 Silver Medal CAN4 Jimmy Chui 10 Honourable Mention CAN5 Byung Kyu Chun 29 Silver Medal CAN6 Mihaela Enachescu 21 Bronze Medal As a team, Canada nished 29 out of the 82 countries. Best of luck to the graduating members of the team, Sabin and Byung, as they take their math skills to the University of Waterloo next year. The remaining four members of the team are eligible to return to Taiwan for next year's IMO. However, it seems that the team should focus more in the area of geometry! Special thanks must also go out to Dr. Graham Wright of the Canadian Mathematical Society for paying all our bills, and Professor Ed Barbeau of the University of Toronto for conducting a year-long correspondence program for all IMO hopefuls. Visiting Argentina was a new experience for many of us, and was definitely a memorable one to all. The food was great, especially the beef and the chocolate alfajores, and the IMO was very well-organized and run. Best of luck to all IMO hopefuls who are working hard to be on the 1998 IMO team, which will compete in Taipei, Taiwan. th

351

Unique Forms Naoki Sato For the mathematician, there is something re-assuring about being able to put an object into a unique form, or a canonical form, such as the prime factorization of a positive integer, or a matrix in Jordan Canonical Form. Such forms make working with these objects much easier and more tractable, as well as allowing a way to get a handle on them. Here, we explore the idea of the unique, canonical form. 1. Let p(x) = an xn + an,1 xn,1 +    + a1 x + a0 , and let c be a real number. Show that there exist unique constants bn , bn,1 , : : : , b1, b0, such that

p(x) = bn(x , c)n + bn,1(x , c)n,1 +    + b1(x , c) + b0: Can you nd the bi explicitly?

What may seem like a lot of (linear) algebra to untangle, comes easily undone with one idea: consider p(x + c). This is a polynomial of degree at most n, so there exist unique bi such that

p(x + c) = bnxn + bn,1xn,1 +    + b1x + b0;

which is equivalent to

p(x) = bn(x , c)n + bn,1(x , c)n,1 +    + b1(x , c) + b0: The expression may look familiar, because it is the n approximant to a power series around x = c, a central object of calculus. Suppose we take the k derivative of both sides. The \constant term" is then k!bk, so evaluating both sides at x = c leads to (k) p(k)(c) = k!bk so that bk = p k!(c) : A quick corollary: If p(x) is a polynomial, then (x , c)k j p(x) if and only if p(c) = p0 (c) = p00 (c) =    = p(k,1) (c) = 0. (Both conditions are equivalent to b0 = b1 =    = bk,1 = 0.) 2. Let N be a non-zero integer. Show that there exist unique positive integers k and a1 < a2 <    < ak such that N = (,2)ak + (,2)ak,1 +    + (,2)a1 : (In other words, show that there is a unique representation of N in base ,2.) th

th

352 We will use strong induction. The statement is true for N = 1 = (,2)0 and for N = ,1 = (,2)1 + (,2)0. Now, assume there is a positive integer n such that the statement is true for N = 1; 2; : : : ; n and N = ,1; ,2; : : : ; ,n. Let N = n + 1. If N is odd, then n is even, and n=(,2) is an integer between ,1 and ,n, so by the induction hypothesis, for some unique distinct ai ,

n=(,2) = (,2)ak + (,2)ak,1 +    + (,2)a1 ;

which is equivalent to

N = n + 1 = (,2)ak+1 + (,2)ak,1+1 +    + (,2)a1+1 + (,2)0: Note that the exponents are still distinct. The same trick works if N is even, and for N = ,(n + 1). Subtracting o 1 = (,2)0 if necessary, and then dividing by ,2, we obtain a form which is simultaneously seen to exist and be unique, by the induction hypothesis. Hence, by strong induction, the statement is proved for all N . Note that this also provides an algorithm for nding the exponents. One may be tempted to isolate the higher exponents, but this tends to get very messy. Instead, one can start with the lower exponents - after seeing if (,2)0 is in the sum, they can then be picked o inductively. 3. Let r be a positive rational number. Show that there exist unique positive integers k and a1  a2      ak such that

r = a1 + a 1a + a a1 a +    + a a a1    a : 1 1 2 1 2 3 1 2 3 k

Note that

r = a1 + a 1a + a a1 a +    + a a a1    a ; 1 1 2 1 2 3 1 2 3 k

which is equivalent to

a1 r , 1 = a1 + a 1a +    + a a 1   a : 2 2 3 2 3 k This suggests nding the right a1 , reducing, and applying the same argument to a1 r , 1. De ne the sequences fai g and frig as follows: set r1 = r, ai = d1=ri e, ri+1 = airi , 1, i  1, and we terminate when ri+1 = 0. For example, if r1 = 3=7, then we obtain a1 = d7=3e = 3, r2 = 3  3=7 , 1 = 2=7, a2 = 4, r2 = 1=7, a3 = 7, and r3 = 0. And we see 3=1+ 1 + 1 : 7 3 34 347

353 We will show that in general, this is the only possible sequence which works. Consider the choice of a1 . Since a1 r  1, a1 must be at least d1=re. However, if a1 was set to d1=re + 1, then

r = a1 + a 1a + a a1 a +    + a a a1    a 1 1 2 1 2 3 1 2 3 k 1 1 1 < a + a2 + a3 +    1

1

1

1

= a , 1 = d11=re  11=r 1 = r;

contradiction. Hence, there is only one choice for a1 , the one we have chosen above. By the above reduction, there is only one choice for a2 , a3 , and so on. Hence, we have uniqueness. We also know terms will be generated, but how do we know the sequence always terminates? Let ri = pi=qi, as a reduced fraction. Then ai = dqi=pie, so l

m

  pi pqii , qi q p i i ri+1 = q p , 1 = qi : i i

However,

   q q i i pi p , qi < pi p + 1 , qi = pi: i i Hence, the numerators of the ri are strictly decreasing. But as they are all 

positive integers, the sequence must terminate at some point. Finally, we must show the sequence fai g is non-decreasing. We have 2

ai+1 = 66 6

pi

l

qmi

qi pi

3

,q

7 7 i7



pi

l

qmi

qi pi

  > pqi > pqi , 1 = ai , 1: i , qi i

Therefore, ai+1  ai . Rider. Does an analogous p result hold for positive irrational numbers? What is the sequence for 2? For e? 4. Let be a root of x3 , x , 1 = 0, and let p, q be polynomials with rational coecients, such that q ( ) 6= 0. Show that there exist unique rationals a, b, and c such that

p( ) = a 2 + b + c: q( )

354 First, let us express p( )=q ( ) as a polynomial in . Note that x3 ,x,1 is irreducible over the rationals. Hence, there exist polynomials u(x) and v(x) with rational coecients such that

u(x)q(x) + v(x)(x3 , x , 1) = gcd(q(x);x3 , x , 1) = 1: Then by substituting x = , u( )q( ) = 1 implies that pq((

)) = p( )u( ) = r( ); where r(x) = p(x)u(x), a polynomial with rational coecients. Now, by the division algorithm for polynomials,

r(x) = (x3 , x , 1)w(x) + ax2 + bx + c; for some rationals a, b, and c (dividing r(x) by x3 , x , 1 leaves, at most, a quadratic remainder). Hence, r( ) = a 2 + b + c, which shows existence. To see uniqueness, suppose r( ) = a0 2 + b0 + c0 implies that (a , a0 ) 2 + (b , b0 ) + (c , c0 ) = 0; that is, is a root of the quadratic (a , a0 )x2 + (b , b0 )x + (c , c0 ) = 0. But as indicated above, the cubic x3 , x , 1 is irreducible; hence, all the coecients of the quadratic must be zero, and a = a0 , b = b0 , c = c0 , showing uniqueness. 5. Let a and b be distinct positive integers, and let d = gcd(a; b). Consider the Diophantine equation ax + by = d. It is well-known that all solutions are given by (x; y ) = (u + bt=d; v , at=d), where t is an integer and (u; v ) is some solution. (i) Show that there exists a unique solution satisfying jxj  b=2d, jyj  a=2d. (ii) Show that the solution in (i) is given by the Euclidean Algorithm. For example, for a = 4 and b = 7, we have

7 = 1  4 + 3; 4 = 1  3 + 1; 3 = 1  3;

so gcd(4,7) = 1, and

1 = 4 , 3 = 4 , (7 , 4) = 2  4 , 7: Hence, the Euclidean Algorithm gives the solution (x; y ) = (2; ,1), which satis es jxj  7=2 and jy j  4=2, We can assume d = 1, as a way of normalizing the equation, since ax + by = d is equivalent to (a=d)x + (b=d)y = 1. It is easy to check that d may be factored back in.

355 (i) Uniqueness is immediate, since the solutions in x form an arithmetic progression with common di erence b=d = b, and similarly for y , with di erence a=d = a. If b = 1, then (x; y ) = (0; 1) serves as a solution. If b = 2, then a is odd, and (x; y ) = (1; (1 , a)=2) serves as a solution. Assume that b > 2. Because the solutions in x form an arithmetic progression with di erence b, there exists a solution (x; y ) with jxj  b=2, or ,b=2  x  b=2. Then y = (1 , ax)=b, so and

b y  1 , ba  2 = 1b , a2 > , a2 ;

b y  1 + ba  2 = 1b + a2 < a +2 1 ; so y  a=2 implies that jy j  a=2.

(ii) We must qualify the statement somewhat | if either a or b is equal to 1, then the Euclidean Algorithm does not explicitly provide a solution. We may stipulate now that the solution is (x; y ) = (1; 0) or (0,1), if a or b is 1, respectively (they cannot both be 1, since they must be distinct). We will use strong induction on max(a; b). The statement is easily veri ed for max(a; b)  2. Assume the statement is true for max(a; b)  n, for some positive integer n. We will prove the statement for max(a; b) = n + 1. Assume b = n + 1. We seek solutions of ax + (n + 1)y = 1. By the division algorithm, there exist unique non-negative integers q and r such that n + 1 = aq + r, 0  r  a , 1. Then re-arranging, the equation becomes ax + (aq + r)y = a(x + qy) + ry = 1. Since a, r  n, by the induction hypothesis, there exist solutions satisfying jx + qy j  r=2, jy j  a=2. By the triangle inequality, jxj  r=2+ jqyj  r=2+ qa=2 = (r + qa)=2 = (n +1)=2. Hence, by strong induction, the statement is proved for all values of max(a; b).

1996 Balkan Mathematical Olympiad 1. Let O and G be the circumcentre and centroid of a triangle ABC respectively. If R is the circumradius and r is the inradius of ABC , show that p

OG  R(R , 2r) : 2. Let p > 5 be a prime number and X = fp , n2 jn 2 Z+; n2 < pg. Prove that X contains two distinct elements x, y , such that x 6= 1 and x divides y.

356 3. Let ABCDE be a convex pentagon. Denote by M , N , P , Q, R the midpoints of the segments AB , BC , CD, DE , EA respectively. If the segments AP , BQ, CR, DN have a common point of intersection, prove that this point also belongs to the segment EN . 4. Show that there exists a subset A of the set f1; 2; 3; : : : ; 21996 , 1g having the following properties: (a) 1 2 A and 21996 , 1 2 A, (b) Every element of Anf1g is the sum of two (not necessarily distinct) elements of A, and (c) The number of elements of A is at most 2012.

Mayhem Problems The Mayhem Problems editors are: Richard Hoshino Mayhem High School Problems Editor, Cyrus Hsia Mayhem Advanced Problems Editor, Ravi Vakil Mayhem Challenge Board Problems Editor. Note that all correspondence should be sent to the appropriate editor | see the relevant section. In this issue, you will nd only solutions | the next issue will feature only problems. We warmly welcome proposals for problems and solutions. With the new schedule of eight issues per year, we request that solutions from the previous issue be submitted by 1 November 1997, for publication in the issue 5 months ahead; that is, issue 4 of 1998. We also request that only students submit solutions (see editorial [1997: 30]), but we will consider particularly elegant or insightful solutions from others. Since this rule is only being implemented now, you will see solutions from many people in the next few months, as we clear out the old problems from Mayhem.

357

High School Problems | Solutions Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario, Canada. L3S 3E8 H211. Proposed by Richard Hoshino, grade OAC, University of Toronto Schools Let p, q , and r be the roots of the cubic equation ax3 , bx2 + cx + d = 0, where a, b, c, and d are real coecients. Given that Arctan p + Arctan q + Arctan r = 4 , where Arctan denotes the principal value, prove that a = b + c + d.  Solutionby Miguel Carrion  Alvarez, Universidad Compluteuse de Madrid, Spain. Let  = Arctan p,  = Arctan q , and = Arctan r. Then we have  +  + = 4 and +tan   + ) + tan = 1tan ,tan  tan  + tan 1 = tan( +  + ) = 1tan( +tan  , tan( + ) tan 1 , 1tan ,tan  tan  tan  + tan  + tan , tan  tan  tan = p + q + r , pqr : = 1tan , tan  tan  , tan  tan , tan  tan 1 , pq , pr , qr

Finally, a(x,p)(x,q )(x,r) = a[x3 ,(p+q +r)x2 +(pq +pr +qr)x,pqr] = ax3 , bx2 + cx + d. From the relationship between the coecients and the roots of this cubic, we get p + q + r = ab , pq + pr + qr = ac , and pqr = , da . Using these relations in the above equation, we have b d 1 = 1a ,+ ca ) 1 , ac = ab + ad ) a = b + c + d : a

Also solved by BOB PRIELIPP, University of Wisconsin-Oshkosh, WI, USA. Prielipp also points out that this problem is similar to problem E2299 (pp. 520-521 of the May 1972 issue of The American Mathematical Monthly.) H212. Given two sequences of real numbers of length n, with the property that for any pair of integers (i;j ) with 1  i < j  n, the i term is equal to the j term in at least one of the two sequences. Prove that at least one of the sequences has all its values the same.  Solutionby Miguel Carrion  Alvarez, Universidad Compluteuse de Madrid, Spain. Let the two sequences be fai g and fbig. Assume that fai g does not have all its values the same. Then, there is a pair (i; j ) such that ai 6= aj , and hence bi = bj . Now, it is impossible that both ak = ai and ak = aj . Hence, either ak 6= ai or ak 6= aj . In either case, bk = bi = bj . This is true for all k, so the sequence of bi 's has all its values the same. th

th

358

H213. Two players begin with one counter each, initially on opposite corners of an n  n chessboard. They take turns moving their counter to an adjacent square. A player wins by being the rst to reach the row opposite from their initial starting row, or landing on the opponent's counter. Who has the winning strategy? (Generalize to an m  n board.)  Solutionby Miguel Carrion  Alvarez, Universidad Compluteuse de Madrid, Spain. On a square board, the rst player always wins by crossing the board to the opposite row. In n , 2 moves, t

d

becomes

?t ?d 

and the rst player wins. If there are more columns than rows, the same strategy works for the rst player. If there are more rows than columns, one of the players wins by \capturing" the opponent. If the parity of both dimensions of the board is the same, both players start from squares of the same colour. In this case, the rst player always moves to the opposite colour and cannot capture the second player. The second player wins by \closing in" on the rst player and capturing the rst player. [Ed: Why is this always possible?] Now if the parity of both dimensions of the board is opposite, the rst player moves towards the colour of the second player, and so can capture the second player using the same strategy.

Advanced Problems | Solutions Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada. M1G 1C3 A190. Find all positive integers n > 1 such that ab  ,1(mod n) implies that a + b  0(mod n). Solution by Q Wai Ling Yee, University of Waterloo, Waterloo, Ontario. Let n = ki=1 p i i where the pi are distinct primes. By the Chinese Remainder Theorem, the given property holds for n if and only if it holds for each p i i ; that is, ab  ,1(mod p i i ), implies that a + b  0(mod p i i ). Also, both a and b must be relatively prime with n.

359 Let Si be the set of positive integers both relatively prime with and less than p i i . Take x, u, and v 2 Si. Note that ux  vx mod p i i is equivalent to x(u , v )  0(mod p i i ), which, in turn, is equivalent to u = v , since x is relatively prime with p i i and u and v are less than p i i . Hence, each element of fx; 2x; : : : ; (pi , 1)x; (pi + 1)x; : : : ; (p i i , 1)xg leaves a unique residue modulo p i i . Also note that the product of two numbers in Si is itself in Si . Therefore, the set fx; 2x; : : : ; (pi , 1)x; (pi + 1)x; : : : ; (p i i , 1)xg taken modulo p i i is a permutation of Si. Since ,1 2 Si, for each x 2 Si, there exists a unique y 2 Si such that xy  ,1(mod p i i ). Case I: pi is odd. Let a = 2. Then there exists a b such that 2b  ,1(mod p i i ). Thus, 2 + b  0(mod p i i ), so that 4  1(mod p i i ). Therefore 3  0(mod pi i ), from which it follows that pi = 3, i = 1. Therefore, the only possible odd prime factor of n is 3, and only one such factor may appear. By a simple check, n = 3 satis es the given property. Case II: pi = 2. Let a = 3. Then there exists a b such that 3b  ,1(mod 2 i ). Thus 3 + b  0(mod 2 i ), so that 9  1(mod 2 i ). Therefore 2 i j 8, from which it follows that i  3. Thus n can have 0, 1, 2, or 3 factors of 2, and again by a simple check, all satisfy the given property. From the above two cases, the solutions for n are 2, 3, 4, 6, 8, 12, and 24. This problem, which generalizes problem B-1 on the 1969 Putnam Competition, is equivalent to nding all n > 1, such that a relatively prime to n implies that a2  1(mod n). A191. Taken over all ordered partitions of n, show that   k1k2    km = m2+mn,,1 1 : k1+k2 ++km =n X

Solution. The problem is particularly suitable for generating functions. As is convention, let [xn ]f (x) denote the coecient of xn in f (x). Consider each ki being contributed from a factor of x + 2x2 + 3x3 +    . Then the number we seek is:

[xn ] |(x + 2x2 +    )(x + 2x2 {z+    )    (x + 2x2 +    )} m

factors

m = [xn ] (1 ,x x)2m = [xn,m ] (1 ,1x)2m        2 m , 1 2 m 2 m + 1 n , m 2 = [x ] 2m , 1 + 2m , 1 x + 2m , 1 x +      m + n , 1 = 2m , 1 :

360

A192. Let ABC be a triangle, such that the Fermat point F lies in the interior. Let u = AF , v = BF , w = CF . Derive the expression 2 2 2 p u2 + v2 + w2 , uv , uw , vw = a + b2 + c , 2 3K:

Solution by Wai Ling Yee, University of Waterloo, Waterloo, Ontario. Using the law of cosines on 4BCF , 4ACF , and 4ABF : a2 = v2 + w2 , 2vw cos 120 = u2 + v2 + vw; (1) 2 2 2  2 2 b = u + w , 2uw cos 120 = u + v + uw; (2) c2 = u2 + v2 , 2uv cos 120 = u2 + v2 + uv: (3)

Adding the areas of 4BCF , 4ACF , and 4ABF :

K = 12 uv sin 120 + 21 uw sin120 + 21 vw sin120 p3 = 4 (uv + vw + uw): (4) p Therefore, u2 + v 2 +p w2 , uv , uw , vw = a2 +b22 +c2 , 2 3K , from 1 [(1) + (2) + (3)] , 2 3(4). 2 Since u2 + v 2 + w2 , uv , uw , vwp 0 for all real u, v , w, it follows from this problem that a2 + b2 + c2  4 3K .

Challenge Board Problems | Solutions Editor: Ravi Vakil, Department of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544{1000 USA

C70. Prove that the group of automorphisms of the dodecahedron is S5, the symmetric group on ve letters, and that the rotation group of the dodecahedron (the subgroup of automorphisms preserving orientation) is A5 . Solution. Assign a number to each of the dodecahedron's edges as shown in Figure 1. (Alternatively, think of it as painting each of the edges one of ve di erent colours.) Notice that if an edge on a face F is numbered n, then the edge from the opposite vertex in F (that isn't an edge of F ) has the same number. Thus the entire numbering can be recovered from the numbering of the edges of a simple face. Notice also that the edges around each face are numbered 1 through 5 (in some order).

361 1 3 5

2 3

4 1

4

5 4

1 5 2

2 4

2 3 5

3

3

2 1

5

5

4

3

2

4

3 1

Figure 1. 



By observation, if a1 2b 3c d4 5e is an element of A5 , then there is a unique face where the edges are numbered (clockwise) a-b-c-d-e. In  1 2 3 4 5 this way each element a b c d e of A5 induces a rotation of the dodecahedron (by sending the face 1-2-3-4-5 to the face a-b-c-d-e, with edge 1 of the rst face sent to edge a of the second). Conversely, each rotation induces an element of A5 . This set map is clearly a group homomorphism, so the group of rotations of the dodecahedron is A5 . Essentially the same argument shows that the group of automorphisms of the dodecahedron (including re ections) is S5 . Comments. 1. What are the rotation and automorphism groups of the icosahedron? How about the other platonic solids? 2. The group A5 comes up often in higher mathematics. For example, because of a simple property of A5 , there are quintic polynomials with integer co-ecients whose roots can't be written in terms of radicals. (This is not true of polynomials of degree less than 5. The proof involves Galois theory.) Surprisingly, the description of A5 as the automorphism group of the dodecahedron comes up in a variety of contexts.

C72. A nite group G acts on a nite set X transitively. (In other words, for any x; y 2 X , there is a g 2 G with g  x = y .) Prove that there is an element of G whose action on X has no xed points.

362 (The problem as stated is incorrect. We need to assume that X has more than one element.) Solution. Fix any element x0 of X . The number of elements of G sending x0 to y is independent of y 2 X : if y1 and y2 are any two elements of X , and

Si = fg 2 G j g  x0 = yig; i = 1; 2; and z is an element of G sending y1 to y2 (z exists by the transitivity of the group action), then zS1 = S2. In other words, the elements of S2 are obtained from the elements of S1 by multiplication on the left by z . In particular, jS1 j = jS2j. Let f (g;x) = 1 if g  x = x and 0 otherwise. Then by the previous comment, for any xed x 2 X , P g2G f (g;x) = 1 : jGj jX j

The average number of xed points (over all elements of the group) is 1:

1

jGj

X

X

g2G x2X

!

f (g;x)

= =

X

x2X X

x2X

= 1:

P

1

g2G f (g;x)



jGj

jX j

One element of the group (the identity) has more than the average (as jX j > 1), so there is an element with less than the average number (and hence zero) xed points. Comments. 1. How does this relate to problem 1 on the 1987 IMO?: Let pn (k) be the number of permutations of the set f1; : : : ; ng, n  1, which have exactly k xed points. Prove that n X

k=0

kpn(k) = n!:

(Hint: Let X be the set f1; : : : ; ng and G the group of permutations on X , acting on X in the natural way.) 2. Can the result be salvaged if the group action is not transitive?

363

PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later than 1 April 1998. They may also be sent by email to [email protected]. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication.

Solutions submitted by FAX There has been an increase in the number of solutions sent in by FAX, either to the Editor-in-Chief's departmental FAX machine in St. John's, Newfoundland, or to the Canadian Mathematical Society's FAX machine in Ottawa, Ontario. While we understand the reasons for solvers wishing to use this method, we have found many problems with it. The major one is that hand-written material is frequently transmitted very badly, and at times is almost impossible to read clearly. We have therefore adopted the policy that we will no longer accept submissions sent by FAX. We will, however, continue to accept submissions sent by email or regular mail. We do encourage email. Thank you for your cooperation.

364

2263.

Proposed by Toshio Seimiya, Kawasaki, Japan. ABC is a triangle, and the internal bisectors of \B, \C , meet AC , AB at D, E, respectively. Suppose that \BDE = 30. Characterize 4ABC . 2264. Proposed by Toshio Seimiya, Kawasaki, Japan. ABC is a right angled triangle with the right angle at A. Points D and E are on sides AB and AC respectively, such that DEkBC . Points F and G are the feet of the perpendiculars from D and E to BC respectively. Let I , I1 , I2 , I3 be the incentres of 4ABC , 4ADE , 4BDF , 4CEG respectively. Let P be the point such that I2 P kI1 I3 , and I3 P kI1 I2 . Prove that the segment IP is bisected by the line BC . 2265. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. Given triangle ABC , let ABX and ACY be two variable triangles constructed outwardly on sides AB and AC of 4ABC , such that the angles \XAB and \Y AC are xed and \XBA + \Y CA = 180. Prove that all the lines XY pass through a common point. 2266. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. BCLK is the square constructed outwardly on side BC of an acute triangle ABC . Let CD be the altitude of 4ABC (with D on AB ), and let H be the orthocentre of 4ABC . If the lines AK and CD meet at P , show that

HP = AB : PD CD

2267. Proposed by Clark Kimberling, University of Evansville, Evansville, IN, USA and Peter Y , Ball State University, Muncie, IN, USA. In the plane of 4ABC , let F be the Fermat point and F 0 its isogonal conjugate. Prove that the circles through F 0 centred at A, B and C meet pairwise in the vertices of an equilateral triangle having centre F . 2268. Proposed by Juan-Bosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Let x, y be real. Find all solutions of the equation s

2xy + x2 + y 2 = pxy + x + y : x+y 2 2

365

2269. Proposed by Cristobal  S a nchez{Rubio, I.B. Penyagolosa, Castellon,  Spain. Let OABC be a given parallelogram with \AOB = 2 (0; =2]. A. Prove that there is a square inscribable in OABC if and only if and

OA  sin + cos sin , cos  OB

sin , cos  OB OA  sin + cos :

B. Let the area of the inscribed square be Ss and the area of the given parallelogram be Sp . Prove that

2Ss = tan2 OA2 + OB 2 , 2Sp : ,



2270.

Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. Given 4ABC with sides a, b, c, a circle, centre P and radius  intersects sides BC , CA, AB in A1 and A2 , B1 and B2 , C1 and C2 respectively, so that

A1A2 = B1B2 = C1C2 =   0: a b c Determine the locus of P .

2271.

Proposed by F.R. Baudert, Waterkloof Ridge, South Africa. A municipality charges householders per month for electricity used according to the following scale: rst 400 units | 4.5cj per unit; next 1100 units | 6.1cj per unit; thereafter | 5.9cj per unit. If E is the total amount owing (in dollars) for n units of electricity used, nd a closed form expression, E (n). 2272?. Proposed by no name on proposal { please identify! Write r - s if there is an integer k satisfying r < k < s. Find, as a function of n (n  2), the least positive integer k satisfying

k - k - k - : : : - k - k: n n,1 n,2 2

[The proposer has not seen a proof, and has veri ed the conjectured solutions for 2  n  600.]

366

2273. Proposed by Tim Cross, King Edward's School, Birmingham, England. Consider the sequence of positive integers: f1, 12, 123, 1 234, 12 345, : : : g, where the next term is constructed by lengthening the previous term at its right-hand end by appending the next positive integer. Note that this next integer occupies only one place, with \carrying" occurring as in addition: thus the ninth and tenth terms of the sequence are 123 456 789 and 1 234 567 900 respectively. Determine which terms of the sequence are divisible by 7. 2274. Proposed by Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA. A. Let m be a non-negative integer. Find a closed form for B. Let m 2 f1; 2; 3; 4g. Find a closed form for

n Y m X

n Y m X

(k + j ).

k=1 j =0

(k + j )2.

k=1 j =0

C?. Let m and j (j = 0; 1; : : : ; m) be non-negative integers. Prove or disprove that

n Y m X

(k + j ) j is divisible by

k=1 j =0

mY +1 j =0

(n + j ).

2275. Proposed by M. Perisastry, Vizianagaram, Andhra Pradesh, India. Let b > 0 and ba  ba for all a > 0. Prove that b = e.

367

SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 1940. [1994: 108; 1995: 107, 205; 1996: 321; 1997: 170] Proposed by Ji Chen, Ningbo University, China. Show that if x; y;z > 0,   (xy + yz + zx) (x +1 y )2 + (y +1 z )2 + (z +1 x)2  94 :

dia.

Comment by Vedula N. Murty, Andhra University, Visakhapatnam, In-

In response to Marcin Kuczma's comment [1997: 170], I present the details of my proof of the assertion:

bc(b , c)2(2a2 , bc) + ca(c , a)2(2b2 , ca) > 0

when

2a2 , bc < 0 < 2b2 , ca  2c2 , ab; where a, b, c are the side lengths of a triangle satisfying 0 < a  b  c. We have

a + b > c:

(1)

bc , 2a2

So, 2 a2 + b2  (a + b)2 > c(a + b) implies that 2 2b , ca < 1; and further that ,



bc bc , 2a2 < bc : (2) ac 2b2 , ca ac c , a  1 implies that (c , a)2  c , a  b ; and further that Thus c ,2 b (c ,2 b)2 c,b a  (c , a)  bc > bc bc , 2a : So we have (c , b)2 ac ac 2b2 , ca bc(b , c)2(2a2 , bc) + ca(c , a)2(2b2 , ca) > 0:

368

2158. [1996: 218] Proposed by P. Penning, Delft, the Netherlands.

Find the smallest integer in base eight for which the square root (also in base eight) has 10 immediately following p the `decimal' point. In base ten, the answer would be 199, with 199 = 14:10673 : : : . Solution by Florian Herzig, student, Perchtoldsdorf, Austria. Let n be a positive integer with the given property. Let m be the integer part of pn and n = m2 + p. The base 8 representation of a real number x is denoted by (x)8. By hypothesis,

9; m + 18 = m + (0:10)8  pn < m + (0:11)8 = m + 64

which is equivalent to m2 + 14 m + 641  n < m2 + 329 m + 64922 : It is easy to see that these inequalities can be replaced by the following stronger ones by taking into account that the expressions on both sides cannot be integers:

m2 + m 4+ 1  n  m2 + 932m : [For example, any integer greater than or equal to m2 + m4 + 641 must be greater than or equal to m2 + m4+1 , since m is an integer.|Ed.] So m + 1  p  9m : 4

32

The lower limit has to be less than the upper limit, hence m  8. For 8  m  10 the lower limit is greater than 2 and the upper limit less than 3 [so no integer value of p exists]. If m = 11 then p = 3; therefore the smallest m is 11 and so the smallest n is m2 + p = 124 = (174)8, where p

(174)8 = (13:10531 : : : )8:

Also solved by CHARLES ASHBACHER, Hiawatha, Iowa, USA; MIGUEL  OCHOA, Logro~no, Spain; THEODORE CHRONIS, student, ANGEL CABEZON Aristotle University of Thessaloniki, Greece; MIHAI CIPU, Instituteof Mathematics, Romanian Academy, Bucharest, Romania; GEORGI DEMIREV, Varna, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Ger many; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Valley Catholic  High School, Beaverton, Oregon, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. The curious fact that the smallest solution for base 8 is 124 = 112 + 3 and for base 10 is 199 = 142 + 3 (both written in base 10) is no coincidence! Geretschlager, Godin, and the proposer consider other bases as well, and

369 show that the smallest (base 10) integer whose square root when written in base b has 10 immediately following the \decimal" point is

b , 1 2 + 1 if b is odd  2 > 3b , 2 2 + 3 if b is even: > : 8  > >
0 is an arbitrary real number. Show that for all k, l  1, there exists an index m = m(k;l) such that xk  xl  xm . Solution by D. Kipp Johnson, Valley Catholic High School, Beaverton, Oregon, USA. Since xn = 1 for all n if t = 1, any index m will suce. We distinguish two other cases. First, let t > 1. By the AM{GM inequality, n,1 p xn = 1 + t + n  + t > n 1  t  : : :  tn,1 p p = n t1++(n,1) = n tn(n,1)=2 = t(n,1)=2: This last quantity increases without bound as n ! 1, so there must be an index m for which xk xl  xm for given indices k; l.

370 Second, let 0 < t < 1. Then

n,1 xn = 1 + t + n  + t < 1 + 1 +n   + 1 = 1; so xk xl < xk and xk xl < xl , and we may choose the index m to equal either k or l.

Also solved by THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; MIHAI CIPU, Institute of Mathematics, Romanian Academy, Bucharest, Romania; GEORGI DEMIREV, Varna, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; THOMAS C. LEONG, City College of City University of New  York, New York, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; and the proposer. Some solvers noted that since x1 = 1 for all t, m = 1 will work for all k and l whenever t < 1. Cipu and the proposer found that m = k + l , 1 works when t > 1.

2160. [1996: 218] Proposed by Toshio Seimiya, Kawasaki, Japan. 

4ABC is a triangle with \A < 90 . Let P be an interior point of ABC such that \BAP = \ACP and \CAP = \ABP . Let M and N be the incentres of 4ABP and 4ACP respectively, and let R1 be the circumradius of 4AMN . Prove that 1 = 1 + 1 + 1 : R1 AB AC AP

Solution by Florian Herzig, student, Perchtoldsdorf, Austria. Let a; b; c denote the sides BC; CA; AB and ; ; the angles of the triangle. Then \APB = 180 , \ABP , \BAP = 180 , \ABP , + \CAP = 180 , and by the cosine law in 4APB

c2 = AP 2 + BP 2 + 2AP  BP cos : 4ABP and 4CAP are by hypothesis similar, whence BP = cAP b and 2 2 2 c2 = AP b2 (b + c + 2bc cos ): By setting x2 = b2 + c2 + 2bc cos 2 AP = bcx and BP = c  bAP = cx :

371 Let X be the point where the incircle of 4APB touches AP . Then

, AB = c  (b + c , x) PX = AP + BP 2 2x 2 , x2 ) c  (( b + c ) = 2x  (b + c + x) 2 sin2 2 = x 2 bc (b + c + x) and since \XPM = 12 \APB = 90 , 2 we get PX = 2bc2 sin 2 : PM = sin x  (b + c + x) 2 PM , and by using the fact Since 4NPM and 4CPA are similar, MN = bAP that \MAN = 2 , we have 1 = 2 sin 2 R1 MN bc 2 = 2  2xbc2sin sin 2

b  x(b+c+x) c+x = b + bc 1 + 1 + 1 : = AB AC AP

 OCHOA, Logro~no, Spain; Also solved by MIGUEL ANGEL CABEZON RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS,  LOPEZ  Ursulinengymnasium, Innsbruck, Austria; MARI A ASCENSION CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.

2161. [1996: 219] Proposed by Juan-Bosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Evaluate 1 1 (2 n , 1)(3 n , 1) : n=1 X

Solution by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA

(modi ed slightly be the editor).

Let S denote the given summation. We show that

p

S = 2 ln2 , 32 ln 3 + 63  0:64527561:

372 Using partial fractions, we have

S =

1 X



n=1 1  X

2

3 2n , 1 , 3n , 1



  1 1 X 2 3 = , + , n=1 2n , 1 2n n=1 n 3n , 1 1 1  1 X X 1 = 2 (,1)n,1 n1 , 3 , n=1 n=1 3n , 1 3n  1 X 1 , 1 : = 2 ln 2 , 3 (1) n=1 3n , 1 3n Since the sequence of functions ffn(x)g, where fn(x) = x3n+1 , x3n+2 , is uniformly convergent on [0; 1], 1  1 X 1  = 1 , 1 + 1 , 1 + 1 , 1 +    , 2 3 5 6 8 9 n=1 3n , 1 3n

= = = = = = =

2

Z 1

0

(x , x2 + x4 , x5 + x7 , x8 +    ) dx

Z 1

x(1 , x)(1 + x3 + x6 +    ) dx 0 Z 1 x(1 , x) dx = Z 1 x dx 0 1 , x3 0 1 + x + x2 Z 1 1 x + 2 dx , 1 Z 1 1 , 2 1 2 0 x + 2 2 + 43 dx 0 1+x+x  1 ln(1 + x + x2 ) , 1  p2 tan,1  2xp+ 1  1 2 2 3 3 0 p 1 ln3 , 3 tan,1 p3 , tan,1 p1  2 3 3 p 1 ln3 , 3   : (2) 2 3 6 p

Substituting (2) into (1), we nd S = 2 ln 2 , 32 ln 3 + 63 , as claimed. Also solved by THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria (jointly); RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Valley Catholic High School, Beaverton, OR, USA; MURRAY S. KLAMKIN, University of Al berta, Edmonton, Alberta; HEINZ-JURGEN SEIFFERT, Berlin, Germany; and the proposer. Two incorrect solutions were also received.

373 About half of the solvers used the 0non{elementary approach of considering the Euler psi-function (x) = ,,((xx)) where ,(x) denotes the gamma function. Both Hess and Sei ert pointed out that the evaluation of the more P 1 general sum 1 n=1 (n+p)(n+f ) (p > ,1, f > ,1, p 6= f ) can be found in Tables of Integrals, Series, and Products by I.S. Gradsteyn and I.M. Ryzhik (Academic Press, 1994, 5th edition). Speci cally, the following formulas can be found on p. 952 and p. 954, respectively:

1  X 1 1  ; x > 0; , (x) = , , n=1 n , 1 + x n   1 = , , 2 ln 2; 2 p   2 = , + 3 , 3 ln 3: 3 6 2 (Here, denotes Euler's constant.)   P 1 1 Since S = 1 from the above n=1 n, 12 , n, 13 , one obtains easily p3 ,2 ,1 3 formulas that S = 3 , 2 = 2 ln2 , 2 ln3 + 6 as in the solution

above.

2162. [1996: 219] Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. In 4ABC , the Cevian lines AD, BE , and CF concur at P . [XY Z ] is the area of 4XY Z . Show that [DEF ] = PD  PE  PF 2[ABC ] PA PB PC

I. Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain

(slightly shortened by the editor).

It is a standard exercise (for example, #13.5.6 in H.S.M. Coxeter, Introduction to Geometry) to show that

AF

[DEF ] =  + 1 [ABC ] ( + 1)( + 1)( + 1)

BD

CE

with  = FB ,  = DC ,  = EA . Since in our case, according to Ceva's Theorem,  = 1, all we have to prove is that

PA  PB  PC = ( + 1)( + 1)( + 1): PD PE PF

374 But it is true because

PA = [PCA] ; PB = [PAB] ; PD [PDC ] PE [PEA] PC = [PBC ] ; PF [PFB ] AF + 1 = [PAF ] + 1 = [PAB] ;  + 1 = FB [PFB ] [PFB ] [PBD] [PBC ]  + 1 = BD DC + 1 = [PDC ] + 1 = [PDC ] ;

and

[PCE ] + 1 = [PCA] : + 1 =  + 1 = CE EA [PEA] [PEA]

II. Comment by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. The corresponding more general problem for a simplex is given in [2] and also with a slightly modi ed proof in [3]. It is shown that if V0 , V1 , : : : , Vn denote0 the0 n +1 vertices of a simplex S in n{dimensional Euclidean space and if V0 , V1 , : : : , Vn0 denote the n + 1 vertices of an inscribed simplex S 0 such that the cevians Vi Vi0 are concurrent at a point P within S , then VOL S 0 n01 : : : n = : VOL S (1 , 0 )(1 , 1 ) : : : (1 , n ) Here the barycentric representation for P is P = 0V0 + 1 V1 + : : : + n Vn where 0 + 1 + : : : + n = 1 and i  0. It is also shown that the volume of S 0 is a maximum if P is the centroid. Also PV 0 =PV = i =(1 , i ). For this and other metric properties of concurrent cevians of a simplex see [1987: 274{275]. Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val OCHOA, Logro~no, Spain; GEORGI ladolid, Spain; MIGUEL ANGEL CABEZON DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; TOSHIO SEIMIYA, Kawasaki, Japan; and the proposer. Janous refers to [4, p. 342, item 1.3] for a closely related identity. Bellot quotes Victor Thebault who reported an 1881 reference for the exercise that began our featured solution. This exercise is often attributed

375 to Routh (1891). For those who read Bulgarian, he also provides a reference concerning interesting properties of a cevian triangle DEF : Hristo Lesov, Projections of noteworthy points of the triangle following the respective cevians (in Bulgarian). Matematyka & Informatyka, 5 (1994) 42{49. He further mentions [1, p. 78] where a problem of Langendonck is solved: Given a 4ABC , nd the probability of selecting a point P inside the triangle such that it is possible to form a triangle whose sides equal the respective distances from P to the sides of the 4ABC . The probability turns out to be [DEF ]=[ABC ]. [1] Heinrich Dorrie, Mathematische Miniaturen, F. Hirt, Breslau, 1943. [2] M.S. Klamkin, A volume inequality for simplexes, Univ. Beograd Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No. 357{380 (1971) 3{4. [3] M.S. Klamkin, International Mathematical Olympiads 1978{1985, Math. Assoc. Amer., Washington, D.C., 1986, pp. 87{88. [4] D.S. Mitrinovic et. al., Recent Advances in Geometric Inequalities, Kluwer Academic Publishers, 1989.

2163. [1996: 219] Proposed by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. Prove that if n;m 2 N and n  m2  16, then 2n  nm. I. Solution by Florian Herzig, student, Perchtoldsdorf, Austria. Let x = pn  m  4. Then we rst prove that x2  2x . This inequality is equivalent to De ne f (x) as x=(ln x). Then

x  2 : ln x ln 2

f 0(x) = ln(lnx x,)21 which is positive for all x  4. Moreover f (x) is continuous and di erentiable in the interval [4; 1) and f (4) = f (2). Hence x2  2x , and so, nm  (x2)x  (2x)x = 2n.

II. Solution by D. Kipp Johnson, Valley Catholic High School, Beaverton, Oregon. We rst show that the inequality is true for any m  4 with the smallest allowable n, namely n = m2, and then we induct on n. For the rst step we let m  4 and n = m2 and consider the following equivalent statements:

2m2  (m2)m () 2m2  22m log2 m () m2  2m log2 m () m=2  log2 m:

376 We establish this last statement by noticing that, for m  4, we have

1

1=2 m,1  2 ,1 1 + m 1, 1  21=2

  log2 1 + m 1, 1  21 log2 m , log2 (m , 1)  12 : This permits us to write the following m , 4 inequalities: log2 5 , log2 4  12 log2 6 , log2 5  12

.. .

log2(m) , log2 (m , 1)  12 ; which, when added, telescope to produce the desired log2 m  m=2.

For the second step we induct on n. We shall assume that for some nn+1  m2  16m we have 2n  nm. Multiplying each side by 2 we obtain 2  2  n . It will suce to show that 2  nm  (n + 1)m , which is equivalent to 2  (1 + 1=n)m. The following sequence of inequalities does the trick:

2  pe 

s 

m2  m  m 1 + m12  1 + m12  1 + n1 ;

and the proof is nished.   University of Sarajevo, Sarajevo, Also solved by SEFKET ARSLANAGIC, Bosnia and Herzegovina; MIHAI CIPU, Romanian Academy, Bucharest, Romania; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; RUSSELL EULER and JAWAD SADEK, NW Missouri State University, Maryville, Missouri; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; FABIAN MARTIN HERCE, student, Universidad de La Rioja, Logro~no, Spain; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New Mexico Highlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; ROBERT P. SEALY, Mount Allison University, Sackville, New  Brunswick; HEINZ-JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH,

377 Mount Royal College, Calgary, Alberta; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; and the proposer. There was 1 incomplete solution. Hess actually proves a somewhat stronger result by replacing the condition n  m2  16 by the 2 conditions n  16 and n  m2.

2164. [1996: 273] Proposed by Toshio Seimiya, Kawasaki, Japan.

Let D be a point on the side BC of triangle ABC , and let E and F be the incentres of triangles ABD and ACD respectively. Suppose that B , C , E, F are concyclic. Prove that

AD + BD = AB : AD + CD AC

Solution by the Con Amore Problem Group, the Royal Danish School of Educational Studies and Informatics, Copenhagen, Denmark. The intersection points of the line EF with AB; AC , and AD will be called G; H and K , respectively, and we put \B = 2y , and \C = 2z , so \GBE = \EBD = y , and \DCF = \FCH = z . Since B; C; E; F are concyclic, \GEB = \FCD = z , and \CFH = \EBD = y ; and so \AGH = y + z = \AHG giving AG = AH: (1) Since AE bisects \A in 4GAK , and AF bisects \A in 4KAH , we have KA KA KF KE KF [using (1)]: KE EG = AG = AH = FH , whence EG + 1 = FH + 1, or

KG = KH : EG FH

(2)

h1 = h2 k1 k2

(3)

k1 = KG = KH = k2 ; r1 EG FH r2

(4)

h1 = h2 : r1 r2

(5)

In 4ABD let the inradius, the altitude from D, and the area be r1 , h1 and T1 , respectively; and in 4ACD let the corresponding quantities be r2; h2 and T2 . In 4AKG, and 4AKH let the altitude from K be k1 and k2 , respectively. Then hh12 = kk12 , whence and [using (2)]

and further, (3) - (4) imply

378 Now so and similarly,

h1  AB = 2T1 = r1  (AB + BD + DA); h1 = AB + BD + DA = 1 + BD + DA ; r1 AB AB

(6)

h2 = AC + CD + DA = 1 + CD + DA ; r2 AC AC

(7)

From (5) - (7) it easily follows that

AD + BD = AB : AD + CD AC

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;  RICHARD I. HESS, Rancho Palos Verdes, California, USA; CRISTOBAL  SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon,  Spain; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer. Bellot Rosado notes that this problem is a special case of problem 1206 [H. Demir, Mathematics Magazine, 58, no. 1, January 1985; solution by V. D. Mascioni, Mathematics Magazine, 59, no. 1, February 1986]

2165. [1996: 273] Proposed by Hoe Teck Wee, student, Hwa Chong Junior College, Singapore. Given a triangle ABC , prove that there exists a unique pair of points P and Q such that the triangles ABC , PQC and PBQ are directly similar; that is, \ABC = \PQC = \PBQ and \BAC = \QPC = \BPQ, and the three similar triangles have the same orientation. Find a Euclidean construction for the points P and Q. Solution by the Con Amore Problem Group, the Royal Danish School of Educational Studies and Informatics, Copenhagen, Denmark. We identify the Euclidean plane with the complex plane, and points with complex numbers. We may assume without loss of generality that B = 0: (1) Suppose such a P and Q exist. Since 4ABC  4PQC , there is a point (complex number) X such that P = XA; (2) and Q = XC; (3)

379 and besides:

P , Q = A: (4) C ,Q C Moreover, if (2) - (4) hold, then P; Q are as desired. Now (2) - (4) imply that XA , XC = A ; C , XC C

or or

and then, by (2) - (3): and

XA , XC = A , XA; X = 2AA, C ;

(5)

2 P = 2AA, C

(6)

Q = 2AAC , C:

(7)

On the other hand, if X; P; Q satisfy (5) - (7) then (2) - (4) hold. This proves the existence and uniqueness of P and Q with the desired properties. To nd a Euclidean construction for P and Q, de ne

R = 2A , C;

and note that

A = R +2 C ; that is, R is the image of C by re ection in A. Also note that Q = A; C R that is, triangles ABR and QBC are similar, and similarly oriented. Finally A + Q = 1  AC + A = A2 = P 2 2 2A , C 2A , C so P is the midpoint of AQ. All this gives us the following construction: (A) Re ect C in A to get R. (B) Construct Q such that A and Q are on opposite sides of the line BC , \CBQ = \RBA, and \QCB = \ARB.

380 (C) Construct P as the midpoint of AQ. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria (two solutions); TOSHIO SEIMIYA, Kawasaki, Japan; and the proposer.

2166. [1996: 273] Proposed by K.R.S. Sastry, Dodballapur, India.

In a right-angled triangle, establish the existence of a unique interior point with the property that the line through the point perpendicular to any side cuts o a triangle of the same area. Solution by Toshio Seimiya, Kawasaki, Japan. Let the right-angled triangle be ABC , with right angle at A. We may assume without loss of generality that AB  AC . P is an interior point of 4ABC . The line through P perpendicular to BC meets BC; AC at D; E respectively (if AB = AC , then E = A); the line through P perpendicular to CA meets BC; CA at F; G respectively, and the line through P perpendicular to AB meets BC; AB at H; K respectively. We assume that the triangles BHK; CDE; CFG have the same area. Since these triangles are similar, they must be congruent. Since 4CDE  4CGF , we have CD = CG, so that 4CDP  4CGP . Thus we have \PCD = \PCG. Therefore, CP is the bisector of \C . Let M be the intersection of AP with BC (if AB = AC; M = D). Since 4BHK  4FCG, we have BH = FC , so that BF = HC: (1) Since PF kAB and PH kAC , we get BF : BM = AP : AM = CH : CM (2) From (1) and (2) we have BM = CM . Therefore, AM is the median of 4ABC . Thus P must be the intersection of the median AM with the bisector of \C . It is interesting to note that the bisector of \B does not pass through P if AB 6= AC . Conversely, let P be the intersection of the median AM with the bisector of \C . Draw the lines DE; FG;HK through P perpendicular to BC , CA, AB respectively, as shown in the gure. Since \PCD = \PCG, we have 4PCD  4PCG, so that CD = CG. As 4CFG is 4CED we get 4CFG  CED. Since PF kAB and PH kAC , we have BF : BM = AP : AM = CH : CM . As BM = CM , we get BF = CH , so that BH = FC . Because 4KBH is 4GFC , we have 4KBH  4GFC . Since 4CDE; 4CGF; 4HKB are congruent, they have the same area. A number of solvers used a coordinate approach. If A = (0; 0), B = (0; c), C = (b; 0) and BC = a, then P = 2bb+2 a ; 2bbc+a .

381 Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; SAM BAETHGE, Nordheim, Texas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CON AMORE PROBLEM GROUP, Royal Danish School of Educational Studies, Copenhagen, Denmark; C. DIXON, Royal Grammar School, Newcastle upon Tyne, England; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT, Franz{ Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California,  USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV  Y,  Ferris State University, Big Rapids, Michigan, USA; D.J. SMEENK, KONECN Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. There were two incorrect solutions. The proposer comments: \I am unable to establish the existence of such a point in a general triangle. If we replace \perpendicular" by \parallel" then the problem is easy and yields the unique point centroid of a triangle."  2167. [1996: 274] Proposed by Sefket Arslanagi c , Berlin, Germany.

Prove, without the aid of the di erential calculus, the inequality, that in a right triangle

a2 (b + c) + b2(a + c)  2 + p2; abc where a and b are the legs and c the hypotenuse of the triangle.

Solution by Mihai Cipu, Institute of Mathematics, Romania Academy, Bucharest, Romania, and CICMA, Concordia University, Montreal Quebec. Since a2 (b + c)+ b2(a + c) = c(a2 + b2 )+ ab(a + b) the given inequality is equivalent to 2 p c(a , b)2  ( 2c , a , b)ab = pab2c(a+,ab+) b :

In an isosceles triangle p this relation holds trivially, while in the general case it is equivalent to ( 2c + a + b)c  ab, a simple consequence of c2  ab. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CON AMORE PROBLEM GROUP, Royal Danish School of Educational Studies, Copenhagen, Denmark; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;  MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV  Y,  Ferris State University, Big Rapids, Michigan, USA; KEE-WAI KONECN LAU, Hong Kong; VEDULA N. MURTY, Andhra University, Visakhapatnam, India; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; BOB

382 PRIELIPP and JOHN OMAN, University of Wisconsin{Oshkosh, Wisconsin,  USA; JUAN-BOSCO ROMERO MARQUEZ, Universidad de Valladolid, Val ladolid, Spain; HEINZ-JURGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; and the proposer. There were four incorrect solutions. From the proof above it is obvious that equality holds if and only if a = b. Both Konecpny and Wang pointed out that a weaker version of this problem with 2 + 2 replaced by  appeared as problem No. 238 in the January, 1983 issue of the College Mathematics Journal. Wang remarked that in the published solution (CMJ 15 (4), 1984; 352{353) Walter Blumberg and Marshall Fraser independently proved the more general result that

a2(b + c) + b2(a + c)  2 + csc(C=2) abc where a, b, and c are the sides of an arbitrary triangle with C being the largest angle. The present problem is the special case when C = =2. Klamkin gave a proof for this more general result. Romero Marquez asked whether the stronger inequality

a2(b + c) + b2(a + c)  p2c abc a+b,c holds. The triangle with a = 3, b = 4, and c = 5 provides a simple counterexample. Janous remarked that the given inequality could be considered as the special case (when n = 1) of the problem of nding the minimum value of 2n n n + b2n(an + cn ) Tn = a (b + c ()abc )n where a, b, c are the legs and hypotenuse of a right triangle and n > 0 is a real number. He proposed the conjecture that Tn  2 + 2(2,n)=2 for all n  n0 where n0 is the solution of the equation 2(n+4)=2 + n = 2.  etokrzyski, 2168?. [1996: 274] Proposed by Jan Ciach, Ostrowiec Swi Poland. Let P be a point inside a regular tetrahedron ABCD, with circumradius R and let R1 ; R2 ; R3 ; R4 denote the distances of P from vertices of the tetrahedron. Prove or disprove that

R1R2 R3R4  34 R4;

and that the maximum value of R1 R2 R3 R4 is attained.

383 Solution by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. [Ed: First, we give Klamkin's proof for a triangle, which was not published as a solution to 2073? [1995: 277; 1996: 282]] Let P be given by the vector P~ = x1 A~1 + x2 A~2 + x3 A~3 , where x1 + x2 + x3 = 1 and x1, x2, x3  0 (barycentric coordinates). Then



2 PA21 = P~ , A~1     2 = x2 A~2 , A~1 + x3 A~3 , A~1 ,  = 3 x22 + x2 x3 + x23 ; p etc. We have taken R = 1, so that the side of the triangle A~2 , A~1 = 3.

The given inequality now takes the form

3; x2 + x2x3 + x23 ,x23 + x3x1 + x21 ,x21 + x1 x2 + x22  64

, 2

or in homogeneous form

64 ,x22 + x2 x3 + x23  ,x23 + x3 x1 + x21  ,x21 + x1 x2 + x22   3 (x1 + x2 + x3)6 (1) where (only) x1 , x2 , x3  0. We assume without loss of generality that x1  x2  x3  0 so that we can let x3 = c, x2 = b + c and x1 = a + b + c with a, b, c  0 and a + b + c > 0. On substituting back into (1), we get 3a6 + 36a5b + 54a5c + 116a4b2 + 348a4bc + 213a4c2 + 160a3b3 +816a3b2c + 1128a3bc2 + 468a3c3 + 80a2b4 + 864a2b3 c +2232a2b2c2 + 2232a2bc3 + 765a2 c4 + 480ab4 c + 2208ab3c2 +3888ab2c3 + 3060abc4 + 918ac5 + 192b5c + 1104b4c2 +2592b3c3 + 3060b2c4 + 1836bc5 + 459c6  0 since all terms on the left side are non-negative (this expansion was con rmed using Mathematica [Ed: and using DERIVE]). There is equality only if a = c = 0, so that if x1 + x2 + x3 = 1, x1 = x2 = 0:5 and x3 = 0, then the point P must be a mid-point of a side. For a regular tetrahedron, we proceed as before: let P be given by the vector P~ = x1 A~1 + x2 A~2 + x3 A~3 + x4 A~4 , where x1 + x2 + x3 + x4 = 1 and x1 , x2 , x3 , x4  0. Then



PA1 = P~ , A~1       = x2 A~2 , A~1 + x3 A~3 , A~1 + x4 A~4 , A~1 ;

384 so that

PA21 = a2 ,x22 + x23 + x24 + x3x4 + x4x2 + x2x3 ; etc., where a is an edge of the tetrahedron. So if we take R = 1, we have a2 = 83 . the desired inequality then becomes (in homogeneous form) 9 (x1 + x2 + x3 + x4 )8  28F1 F2F3 F4; (2) where

F1 = x22 + x23 + x24 + x3x4 + x4 x2 + x2x3 ; and the other Fi's are obtained by cyclic interchange of the indices 1, 2, 3, 4. Assuming without loss of generality that x1  x2  x3  x4 , we take x4 = d, x3 = d + c, x2 = d + c + b, x1 = d + c + b + a, where a, b, c, d  0 and x1 > 0. On substituting back into (2) and expanding out (using Mathematica), we get a polynomial [Ed: we have omitted this polynomial, since it would cover two whole pages of CRUX with MAYHEM] which is non-negative since all the terms are non-negative. There is equality only if a = c = d = 0. Hence x1 = x2 = 0:5 and so P must be the mid-point of an edge. I also conjecture for the correponding result for a regular n{dimensional simplex: the product of the distances from a point within or on a regular simplex to the vertices is a maximum only when the point is a mid-point of an edge. Analogously to inequality (2), the conjecture is that

3n,1 (x0 + x1 + : : : + xn)2n+2  22n+2 F0F1 : : : Fn (x0 , x1 , : : : , xn  0), where Fi is the complete symmetric homogeneous polynomial of degree 2 with unit coecients of all the variables x0 , x1 , : : : ,

xn , except xi, and there is equality only if two of the variables are equal and

the rest are zero. Also solved by CON AMORE PROBLEM GROUP, Royal Danish School of Educational Studies, Copenhagen, Denmark; and RICHARD I. HESS, Rancho Palos Verdes, California, USA.

Crux Mathematicorum

Founding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell Editors emeriti / Redacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem

Founding Editors / Redacteurs-fondateurs: Patrick Surry & Ravi Vakil Editors emeriti / Redacteurs-emeriti: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

385

THE ACADEMY CORNER No. 14

Bruce Shawyer All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7 Christopher Small writes: I notice that in the Hints { 2 section of the Bernoulli Trials, the statement that the hint does not work for question 9 has been interpolated. It seems to me that the statement is correct as I originally gave it to you. A counterexample is easily found with a = c. For example, a = c = 1 and b = 1=2 is a counterexample. This month, we present solutions to some of the problems in a university entrance scholarship examination paper from the 1940's, which appeared in the April 1997 issue of CRUX with MAYHEM.

1. Find all the square roots of

p

1 , x + 22x , 15 , 8x : 2

Solution. First we note that the only way to generate the given expression as a square, is to square p p

ax + b + cx + d: There are, of course, two square roots, being  this quantity. So we have p

p

2

ax + b + cx + d p = (a + c)x + r (b + d) + 2 acx + (ad + bc)x + bd 15 : = ,x + 1 + 2 ,2x + 11 x , 2 4 2

2

386 Since this is an identity, we equate coecients, yielding

a + c = ,1; (1) b + d = 1; (2) ac = ,2; (3) (4) ad + bc = ; bd = , : (5) Solving (1) and (2) for c and d respectively, and substituting in (3) and (5) 11 2 15 4

gives

a + a , 2 = (a + 2)(a , 1) = 0; b , b , = (b , )(b + ) = 0: so that a = ,2; 1 and b = ; , , giving the corresponding values of c = 1; ,2 and d = , ; . 2

2

15 4

3 2

5 2

5 2

5 2

3 2

3 2

We must now substitute these into (4), and this leads to the solution that the square roots of the given expression are



q

q

,2x + + x , 5 2

 3 2

:

2. Find all the solutions of the equations:

x + y + z = 2; x + y + z = 14; xyz = ,6: 2

2

2

Solution. First we recall the expressions involving the roots of a cubic:

(P , x)(P , y)(P , z) = P , (x + y + z )P + (xy + yz + zx)P , xyz: 3

2

We also note that

(x + y + z) , (x + y + z ) = xy + yz + zx: 2 2

2

2

2

So, from two of the given three expressions, we get

xy + yz + zx = ,5: Thus, the solution of the given equations is the set of roots of the cubic equation

P , 2P , 5P + 6 = 0: 3

2

387 It is easy to check that P = 1 is a root: so it is easy to factor into

(P , 1)(P + 2)(P , 3) = 0; giving that the other two roots are P = ,2 and P = 3.

Thus, the solution set of the given equations is any permutation of 1; ,2; 3. 3. Suppose that n is a positive integer and that Ck is the coecient of xk in the expansion of (1 + x)n . Show that

, 1)! : (k + 1)Ck = (n +n!2)(n(2,n1)!

n X

2

k=0

Solution. n ! X Consider (k + 1)Ck xn. Since Cn,k = Ck , we can write this as 2

k=0

n , X

(k + 1)Ck xk  ,Cn,k xn,k  :

k=0

This is the term in xk in the product n X

(k + 1)Ckxk

k=0

!

n X

Ck xk

k=0 (1 + x)n .

The right member of this product is determine the left member: for example,

x(1 + x)n =

n X k=0

!

:

There are several ways to

Ckxk ; +1

so that, on di erentiating, we have that

(1 + x)n + nx(1 + x)n, = 1

n X k=0

(k + 1)Ckxk :

Thus, the product above is

(1 + x)n, (1 + (n + 1)x)  (1 + x)n = (1 + (n + 1)x)(1 + x) n, : 1

The coecient of xn in this product is

(2n , 1)! (n + 2) (2n , 1)! (2n , 1)! (n , 1)! n! (n + 1) + n! (n , 1)! = n! (n , 1)! :

2

1

388

THE OLYMPIAD CORNER No. 185

R.E. Woodrow All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. To begin this number, we give the ten problems of the Seventh Irish Mathematical Olympiad written May 7, 1994. My thanks go to Richard Nowakowski for collecting the problems when he was Canadian Team leader at the IMO in Hong Kong.

IRISH MATHEMATICAL OLYMPIAD 1994 Second Paper | 7 May 1994 Time: 3 hours

1. Let x, y be positive integers with y > 3 and

x + y = 2 (x , 6) + (y + 1)  : 2

4

2

2

Prove that x + y = 1994. 2. Let A, B, C be three collinear points with B between A and C . Equilateral triangles ABD, BCE , CAF are constructed with D, E on one side of the line AC and F on the opposite side. Prove that the centroids of the triangles are the vertices of an equilateral triangle. Prove that the centroid of this triangle lies on the line AC . 3. Determine with proof all real polynomials f (x) satisfying the equation 2

4

f (x ) = f (x)f (x , 1): 4. Consider the set of m  n matrices with every entry either 0 or 1. 2

Determine the number of such matrices with the property that the number of \1"s in each row and in each column is even. 5. Let f (n) be de ned on the set of positive integers by the rules

f (1) = 2 and f (n + 1) = (f (n)) , f (n) + 1; n = 1; 2; 3; : : : : Prove that for all integers n > 1 2

1 + 1 +  + 1 < 1, 1 : 1 , 2 1n, < f (1) f (2) f (n) 2n 2

1

2

389

6. A sequence xn is de ned by the rules x =2 1

and

nxn = 2(2n , 1)xn, ; n = 2; 3; : : : : Prove that xn is an integer for every positive integer n. 7. Let p, q, r be distinct real numbers which satisfy the equations q = p(4 , p) r = q(4 , q) p = r(4 , r): Find all possible values of p + q + r. 8. Prove that for every integer n > 1 n   X n((n + 1) =n , 1) < 2ki+ 1 < n 1 , n, = n, + 4: i 1

2 (

2

1)

2

=1

9. Let w, a, b, c be distinct real numbers with the property that there exist real numbers x, y , z for which the following equations hold: x+y+z xa + yb + zc xa + yb + zc xa + yb + zc Express w in terms of a, b, c. 2

2

2

3

3

3

4

4

4

= = = =

1

w w w: 2 3 4

10. If a square is partitioned into n convex polygons, determine the maximum number of edges present in the resulting gure. [You may nd it helpful to use a theorem of Euler which states that if a polygon is partitioned into n polygons, then v , e + n = 1 where v is the number of vertices and e is the number of edges in the resulting gure]. Next we turn to the \ocial" results of the 38th IMO which was written in Mar del Plato, Argentina, July 24 and 25, 1997. My source this year was the contest WEB site. I hope that I have made no serious errors in compiling the results and transcribing names. This year a total of 460 students from 82 countries took part. This is somewhat up from last year. Sixty- ve countries sent teams of 6 (the number invited to participate in recent years). But there were 13 teams of smaller size, 3 of ve members, 2 of size four, 7 of size three, and 1 with two members.

390 The contest is ocially an individual competition and the six problems were assigned equal weights of seven marks each (the same as the last 16 IMO for a maximum possible individual score of 42 and a total possible of 252 for a national team of six students). For comparison see the last 16 IMO reports in [1981: 220], [1982: 223], [1983: 205], [1984: 249], [1985: 202], [1986: 169], [1987: 207], [1988: 193], [1989: 193], [1990: 193], [1991: 257], [1992: 263], [1993: 256], [1994: 243], [1995: 267] and [1996: 301]. There were 4 perfect scores. The jury awarded rst prize (Gold) to the thirty-nine students who scored 35 or more. Second (Silver) prizes went to the seventy students with scores from 25 to 34, and third (Bronze) prizes went to the one hundred and twenty-two students with scores from 15 to 24. Any student who did not receive a medal, but who scored full marks on at least one problem, was awarded honourable mention. This year there were seventy-eight honourable mentions awarded. The median score on the examination was 14. Congratulations to the Gold Medalists, and especially to Ciprian Manolescu who scored a perfect paper last year as well. Name

Eftekhari, Eaman Manolescu, Ciprian Bosley, Carleton Do Quoc Anh Frenkel, Peter Pap, Gyula Ivanov, Ivan Terpai, Tamas Maruoka, Tetsuyuki Curtis, Nathan Najnudel, Joseph Dourov, Nikolai Rullgard, Hans Gyrya, Pavlo Lam, Thomas Han, Jia Rui Ni, Yi Zou, Jin Woo, Jee Chul Hornet, Stefan Laurentiu Summers, Bennet Tchalkov, Rayko An, Jin Peng Sun, Xiao Ming Salmasian, Hadi Leptchinski, Mikhail Tcherepanov, Evgueni

Country

Iran Romania United States of America Vietnam Hungary Hungary Bulgaria Hungary Japan United States of America France Russia Sweden Ukraine Australia China China China Republic of Korea Romania United Kingdom Bulgaria China China Iran Russia Russia

Score

42 42 42 42 41 41 40 40 40 40 39 39 39 39 38 38 38 38 38 38 38 37 37 37 37 37 37

391 Name

Country

Herzig, Florian Lippner, Gabor Bahramgiri, Mohsen Battulga, Ulziibat Farrar, Stephen Zheng, Chang Jin Podbrdsky, Pavel Holschbach, Armin Bayati, Mohsen Merry, Bruce Grechuk, Bogdan Kabluchko, Zajhar

Austria Hungary Iran Mongolia Australia China Czech Republic Germany Iran Republic of South Africa Ukraine Ukraine

Score

36 36 36 36 35 35 35 35 35 35 35 35

Next we give the problems from this year's IMO Competition. Solutions to these problems, along with those of the 1997 USA Mathematical Olympiad will appear in a booklet entitled Mathematical Olympiads 1997 which may be obtained for a small charge from: Dr. W.E. Mientka, Executive Director, MAA Committee on HS Contests, 917 Oldfather Hall, University of Nebraska, Lincoln, Nebraska, 68588, USA.

38th INTERNATIONAL MATHEMATICAL OLYMPIAD

July 24{25, 1997 (Mar del Plata, Argentina) First Day | Time: 4.5 hours

1. In the plane the points with integer coordinates are the vertices of unit squares. The squares are coloured alternately black and white (as on a chessboard). For any pair of positive integers m and n, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths m and n, lie along edges of the squares. Let S be the total area of the black part of the triangle and S be the total area of the white part. Let 1

2

f (m;n) = jS , S j : 1

2

(a) Calculate f (m;n) for all positive integers m and n which are either both even or both odd. (b) Prove that f (m;n)  maxfm; ng for all m and n. (c) Show that there is no constant C such that f (m;n) < C for all m and n. 2. Angle A is the smallest in the triangle ABC . The points B and C divide the circumcircle of the triangle into two arcs. Let U be an interior point of the arc between B and C which does not contain A. 1 2

392 The perpendicular bisectors of AB and AC meet the line AU at V and W , respectively. The lines BV and CW meet at T . Show that

AU = TB + TC:

3. Let x ; x ; : : : ; xn be real numbers satisfying the conditions: 1

2

jx + x +    + xnj = 1 1

2

and

jxi j  n +2 1 for i = 1; 2; : : : ; n: Show that there exists a permutation y ; y ; : : : ; yn of x ; x ; : : : ; xn such that jy + 2y +    + nynj  n + 1 : 1

1

2

1

2

2

2

.

Second Day | Time: 4.5 hours

4. An n  n matrix (square array) whose entries come from the set S = f1; 2; : : : ; 2n , 1g, is called a silver matrix if, for each i = 1; : : : ; n, the ith row and the ith column together contain all elements of S. Show that (a) there is no silver matrix for n = 1997; (b) silver matrices exist for in nitely many values of n. 5. Find all pairs (a; b) of integers a  1, b  1 that satisfy the equation a b = ba : 6. For each positive integer n, let f (n) denote the number of ways of representing n as a sum of powers of 2 with nonnegative integer exponents. Representations which di er only in the ordering of their summands are considered to be the same. For instance, f (4) = 4, because the number (

2

)

4 can be represented in the following four ways: 4; 2 + 2; 2 + 1 + 1; 1 + 1 + 1 + 1: Prove that, for any integer n  3: 2n = < f (2n) < 2n = : 2

4

2

2

As the IMO is ocially an individual event, the compilation of team scores is unocial, if inevitable. These totals and the prize awards are given in the following table.

393 Rank

1. 2. 3. 4.{5. 4.{5. 6. 7.{8. 7.{8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.{21. 20.{21. 22.{23. 22.{23. 24. 25. 26. 27. 28. 29. 30.{31. 30.{31. 32.{33. 32.{33. 34.{35. 34.{35. 36. 37.{38. 37.{38. 39. 40. 41.{42. 41.{42. 43. 44. 45.

Country

China Hungary Iran Russia United States of America Ukraine Bulgaria Romania Australia Vietnam Republic of Korea Japan Germany Republic of China (Taiwan) India United Kingdom Belarus Czech Republic Sweden Poland Yugoslavia Israel Latvia Croatia Turkey Brazil Colombia Georgia Canada Hong Kong Mongolia France Mexico Armenia Finland Slovakia Argentina The Netherlands Republic of South Africa Cuba Belgium Singapore Austria Norway Greece

Score Gold Silver Bronze Total

223 219 217 202 202 195 191 191 187 183 164 163 161 148 146 144 140 139 128 125 125 124 124 121 119 117 112 109 107 106 106 105 105 97 97 96 94 94 93 91 88 88 86 79 75

6 4 4 3 2 3 2 2 2 1 1 1 1 { { 1 { 1 1 { { { { { { { { { { { 1 1 { { { { { { 1 { { { 1 { {

{ 2 2 2 4 3 3 3 3 5 4 3 3 4 3 2 2 2 3 2 2 1 1 1 1 1 { 1 2 { { { 1 { 1 { 2 { 1 { { { { 1

{ { { 1 { { 1 1 1 { 1 1 2 2 3 2 4 2 { 2 3 5 4 4 4 4 6 3 2 5 3 1 3 3 4 2 3 { 2 2 3 4 1 3 {

6 6 6 6 6 6 6 6 6 6 6 5 6 6 6 5 6 5 4 4 5 6 5 5 5 5 6 4 4 5 4 2 4 3 4 3 3 2 3 3 3 4 2 3 1

394 Rank

Country

46.{47. Kazakhstan 46.{47. Former Yugoslav Republic of Macedonia 48.{49. Italy 48.{49. New Zealand 50. Slovenia 51. Lithuania 52. Thailand 53.{54. Estonia 53.{54. Peru 55. Azerbaijan 56. Macao 57.{59. Denmark 57.{59. Moldova (team of 3) 57.{59. Switzerland (team of 5) 60.{61. Iceland 60.{61. Morocco 62. Bosnia and Herzegovina (team of 5) 63. Indonesia 64. Spain 65. Trinidad and Tobago 66. Chile 67. Uzbekistan (team of 3) 68. Ireland 69.{70. Malaysia 69.{70. Uruguay 71.{72. Albania (team of 3) 71.{72. Portugal (team of 5) 73. Philippines (team of 2) 74. Bolivia (team of 3) 75. Kyrgyztan (team of 3) 76.{78. Kuwait (team of 4) 76.{78. Paraguay 76.{78. Puerto Rico 79. Guatemala 80. Cyprus (team of 3) 81. Venezuela (team of 3) 82. Algeria (team of 4)

Score Gold Silver Bronze Total

73 73

{ {

{ {

1 3

1 3

71 71 70 67 66 64 64 56 55 53 53 53 48 48 45

{ { { { { { { { { { { { { { {

{ { { 1 { { { { { { { { 1 { {

1 2 2 1 1 2 2 1 { 1 2 2 { { 1

1 2 2 2 1 2 2 1 { 1 2 2 1 { 1

44 39 30 28 23 21 19 19 15 15 14 13 11 8 8 8 7 5 4 3

{ { { { { { { { { { { { { { { { { { { {

{ { { { { { { { { { { { { { { { { { { {

{ { { { { { { { { { { { { { { { { { { {

{ { { { { { { { { { { { { { { { { { { {

395 This year the Canadian Team slid to 29th place from 16th last year and 19th the previous year. The Team members were: Byung Kyu Chun Adrian Chan Mihaela Enachescu Sabin Cautis Jimmy Chui Adrian Birka

29 25 21 16 10 6

Silver Silver Bronze Bronze Honourable Mention

The Canadian Team Leader was Richard Nowakowski, of Dalhousie University, and the Deputy Team Leader was Naoki Sato, currently a student at the University of Toronto and former Canadian Silver medalist. The Chinese Team placed rst this year. Its members were: Jia Rui Han Yi Ni Jin Zou Jin Peng An Xiao Ming Sun Chang Jin Zheng

38 38 38 37 37 35

Gold Gold Gold Gold Gold Gold

Congratulations to the Chinese Team!! Now we turn to solutions to problems of the Czechoslovakia Mathematical Olympiad 1993 [1996: 109].

CZECHOSLOVAK MATHEMATICAL OLYMPIAD 1993 Final Round 1. Find all natural numbers n for which 7n , 1 is a multiple of 6n , 1.

Solutions by Mansur Boase, student, St. Paul's School, London, England; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. No such n exists. Suppose n is a natural number for which 6n , 1 j 7n , 1. Since 6n  1 mod 5 we have 5 j 7n , 1. From 7  2; 7  ,1, 7  3 and 7  1 mod 5, we see that 7n , 1  0 mod 5 if and only if 4 divides n. In particular, n must be even, which implies, since 6  1 mod 7 that 6n  1 mod 7; that is, 7 j 6n , 1. However, obviously 7 6 j 7n , 1, which is a contradiction. 2. A 19  19 table contains integers so that any two of them lying on neighbouring elds di er at most by 2. Find the greatest possible number of mutually di erent integers in such a table. (Two elds of the table are considered neighbouring if they have a common side.) 2

2

3

4

396 land.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

We shall rst prove that the value 71 for the number of di erent integers is attainable. We can assume, without loss of generality, that the number in the top left hand square is 0. We can then form the following array:

0 2 4 6 .. .

1 3 5 7 .. .

3 5 7 9 .. .

5 7 9 11 .. .

32 33 35 37 34 35 37 39 35 37 39 41

   

31 33 35 37 .. .

33 35 37 39

.. .

34 36 38 40 .. .

63 65 66 65 67 68 67 69 70

The di erence in the values of opposite corners of the array cannot exceed twice the number of moves to get from one to the other passing through adjacent squares. This equals

2  36 = 72: Thus there can be at most 73 di erent values since every square in the array can be reached from the top left square in at most 36 moves. We need to prove that there is no array with 72 or 73 di erent numbers. The value 73 can only occur if the di erence between every pair of adjacent squares is exactly 2, for otherwise one of the shortest paths from opposite corners can pass through the pair of adjacent squares whose value is not 2, giving a contradiction. If the number of distinct values is 72, then the bottom right hand square must have value 71, so in any shortest path, there is a pair of adjacent squares with di erence only 1. To have 72 values, all the even and odd numbers less than or equal to 71 must be in the table. There is a (possibly broken) dividing line separating the odd and even numbers, all numbers on the side of 71 being odd. If there is a gap on the line, then there is a shortest path from 71 to 0 without crossing the line, which is impossible. Considering 1 and 71 we see that the shortest distance must be 35, so that one of the squares adjacent to 0 must contain 1. The broken line separating even and odd entries cannot have the form

397

or

for otherwise these would be a shortest path from 0 to 71 with more than one di erence of 1. It follows that the even entries are con ned to the rst row or the rst column, contradicting that there are 36 of them. 4. A sequence fang1n of natural numbers is de ned recursively by a = 2 and an = the sum of 10th powers of the digits of an , for all n1 1. Decide whether some numbers can appear twice in the sequence fan gn . Solution by Mansur Boase, student, St. Paul's School, London, England. Let an have dn digits, so dn  1 + log an . Then an  dn 9  (1 + log an )9 < an for an suciently large, as x ! 1 as x ! 1. Thus an < an if an x an > 9 . Let K be the rst natural number for which K K > 9 . Then if ai  K , the sequence decreases until there is a term < K . Hence the sequence has in nitely many terms < K , and there must be a term repeated (in nitely often). 5. Find all functions f : Z! Zsuch that f (,1) = f (1) and =1

1

+1

=1

+1

1+log

10

10

10

+1

10

1+log10

10

1+log10

10

f (x) + f (y) = f (x + 2xy) + f (y , 2xy) for all integers x, y . land.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

f (1) + f (y) = f (1 + 2y) + f (,y) (1) f (y) + f (,1) = f (,y) + f (,1 + 2y) (2) f (1) = f (,1) and f (y) = f (y), so equating (1) and (2) f (1 + 2y) + f (,y) = f (,y) + f (,1 + 2y): Hence f (2y + 1) = f (2y , 1) for all integers y . (?) Also f (y ) + f (1) = f (3y ) + f (1 , 2y ) and since f (1) = f (1 , 2y ) (making use of (?)) f (y ) = f (3y ). Also, equating this with equation (1)

f (3y) + f (1 , 2y) = f (1 + 2y) + f (,y);

398 and since by (?) f (1 , 2y ) = f (1 + 2y )

f (3y) = f (,y): Thus f (y ) = f (3y ) = f (,y ) and f (y ) = f (,y ) for all y . If y is odd, then f (y ) = f (y , 2xy ) so f (x) = f (x + 2xy ). Thus f (2a) = f (2a(1 + 2y )).

Thus an odd multiple of a number and that number give the same value. Therefore if n = 2k a, a odd, then f (n) = f (2k). Therefore all functions must be such that f (2ka) = f (2k), k  0 where a is odd and f (2k) for each k can take any value as can f (0). These functions always satisfy the conditions since f (,1) = f (1) and

f (x + 2xy) + f (y , 2xy) = f (x(1 + 2y)) + f (y(1 , 2x)) = f (x) + f (y ):

For the remainder of this number of the Corner we turn to readers' solutions to 10 of the problems of the Baltic Way | 92 Contest given in the May 1996 number of the Corner [1996: 157{159].

MATHEMATICAL TEAM CONTEST \BALTIC WAY | 92" Vilnius, 1992 | November 5{8 1. Let p, q be two consecutive odd prime numbers. Prove that p + q is a product of at least 3 positive integers > 1 (not necessarily di erent).  Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; by Michael Selby, University of Windsor, Windsor, Ontario; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Since p and q are odd, p + q is even and has a factor of 2. The corresponding factor (p + q ) lies strictly between p and q and is not prime since p and q are consecutive prime numbers. It must therefore have two factors at least. Hence p + q is a product of at least 3 positive integers greater than 1. 2. Denote by d(n) the number of all positive divisors of a positive integer n (including 1 and n). Prove that there are in nitely many n such that d nn is an integer. 1 2

(

)

399 Solutions by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; by Michael Selby, University of Windsor, Windsor, Ontario; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Bradley's solution. Let p be a prime. Then pp, has factors 1; p; p ; : : : ; pp, , which are p in number. Hence, for such n, pp, =d(pp, ) = pp, which is integral. Hence there are in nitely many n such that n=d(n) is an integer. 3. Find an in nite non-constant arithmetic progression of positive integers such that each term is neither a sum of two squares, nor a sum of two cubes (of positive integers). Solutions by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by Michael Selby, University of Windsor, Windsor, Ontario; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Godin's solution. Looking at squares mod 4 we see 1

2

1

1

1

2

x x mod 4 0 1 2 3

0 1 0 1

2

thus if x  3 mod 4, it is impossible to express x as a sum of two squares. Similarly, looking at cubes mod 7 we see

x x mod 7 0 1 2 3 4 5 6

0 1 1 6 1 6 6

3

and again if x  3 mod 7 it is impossible to express x as a sum of 2 cubes. Thus the sequence s = 3, sn = sn + 28 has all sn  3 mod 4 and sn  3 mod 7 so all the terms are not the sum of 2 squares or cubes. 4. Is it possible to draw a hexagon with vertices in the knots of an integer lattice so that the squares of the lengths of the sides are six consecutive positive integers? Solutions by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, Clifton College, Bristol, UK; and by Shawn 0

+1

400 Godin, St. Joseph Scollard Hall, North Bay, Ontario. We give Bradley's solution. No. It is not possible. Six consecutive positive integers must contain three odd numbers and three even numbers. To form them as the sums of squares must involve an odd number of odd squares. The sum of all the numbers whose squares are made is therefore odd. But to move from (0; 0; 0) to (0; 0; 0) involves a total number which is even (going out and coming back to put it loosely). [A pentagon, on the other hand, can be made, for example,

(0; 0; 0) ! (0; 1; 1) ! (1; 2; 2) ! (1; 2; 4) ! (1; 1; 2) ! (0; 0; 0) involving sides whose squares are 2; 3; 4; 5; 6 | with two odd numbers.] 5. Given that a + b + (a + b) = c + d + (c + d) , prove that a + b + (a + b) = c + d + (c + d) . 2

4

4

4

2

4

2

4

2

2

2

4

 Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by Michael Selby, University of Windsor, Windsor, Ontario; by Panos E. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's solution. This follows immediately from the fact that

x + y + (x + y) = 2 ,x + 2x y + 3x y + 2xy + y    ,  = 2 x + xy + y = 21 x + y + (x + y ) for all x, y . 6. Prove that the product of the 99 numbers kk , , k = 2; 3; : : : ; 100, 4

4

4

4

3

2

2

3

2 2

2

2

4

2

2 2

3

1 3 +1

is greater than .  Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by Michael Selby, University of Windsor, Windsor, Ontario; by Panos E. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Wang's solution and comment. Q Let P (n) = nk kk , where n  2. We show that in general 2 3

3

=2

1 3 +1

P (n) = 2(3nn(+n n+ +1)1) 2

from which it follows that P (n) > . Since 2 3

(k + 1) , (k + 1) + 1 = k + k + 1; 2

2

401 we have

(k , 1)(k + k + 1) k (k + 1)(k , k + 1) Qn, Qn ( k + 1) k = Qn (k + 1)  Qkn, (k + k + 1) k k k +k+1 = n(1n +2 1)  n +3n + 1 = 2(3nn(+n n+ +1)1) :

P (n) =

n Y

2 2

=2

2 =0

=2 1 =1

=2

2

2

2

2

Remark: It is easy to see that the sequence fP (n)g, n = 2; 3; : : : is strictly decreasing and from the result established above we see that limn!1 P (n) = . This result is well known and can be found in, for example, Theory and Application of In nite Series, by K. Knopp (Ex. 85(2) on p. 28). 7. Let a = p1992. Which number is greater: 2 3

1992

q

a

a

a

q q

a9 > > > > = > > > > ;

1992 or 1992?

Solutions by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by Michael Selby, University of Windsor, Windsor, Ontario; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Godin's solution. Let f (x) = ax . Clearly p > q if and only if f (p) > f (q ); that is, f is strictly increasing. Similarly if f n (x) is de ned by f (x) = f (x) and f n (x) = f (f n(x)), so f n(x) = f| f {z: : : f}(x) then p > q if and only 1

+1

n

if f n(p) > f n(q ) for all n. Now 1992 > a. Thus f (1992) > f (a). But f (1992) = 1992, so f (1992) = 1992. Now f (a) is the expression in question so 1992 is the larger. Editor's Note. Selby points out that if xk = f k(a) then limk!1 xk = L where a = L =L or L = 1992. 8. Find all integers satisfying the equation 1992

1992

1992

1992

1

2x  (4 , x) = 2x + 4:

 Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by Shawn Godin,

402 St. Joseph Scollard Hall, North Bay, Ontario; by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; by Michael Selby, University of Windsor, Windsor, Ontario; by Panos E. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Tsaoussoglou's solution.

2x = 2(4x,+x2) > 0; so , 2 < x < 4:

Checking

x = ,1 x=0 x=1 x=2 x=3

1 6= 2(,1) + 4 = 2 2 4 , (,1) 5 + 2) is a solution 1 = 2(0 4,0 2(1 2 = 4 ,+12) is a solution + 2) is a solution 4 = 2(2 4,2 2(3 8 = 4 ,+32) is not a solution

9. A polynomial f (x) = x

3

+ ax + bx + c is such that b < 0 and 2

ab = 9c. Prove that the polynomial has three di erent real roots.

Solutions by Christopher J. Bradley, Clifton College, Bristol, UK; by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; and by Michael Selby, University of Windsor, Windsor, Ontario. We give Selby's solution. Suppose a = 0. Then p c = 0 andpf (x) = x + bx. Therefore there are three roots: x = 0, x = ,b, x = , ,b. If a > 0, then c = ab < 0, since b < 0. Now f (0) = c < 0 and since limx!1 f (x) = 1, there is some r in (0; 1) such that f (r ) = 0. Also f (,a) = ,a + a , ab + c = , ab > 0. Since f (0) < 0 and f (,a) > 0, there is a root r in the interval (,a; 0). Finally, since f (x) ! ,1 as x ! ,1 and f (,a) > 0 there must be a root r in (,1; ,a). If a < 0, c = ab > 0. Hence f (0) = c > 0 while f (x) ! ,1 as x ! ,1. Therefore we have a root t in (,1; 0). f (,a) = ,a + a , ab + c = , ab < 0. Hence there must be a root t in (0; ,a). Finally f (x) ! 1 as x ! 1. Since f (,a) < 0 there need be a root t in (,a; 1). In all cases we have three distinct roots. 3

9

1

1

3

8 9

3

2

3

9

1

3

2

3

3

8 9

403

10. Find all fourth degree polynomials p(x) such that the following four conditions are satis ed: (i) p(x) = p(,x), for all x, (ii) p(x)  0, for all x, (iii) p(0) = 1, (iv) p(x) has exactly two local minimum points x and x such that jx , x j = 2. Solutions by Christopher J. Bradley, Clifton College, Bristol, UK; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; and by Michael Selby, University of Windsor, Windsor, Ontario. We give Selby's solution. Condition (i) implies p(x) is even. 1

1

2

2

p(x) = ax + bx + cx + dx + e; p(,x) = ax , bx + cx , dx + e: Hence we have 2bx + 2dx = 0 for all x. Thus b = d = 0. Also p(0) = 1. Therefore p(x) has the form p(x) = ax + cx + 1: Now p0 (x) = 4ax + 2cx. The critical points are x = 0, x = ,ac . Hence we must have ,ac > 0. p00 (x) = ,12c ax0 +2c. Since, ,we  want exactly two local minima, we must have at x = a , p (x) = 12a ac + 2c = ,4c > 0. Therefore c < 0 and a > 0. Further we want p(x)  0. Since p(x) ! 1 as jxj ! 1 then since p(0) = 1, we need at x q= ,ac , p(x) =qa ca +c , ,ac +1  0. So q,ca +1  0 or 4a  c . Also x = ,ac , x = , ,ac , jx , x j = 2 ) ,ac = 1 or ,c = 2a. Since 4a  c , we have 4a  4a or a(1 , a)  0. Therefore since a > 0, 1 , a  0 and we have 0 < a  1, c = ,2a. The polynomials are of the form ax , 2ax + 1 where 0 < a  1. 4

3

4

2

3

2

3

4

2

3

2

2

2

2

2

2

2

2

2

2

1

2

2

4 2

2

2

1

2

2

2

4

2

2

2

4

2

That completes the Corner for this issue. Send me your nice solutions and contest materials.

404

BOOK REVIEWS Edited by ANDY LIU

144 Problems of the Austrian-Polish Mathematics Competition 1978{1993, compiled by Marcin Emil Kuczma, University of Warsaw, published by the Academic Distribution Center, 1993, 1216 Walker Road, Freeland, Maryland 21053, USA, ISBN# 0-9640959-0-4, softcover, 157+ pages, US$20 plus handling. 40th Polish Mathematics Olympiad 1989/90, edited by Marcin Emil Kuczma, University of Warsaw, published by Science, Culture, Technology Publishing, AMK Central Post Oce, Box 0581, Singapore 915603, no ISBN#, softcover, 49+ pages, US$10 plus handling. Reviewed by Andy Liu, University of Alberta. Both books under review are by Marcin Kuczma, one of the world's leading problem proposers. In IMO95 in Canada, two of his problems made it and a third nearly did. He is responsible for a substantial number of problems in these two books, which come with the highest recommendation. While originality can never be guaranteed since great minds tend to think alike across space and time, the author strived to come up with something new and exciting, and has largely succeeded. The rst book contains all the problems of the Austrian-Polish Mathematics Competition since its inception in 1978, up to 1993. It is one of the world's oldest regional mathematics competitions, attesting to the close friendship of the mathematical communities in these two nations. Apart from six problems given over two days in the IMO format, there is a team competition in which the six members collaborate to solve three problems in three hours. These students are usually those ranked immediately below the IMO team members. While the APMC may be considered a consolation for a graduating high school student, it is a wonderful experience and excellent preparation for those who have further aspirations in making the IMO team in future years. The second book is somewhat of an anomaly in publishing, in that it contains only the problems of one competition in one year. There are 24 problems in all, consisting of 12 from the First Round of the 1989/90 Polish Mathematics Olympiad, and 6 from each of the Second and the Third Rounds. Discounts are o ered for orders of multiple copies. Contact the publishers for details.

405 The Introduction in each of these two books provides valuable information about the respective competitions, but of course the main value lies in the problems. We give below a sample from each book. Both should be on the shelf of anyone serious about mathematics competitions. Problem 5, Austrian-Polish Mathematics Competition, 1985. We are given a certain number of identical sets of weights; each set consists of four di erent weights expressed by natural numbers (of weight units). Using these weights we are able to weigh out every integer mass up to 1985 (inclusive). How many ways are there to compose such a set of weights given that the joint mass of all weights is the least possible? Problem 21, Third Round, Polish Mathematical Olympiad, 1989/90. The edges of a cube are numbered 1 through 12. (a) Show that for every such numbering, there exist at least eight triples of integers (i;j; k), with 1  i < j < k  12, such that the edges labelled i, j and k are consecutive sides of a polygonal line. (b) Give an example of a numbering for which a ninth triple with these properties does not exist.

406

MORE UNITARY DIVISOR PROBLEMS K.R.S. Sastry A (positive) integral divisor d of a (natural) number n is calleda unitary  din n visor of n if and only if d and are relatively prime; that is d; = 1.

d

d

18  = (9; 2) = 1. For example, 9 is a unitary divisor of 18 because 9; 9 ,  But, 3 is not a unitary divisor of 18 because 3; = (3; 6) = 3 6= 1. This 

18 3

novel de nition of divisibility produces interesting analogues and contrasts with the results of ordinary divisibility. See [2, 3, 4]. In this paper we consider the solutions of two more new unitary divisor problems. BACKGROUND MATERIAL: Let n = p p    p k k denote the prime decomposition of n. Let d (n) denote the number of unitary divisors of n. Then, see [2, 3, 4], we have (1) = 1; d(n) = 2k for n > 1: (1) The Euler  function counts the number (n) of natural numbers that are less than and relatively prime to n. Then, see [1], we have 1

(1) = 1; (n) =

k Y

1

2

2

(pi , 1)p i i, for n > 1:

(2)

1

i=1

The rst problem asks us to determine the positive integers n for which d (n) = (n). The second asks us to establish a curious result concerning the number of unitary divisors of each unitary divisor of n. THE FIRST PROBLEM: Determine the set S = fn : d (n) = (n)g.

Solution: In view of equations (1) and (2), we have to solve the equation (p , 1) p , (p , 1) p ,    (pk , 1) p k k , = 2k : (3) The right hand side of equation (3) is a power of 2, so each factor on its left hand side is to be necessarily a power of 2. Two cases arise because of the even prime 2. (I) p = 2 and pi are distinct odd primes for 2  i  k. Then , 1 = , 1 =    = k , 1 = 0 and p , 1 = 2 ; p , 1 = 2 ;    ; pk , 1 = 2 k . Thus pi = 2 i + 1 are distinct odd primes for 2  i  k: (4) (II) pi are all odd primes for 1  i  k. Then as in the preceding, i = 1 and pi = 2 i + 1 are distinct odd primes for 1  i  k: (5) 1

1

1

1

2

2

2

1

1

1

3

2

2

2

3

3

407 First we note the obvious solution n = 1 that follows from d (1) = 1 and (1) = 1. (B) k = 1. This implies that n is formed from a single prime. (I) k = 1, p = 2. So n = 2 . This yields 2 , = 2 ; = 2. Hence n = 2 = 4. It is easily veri ed that d (4) = (4) = 2. (II) k = 1, p an odd prime. This yields (p , 1) = 2 ; p = 3. Hence n = 3. Again d(3) = (3) = 2. (C) k = 2. This implies that n is formed from powers of two distinct primes p ;p . (I) k = 2, p = 2 and p an odd prime. Hence n = 2 p . This yields 2 , 2 = 2 , ( , 1) + = 2. This equation has two solutions: ,  (C ) = 1; = 2 =) n = 2 2 + 1 = 10; ,  (C ) = 2; = 1 =) n = 2 2 + 1 = 12: (II) k = 2, p , p are both odd primes. Hence n = p p , 2 2 = 2 , + = 2. Now p and p are distinct odd primes so + = 2 are distinct natural numbers. Hence + > 2 and there is no solution in this (A)

1

1

2

1

1

1

1

1

1

1

1

1

2

1

1

1 2

2

1

2

2

1

1

2

2

1

2

1

1

1

2

1

2

1

2

2

1

2

1

2

2

1

1

1

2

2

2

2

case. (D) k = 3. In this case, n is the product of powers of three distinct primes. (I) k = 3, p = 2, and p , p are distinct odd primes. Hence n = 2 p p , 2 , 2 2 = 2 , ( , 1) + + = 3. Since is at least 1 and only one of and can equal, 1, we have the single solution = 1, = 1, = 2. This gives n = 2 2 + 1 ,2 + 1 = 30. (II) k = 3, p , p , p are distinct odd primes. This leads to the equation + + = 3 that has no solution in natural numbers, as explained above. (E) k > 3, n = p p    p k k . (I) k > 3, p = 2 yields ( , 1) + + +    + k = k, which has no solution for the reason explained above in (D). (II) k > 3 yields + +    + k = k, and this equation too does not have a solution. Combining (A),    , (E) we have determined S = f1; 3; 4; 10; 12; 30g. The veri cation that if n 2 S then d (n) = (n) is left to the reader. THE SECOND PROBLEM: The fascinating property of the natural numbers that 1 + 2 +    + n = (1 + 2 +    + n) motivates us to look for a collection of integers with the similar property. For instance the collection of the integers 1, 2, 2, 2, 4, 4, 4, 8 exhibit that property: 1

1

2

1 1

3

2

2

3

2 3

2

3

2

1

2

1

2

1

3

1

2

3

3

1

1

2

2

1

1

1

3

3

3

1

1

1

2

3

2

3

2

3

1 + 2 + 2 + 2 + 4 + 4 + 4 + 8 = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8) : 3

3

3

3

3

3

3

3

2

408 How to generate collections of such integers exhibiting the above property? Here is an algorithm: (A ) Start with any integer n. Say n = 30. (A ) Write down the unitary divisors of n. The unitary divisors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. (A ) Now write down the number of unitary divisors of each unitary divisor of n. In view of (1) each of these integers will be a power of 2. In the present case we have respectively 1 2

3

1; 2; 2; 2; 4; 4; 4; 8:

Lo and behold! We have generated a collection of integers having the mentioned property. Intrigued? Try this on n = 24 or n = 420 or any other value. In general we have the following result: Let a , a ,    , a denote the unitary divisors of an integer n. Let d (ai) denote the number of unitary divisors of ai. Then 1

2

 X

[d(ai)]3 =

i=1

"

 X

i=1

#2

d(ai)

:

Proof: Let n = p p    p k k denote the prime decomposition of n. Then its 2k =  unitary divisors are: 1

1

2

2

a ; a ;    ; a = 1;k p ; p ;    ; pk ; k, p p ; p p ;    ; pk, pk k ; 1

1

.. .

1

2

1 1

2

2 2

2

1

1

3

1 1

3

p p    p k k : 1

1

2

2

Observe that the unitary divisors have been put together, for convenience, in collections containing

k ; k ; k ;    ; k 0 1 2 k

     

terms respectively. Here kr = k k, r k,r denotes the binomial coecient for 0  r  k. We now use (1) to write down d (ai ) for 1  i  2k : , 

(

1)

( !

+1)

1; 2; 2;    ; 2; 2 ; 2 ;    ; 2 ; 2 ; 2 ;    ; 2 ;    ; 2k: 2

2

2

3

3

3

, 

Observe that these collections contain, respectively,, k term equalling 1: ,k ,k terms equalling 2; terms equalling 2 ;    ; kk term equalling 2k . Hence     X k  [ (ai)] = 1 + 2 + 2 +    terms 1

2

2

3

3

3

3

0

1

 k + 2 + 2 +    2 terms +    + 2 k  : 

 

6

6

3

409 This is nothing but the expansion of ,

1+2

Also X

3

k

= 9k = 3 k: 2

    (ai) = 1 + 2 + 2 +    k1 terms     + 2 + 2 +    k terms +    + 2k 2

=

2

(1 + 2)k = 3k:

2

It is easy to see that the assertion follows. Finally we invite the reader to state and prove an analogous result for the ordinary (positive integral) divisors of (positive) integers n. For other problems on cubes of natural numbers see [5, 6]. References [1] L.E. Dickson, History of the Theory of Numbers, Vol. I, Chelsea Publications, (1971), pp. 113{114. [2] R.T. Hanson and L.G. Swanson, Unitary Divisors, Mathematics Magazine, 52 (September 1979), pp. 217{222. [3] K.R.S. Sastry, Unitary Divisor Problems, Crux Mathematicorum, 22 (February 1996), pp. 4{9. [4] K.R.S. Sastry, Unitary Divisor Analogues, Mathematics and Computer Education, 30 (Fall 1996), pp. 300{311. [5] K.R.S. Sastry, Fermat-Pell: An Observation and an Application, Mathematics and Computer Education, 28 (Winter 1994), pp. 24{35. [6] K.R.S. Sastry, Cubes of Natural Numbers in Arithmetic Progression, Crux Mathematicorum, 18, (1992), pp. 161{164. K.R.S. Sastry Jeevan Sandhya Doddakalsandra Post Raghuvanahalli Bangalore 560 062, India

410

THE SKOLIAD CORNER No. 25

R.E. Woodrow This number we give the problems of the 1997 Alberta High School Mathematics Prize Exam, second round, written last February by students invited on the basis of the rst round results from November. My thanks go to Ted Lewis, Chair of the Alberta High School Mathematics Prize Exam Board for forwarding the contest, which is partially supported by the Canadian Mathematical Society.

ALBERTA HIGH SCHOOL MATHEMATICS COMPETITION February 11, 1997

Second Round 1. Find all real numbers x satisfying jx , 7j > jx + 2j + jx , 2j. Remark. Note that jaj is called the absolute value of the real number a. It has the same numerical value as a but is never negative. For example, j3:5j = 3:5, while j , 2j = 2. Of course, j0j = 0. 2. Two lines b and c form a 60 angle at the point A, and B is a point on b. From B , draw a line perpendicular to the line b meeting the line c at the point C . From C draw a line perpendicular to c meeting the line b at B . Continue in this way obtaining points C , B , C , and so on. These points are the vertices of right triangles AB C ; AB C ; AB C ; : : : . If area (AB C ) = 1, nd 1

1

1

1

2

2

1

1

1

3

2

3

2

3

3

1

area (AB1C1 )+ area (AB2C2 )+ area (AB3C3 )+    + area (AB1997C1997):

3. A and B are two points on the diameter MN of a semicircle. C , D, E and F are points on the semicircle such that \CAM = \EAN = \DBM = \FBN . Prove that CE = DF . 4. (a) Suppose that p is an odd prime number and a and b are positive integers such that p divides a + b and p also divides a(a + b) . Prove that p also divides a(a + b). (b) Suppose that p is an odd prime number and a and b are positive integers such that p divides a + b and p also divides a(a + b) . Show by an example that p does not necessarily divide a(a + b). 4

2

2

4

2

4

5

2

2

5

2

5

5. The picture shows seven houses represented by the dots, connected by six roads represented by the lines. Each road is exactly 1 kilometre long. You live in the house marked B . For each positive integer n, how many

411 ways are there for you to run n kilometres if you start at B and you never run along only part of a road and turn around between houses? You have to use the roads, but you may use any road more than once, and you do not have to nish at B . For example, if n = 4, then three of the possibilities are: B to C to F to G to F ; B to A to B to C to B; and B to C to B to A to B.

D A

B

s

E

s

s

C

s

s

F

G

s

s

Next we give the solutions to the 1995 Concours Mathematique du Quebec which comes to us from Therese Ouellet, secretary to the contest. We give the solutions in French, the language of the competition.

  CONCOURS MATHEMATIQUE DU QUEBEC 1995 February 2, 1995 Time: 3 hours

1. LA FRACTION A SIMPLIFIER Simpli ez la fraction Solution.

1 358 024 701 : 1 851 851 865

1 358 024 701 = 11  123 456 791 = 11 : 1 851 851 865 15  123 456 791 15

 2. LA FORMULE MYSTERE

Considerons les e quations suivantes

xy = p; x + y = s; x + y = t; x + y = u: En faisant appel aux lettres p, s, t, u mais pas aux lettres x, y , donnez une 1993

formule pour la valeur

x

1995

1993

+y

1994

:

1995

1994

412 Solution. On a successivement

su = (x + y)(x + y ) = x + xy + x y + y = x + y + xy  (x + y = x + y + pt: 1994

1995

D'ou l'on tire

1994

1995

1995

1995

1995

x

1994

1995

+y

1995

1994

1995

1993

)

1993

= su , pt:

 3. LA DIFFE RENCE ETONNANTE

Lorsque la circonference d'un cercle est divisee en dix parties e gales, les cordes qui joignent les points de division successifs forment un decagone regulier convexe. En joignant chaque point de division au troisieme suivant, on obtient un decagone regulier e toile. Montrer que la di erence entre les c^otes de ces decagones est e gale au rayon du cercle. Solution. On considere le cercle divise en dix parties e gales. On joint les points CD et DE , par des segments e gaux aux c^otes du decagone convexe, les points BE et CF , par des segments e gaux aux c^otes du decagone e toile, et les diametres AF et BG. Alors CD k BE k AF et DE k CF k BG. En vertu des proprietes des parallelogrammes, on a

BE , CD = BE , HE = BH = OF = r: D E s

C B

H

s

s

O

s

A

s

r

F s

s

G

s

s

s

4. LE TERRAIN EN FORME DE CERF-VOLANT

Abel Belgrillet est membre du Club des aerocervidophiles (amateurs de cerfs-volants) du Quebec. Il dispose de quatre troncons de cl^oture rectilignes AB, BC , CD, DA pour delimiter un terrain (en forme de cerf-volant, voir gure) sur lequel il s'adonnera a son activite favorite cet e te. Sachant que les

413 troncons AB et DA mesurent 50 m chacun et que les troncons BC et CD mesurent 120 m chacun, determinez la distance entre les points A et C qui maximisera l'aire du terrain. Solution. Decoupons le terrain selon AC : D D l'aire de A

120

50 50

= l'aire de A

C

120

B

120

50 120

B

C 50

Mais, l'aire d'un parallelogramme de c^otes donnes est maximale lorsque ce parallelogramme est un rectangle (car dans ce cas, la hauteur sera maximale). On doit donc avoir que l'angle Bb = 90 . Le triangle ABC est donc rectangle en B . Ainsi

AC = AB + BC = p2500 + 14400 = 130: q

2

2

 5. L'INEGALIT E MODIFIE E D'AMOTH DIEUFUTUR

(a) (2 points) L'inegalite x + 2y  3xy est-elle vraie pour tous les entiers? (b) (8 points) Montrez que l'inegalite x + 2y  xy est valide pour tous reels x et y . Solution. (a) Non. Voici quelques couples qui fournissent un contreexemple: 2

2

2

14 5

2

x y 3 4 5 5 6

2 3 3 4 4

(b) 1e re solution. Considerons l'inegalite (x , y )  0, qui est vraie pour tout couple de reels x, y . Elle est e quivalente a x + y  2 xy . p p En posant = 2, on trouvepx +2y  2 2 xy = (2; 828 : : : )xy  xy . On peut aussi montrer que 2 2  en e levant au carre de chaque c^ote. 2ieme solution. Montrer que x + 2y  xy revient a demontrer l'inegalite 5x + 10y , 14xy  0. Or, 2

2

2

2

2

14 5

2

14 5

2

2

14 5

2

2

5x + 10y , 14xy = x , 2xy + y + 4x , 12xy + 9y = (x , y ) + (2x , 3y )  0 2

2

2

2

2

2

2

2

414 la derniere inegalite resultant du fait qu'une somme de carres est toujours superieure ou e gale a zero.  6. L'ECHIQUIER COQUET Trouvez l'unique facon de colorier les 36 cases d'un e chiquier 6  6 en noir et blanc de sorte que chacune des cases soit voisine d'un nombre impair de cases noires. (Note: deux cases sont voisine si elles se touchent par un c^ote-ou par un coin). On ne demande pas de demontrer que la solution est unique. Solution. Ayant une solution, chacune des quatre symmetries possibles (les deux diagonales, la verticale et l'horizontale) engendre une nouvelle solution. Or, puisque la solution est unique, elle doit necessairement e^ tre symetrique selon ces quatre axes de symetries. Il s'ensuit que les cases numerotees 1 ci-dessous sont toutes de la m^eme couleur. De m^eme pour les cases numerotees 2, numerotees 3; : : : ; numerotees 6.

1 2 3 2 1

2 4 5 4 2

3 5 6 5 3

3 5 6 5 3

2 4 5 4 2

1 2 3 2 1

Pour que les cases 1 touchent a un nombre impair de cases noires, les cases 4 doivent necessairement e^ tre noires, puisque chacune d'elles est voisine de deux cases 2 et d'une case 4. De plus, les cases 4 e tant voisines d'une case 1, d'une case 6, et de deux cases 2, 3 et 5, il est clair que les cases 1 et 6 doivent e^ tre de couleurs di erentes. De plus, les cases 4 e tant noires, les cases 6 sont necessairement blanches, puisqu'elles sont voisines chacune d'une case 4, de deux cases 5 et de trois cases 6. Les cases 1 sont donc noires, puisque nous avons deja e tabli qu'elles sont de couleurs di erentes que les cases 6. En considerant maintenant les cases 3, et ce qu'on sait deja, on conclut que les cases 2 et 3 doivent e^ tre de m^eme couleur. Maintenant, en considerant les cases 2, sachant que 1 et 4 sont noires et que 2 et 3 sont de m^eme couleur, on deduit que 5 doit e^ tre noir. En considerant la position 5, maintenant, on conclut nalement que les cases en 2 et en 3 sont blanches, ce qui nous donne nalement:

415

7. LA FRACTION D'ANNE GRUJOTE

 En base 10, = 0:333 : : : . Ecrivons 0:3 pour ces decimales repetees. Comment e crit-on dans une base b, ou b est de la forme (i) b = 3t (trois points), (ii) b = 3t + 1 (trois points), (iii) b = 3t + 2 (quatre points), ou t est un entier positif quelconque? Solution. (i) Notons = tt , d'ou = 0:t en base 3t. 1 3

1 3

1 3

1 3

3

(ii) Supposons que = 0:d d d : : : . Alors t est e gal a la partie entiere de t . Donc d = t. 1 3

1 2 3 +1 3

3 +1 3

3

= d :d d : : : et d 1

2

3

1

1

Soustrait d = t de t . On obtient 3 +1 3

1

3t + 1 , t = 3t + 1 , 3t = 1 = 0:d d : : : : 3 3 3 3 2

3

Le procede donne t = d = d = : : : . Donc = 0:t. 2

1 3

3

(iii) Supposons encore = 0:d d d : : : . Alors t La partie entiere de t est t. Donc d = t. 3 +2 2

1 3

1

2

3 +2 3

3

= d :d d : : : : 1

2

3

1

3t + 2 , 3t = 2 = 0:d d : : : : 3 3 3 2

3

On multiplie encore par 3t + 2. On obtient d :d d : : : = t = t , dont la partie entiere est 2t +1. Donc d = 2t +1. On soustrait d = 2t +1 et on obtient 2

3

2(3 +2) 3

4

2

6 +4 3

2

6t + 4 , 3(2t + 1) = (6t + 4) , (6t + 3) 3 3 3 1 = 3 = 0:d d : : : : 3

4

On voit que t = d = d = d = : : : et 2t +1 = d = d = d = : : : : Donc 1

3

5

1 = 0:t (2t + 1): 3

2

4

6

416 Veri cation

0:t (2t + 1) = = = = =

(3t + 2)t + (2t + 1) 1 + 1 + 1 +     (3t + 2)( (3t + 2) (3t + 2) ) 1 3t + 4t + 1 (3t + 2) 1, t  3t + 4t + 1 (3t + 2)  (3t + 2) (3t + 2) , 1 3t + 4t + 1 9t + 12t + 3 1: 3 2

2

4

2

2

1 (3 +2)2

2

2

2

2

2

2

That completes the Skoliad Corner for this issue. Please send me suitable contest materials at this level, as well as your comments and suggestions for future columns.

417

MATHEMATICAL MAYHEM Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. All material intended for inclusion in this section should be sent to the Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University, PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic address is still [email protected] The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto). The rest of the sta consists of Richard Hoshino (University of Waterloo), Wai Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices Factorial Fanaticism If you've seen one too many math contests, then the following factorial problem will be old hat: \How many factors of 2 are there in 100!?" For those who don't recognize \!" as a mathematical symbol, here is its meaning: n! is the product of all the integers from n down to 1, or

n! = n  (n , 1)  (n , 2)      3  2  1: For example, 5! = 5  4  3  2  1 = 120.

So how many factors of 2 are there in 100!? Well, there are = 50 even numbers giving 50 factors of 2. But every multiple of 4 gives = 25 of them. Continuing, each an additional factor of 2. There are multiple of 8 gives another factor of 2, and there are b c = 12 of them, and so on for multiples of 16, 32, and 64. This is enough, as the next power of 2, that is 128, is greater than 100, so there are no multiples of it which are less than 100. The total number of factors of 2 is then 50 + 25 + 12 + 6 + 3 + 1 = 97: 100 2

100 4

100 8

[Ed.: Coincidence?] Now try and nd how many factors of 5 there are 100!. Find a general formula for nding the number of primes p in n! for given integers p and n. A similar problem to the one above is to calculate the number of consecutive zeros at the end of 100!. It is equal to the number of factors of 5

418 in 100! (Why?). Once you have braved this question, you can easily gure out a method for nding the number of factors of m in n! for given integers m and n. In fact, it has been shown (and is an excellent exercise for the reader to show) that the number of factors of a prime number p in n! is

n , sn ; p,1 where sn is the sum of the digits of n when expressed in base p.

A natural question to ask, after knowing all the zeros at the end of n!, is \What is the last non-zero digit in n!?" One way, perhaps not the most elegant, is to write n! in the form p p    p r r , where the pi are all the distinct primes dividing n. The , : : : , r , are calculated by the method above. Since we don't care about the trailing zeros, say k of them, remove a factor of 10k and then take the quotient modulo 10. For those who are not familiar with modular arithmetic, all this does is nd the units digits. For example, 6! = 720 = 2  3  5. Removing a factor of 10 to get rid of the trailing zeros gives 72 = 2  3 . Then 2  3 is congruent to 2 modulo 10, that is, the units digit is 2. Thus, the last non-zero digit of 6! is 2. Now for some more interesting questions. 1. How many digits does n! have? 2. What is the initial digit of n!, or more generally its rth digit? 3. How many digits are zeros in n!? How many are 1's? Likewise for the other digits. 1

1

2

2

1

4

2

3

2

3

2

419

Solving the Quartic Cyrus Hsia Most senior high school students are familiar with the general solution to the quadratic equation ax + bx + c = 0. Either it is derived or given as a mere fact in class, and the notorious solution is 2

p x = ,b  2ba , 4ac : 2

It is rarely the case, however, that the solution to the general cubic equation ax + bx + cx + d = 0 is ever derived much less even mentioned. Here we do not tread this path, but refer the reader to [1] and [2] for its treatment. Instead, we go one step further and give the process of solving the general quartic (fourth degree) equation ax + bx + cx + dx + e = 0. Readers are encouraged to read [1] and [2] to learn or refresh their memories on solving the cubic equation. This is because we wish to reduce the quartic equation into that of a cubic equation. The process we give here is a variant of the solution credited to Lodovico Ferrari [3]. Motivations Readers who are familiar with solving cubic equations know that the rst step is to reduce the general cubic into the depressed cubic equation x + px + q = 0. This may be of some value in the quartic case as well. Also, if the quartic equation could be written as a di erence of squares, that is (x + p) , (qx + r) = 0, then solving the original quartic becomes an exercise in solving two quadratic equations | namely, solving x + p + qx + r = 0 and x + p , qx , r = 0: To this end, we try to convert the general quartic equation into this form somehow. Supposing we have reduced ax + bx + cx + dx + e = 0 into the form x + ux + vx + w = 0. (How do we do this? See the method below.) Then we compare coecients in 3

2

4

3

4

3

2

3

2

2

2

2

2

4

2

2

x + ux + vx + w = (x + p) , (qx + r) = x + (2p , q )x + (,2qr)x + (p , r ) to solve for p, q , and r in terms of the known u, v , and w. (How is this done? 4

2

2

2

2

4

2

2

2

2

See the method below.) Voila! The solution falls apart. Method for Solving the General Quartic Equation  Step 1: Reduce ax + bx + cx + dx + e = 0 to the form x + ux + vx + w = 0. To do this, simply divide through by a. (We must have a 6= 0; else the equation would no longer be a quartic.) Next, make the substitution x = t , ba to get t + ut + vt + w = 0. 4

4

3

2

2

4

4

2

420 

Step 2: Convert the reduced quartic from Step 1 to a di erence of squares as described above. To do this, we need to solve for p, q , and r in terms of the known u, v , and w. By comparing coecients, we have three equations in three unknowns:

2p , q = u ,2qr = v p ,r = w Solving for p, we have v = 4q r = 4(2p , u)(p , w). This is a cubic 2

2

2

2

2 2

2

equation in p, which can be solved.  Step 3: Take the di erence of squares found in Step 2 and solve for the two quadratic equations to get the four solutions of the quartic equation t + ut + vt + w = 0. Remember,bthese are not the desired solutions. Make the substitution x = t , a for each of the four solutions to get the four solutions of the original quartic equation. (Why didn't we multiply by a?) Afterword This seems quite tedious, and is thus one reason why it is not taught in classrooms. However, note how simple solving the quartic is if a black-box was given to automatically solve a cubic. The quartic is solved by using this black-box once and solving an additional two quadratic equations. The best way to learn a technique is to use it. Here is a problem where the reader is encouraged to apply the above algorithm. Although there are more elegant ways to solve this, by using Ferrari's method the reader will gain a greater understanding of solving polynomial equations. Solve the equation 4

2

4

x + (x +x 1) = 3: 2

2

2

(1992 CMO Problem 4) We end with a note that polynomial equations of degree ve or higher cannot be solved, so readers can be relieved that there are not longer processes which reduce polynomials one degree at a time to reach a solution. References [1 ] Hlousek, Daniela and Lew Dion, \Cardano's Little Problem", Mathematical Mayhem, Volume 6, Issue 3, March/April 1994, pp. 15-18. [2 ] Dunham, William, Journey Through Genius, Penguin Books, New York, 1990, pp. 142-149. [3 ] Dunham, William, Journey Through Genius, Penguin Books, New York, 1990, p. 151.

421

Pizzas and Large Numbers Shawn Godin The following is an excerpt from a television commercial for a certain unnamed pizza chain: Customer: So what's this new deal? Pizza Chef: Two pizzas. Customer: [Towards four-year-old Math Whiz] Two pizzas. Write that down. Pizza Chef: And on the two pizzas choose any toppings - up to ve [from the

list of 11 toppings]. Older Boy: Do you... Pizza Chef: ...have to pick the same toppings on each pizza? No! Math Whiz: Then the possibilities are endless. Customer: What do you mean? Five plus ve are ten. Math Whiz: Actually, there are 1 048 576 possibilities. Customer: Ten was a ballpark gure. Old Man: You got that right. [At this point, the camera fades to a picture of hot pizzas; the announcer's voice and a very small Roman who says \Pizza! Pizza!"]

This simple advertisement caused an uproar in the mathematics community ([1], [2], [3], [4], [5]). The question is, should a four year old be trusted with higher mathematics? To answer the question, we should rst do a little analysis. Before jumping into the problem, let us state some assumptions: (i) Each pizza may have up to ve toppings on it (given). (ii) Both pizzas do not have to have the same toppings (given). (iii) A pizza may have no toppings (although a boring choice, it seems reasonable since most pizza places include a price for \sauce and cheese" as the basic \no item" pizza). (iv) The order that the items are put on the pizza does not matter. So a pizza with ham and bacon is the same as a pizza with bacon and ham (it seems obvious, but this decision will a ect our solution). (v) The order that you order the pizzas, or the order that they are put in the box, or handed to you, does not matter. That is, ordering a pizza with ham and a second pizza with mushrooms is the same as ordering a pizza with mushrooms and another with ham (again, it seems obvious, but this decision will a ect our solution).

422 (vi) Double items are not allowed (although it is allowed in most pizza places, we will consider this simpler problem rst then extend it). Now o to the solution. First, we will look at the number of ways that one pizza can be made. Our pizza may have zero, one, two, three, four, or ve toppings, so the total number of pizzas available to us is: 

11 + 11 +    + 11 = 1024: 5 4 0

Thus, if we are to choose two pizzas we have two possibilities: Case 1: Both pizzas are the same. In this case, there are 1024 ways to pick the rst pizza and, having picked the rst one, we are locked in for our second one (it must be the same as the rst) so there are 1024 ways to pick two pizzas if they must be the same. Case 2: Both pizzas are di erent. In this case, there are 1024 pizzas to choose from and I want to pick two, so the number of ways to do this is: 

1024 = 523 776: 2

So the total number of ways to pick the two pizzas is:

1024 + 523 776 = 524 800; which does not agree with the four year old. Now, as luck would have it, if we disregard our assumption (v), the number of ways to create pizzas in Case 2 is P (1024; 2) = 1 047 552, giving a total of 1024 + 1 047 552 = 1 048 576, which is the four year old's claim. It would seem that the child regards the order that the pizzas are ordered in as important. Indeed, Jean Sherrod of Little Caesars Enterprises explained in a letter to Mathematics Teacher ([3]) that that is indeed what they assumed. If they are assuming this, that is disregarding our assumption (v), would not it also make sense to regard the order that the toppings were asked for as important? How many pizzas would be possible if we rewrote assumption (iv) so that order was important? Let us pursue the truth even further. As is well known to pizza connoisseurs, and as is admitted by Sherrod ([3]), a pizza may have a multiple selection of an item. Double pepperoni would count as two choices, both of the same topping. Weber and Weber ([4]) consider the multiple topping question with an interesting technique. Suppose our 11 toppings are as follows: pepperoni (P), bacon (B), sausage (S), ham (H), anchovies (A), mushrooms (M), green peppers (GP), tomatoes (T), green olives (GO), black olives (BO), and hot peppers (HP). When you order a pizza, the order is processed on an order form that lists all

423 possible toppings separated by vertical lines. The order form may look like this: P j B j S j H j A j M j GP j T j GO j BO j HP and the person taking the order just marks an X over the symbol for each topping you order. If you wanted double pepperoni, two X's would be placed in the P slot, counting as two toppings. If we allow multiple toppings, we can calculate the total number of pizzas possible by considering the total number of ways that the order form can be lled out. For example, if you ordered your own favourite pizza, the \Cardiac Conspiracy", which has double bacon, double pepperoni, and ham, the order form (if we just look at the lines and X's) would look like this: XX j XX j j X j j j j j j j Similarly, if you ordered a \Trip To The Sea" pizza (quintuple anchovies), the order form would look like this: j j j j XXXXX j j j j j j Thus, each 5 item pizza will be represented by an arrangement of 5 X's and 10 lines. The number of ways to arrange these 15 symbols is:

15! 5!10! = 3003:

This is in fact the binomial coecient C (15; 5), or C (15; 10). This result makes more sense if we think of the problem as picking 5 of the 15 positions to mark with an X (or alternately, pick 10 of the 15 spaces to mark with a line). To account for all possible pizza choices, we need to consider pizzas with zero, one, two, three, four, or ve toppings. If we do this we get:

15! 14! 10! 10!5! + 14!4! +    + 10!0! = 4368: So, the number of choices for two pizzas is then C (4368; 2) + 4368 = 9 541 896. So now we have a lot of numbers to deal with. In our rst calculation, we saw that there are 1024 possible pizzas allowing no multiple toppings. In our latest calculation, we saw that there are a total of 4368 pizzas if multiple toppings are allowed. Thus, since the original 1024 has been accounted for in the 4368, it would seem that 4368 , 1024 = 3344 pizzas have multiple toppings. Our next problem to consider is to classify these 3344 pizzas. Consider the case where a double topping is allowed and is used. Then a ve item pizza could be made up of a double item and three singles, or two double items and a single. The method of Weber and Weber for enumerating pizza choices here becomes too cumbersome to work with. For example, if I marked a double item with a D and a single with an X, then arranging 10

424 lines, a D and 3 X's to get our rst case does not take into account X's beside the D (more than a double) or even two or three X's together (which means we recount later cases). So to solve the problem, we will have to go back to the fundamental counting theorem. First, we will consider the pizza made up of a double item and three singles. Our rst task is to select the item that will be double. This can be done in C (11; 1) = 11 ways. Next, pick the three singles from the other remaining toppings. This can be done in C (10; 3) = 120 ways. So the total number of pizzas with a double item and three singles is C (11; 1)C (10; 3) = 1320. Continuing in this manner, we can account for all pizzas with at least one double topping and no topping is allowed more than twice (can you identify each term with the case it represents?):

1110 + 119 + 1110 + 11 + 1110 + 11 = 2486: 1 3 2 1 1 2 2 1 1 1 Now we have accounted for 1024 + 2486 = 3510 of the total 4368



possible single pizzas. At this point, the reader should have enough ammunition to attack the remaining problem. The following questions are still unanswered. 1. How many pizzas are possible if triple (quadruple, quintuple) toppings are allowed? (pick your cases carefully!) 2. How many choices of two pizzas do you have if you are allowed to pick an item at most twice (three, four, ve times)? 3. Using local pizza company choices, how many pizzas (or pairs of pizzas) are possible from each pizza place? 4. Who in town makes the best pizza? Have fun trying to solve these and other problems of your creation. Mathematics is a demanding endeavour, so if you feel hungry you may want to pick up the phone... References 1. N. Kildonan, problem 1946, Crux Mathematicorum, 20:5 (1994) 137. 2. M. Woltermann, \Pizza Pizza", Mathematics Teacher, 87:6 (1994) 389. 3. J. Sherrod, \Little Caesars Responds", Mathematics Teacher, 87:6 (1994) 474. 4. G. Weber and G. Weber, \Pizza Combinatorics", The College Mathematics Journal, 26:2 (1995) 141-143. 5. S. Godin, Solution to problem 1946, Crux Mathematicorum, 21:4 (1995) 137-139.

425

Minima and Maxima of Trigonometric Expressions Nicholae Gusita Problem 1. Find the minimum and maximum values of

y = sin x + cos x; 1

for 0  x < 360. Solution. We write

p y = sin x + sin(90 , x) = 2 sin45 cos (x , 45) = 2 cos (x , 45) : p Then the minimum of , 2 occurs when x , 45 = 180 , so that p  x = 225 , and the maximum of 2 occurs when x , 45 = 0, so that  x = 45 . 1

Problem 2. Find the minimum and maximum values of

y = 3 sin x + 4 cos x; 2

for 0  x < 360. Solution. If we put  = tan,

1

,4 3

, then

 sin  cos x = 3 sin(x + ) : y = 3 (sin x + tan  cos x) = 3 sin x + cos  cos  2

But

1 9; cos  = 1 + tan =  25 so that cos  = 35 ; since 0 <  < 90 . Then y = 5 sin (x + ). It is clear now that the minimum of ,5 occurs when x +  = 270 , so that x = 270 , , and the maximum of 5 occurs when x +  = 90 , so that x = 90 , . 2

2

2

Problem 3. Find the minimum and maximum values of

y = a sin x + b cos x; for 0  x < 360, a, b 2 Rnf0g. ,  Solution. Let  = tan, ab , ,90 <  < 90 . Assume a > 0. Then   sin  (x + ) : y = a (sin x + tan  cos x) = a sin x + cos  cos x = a sincos  3

1

3

426 But as in Problem 2,

1 a cos  = 1 + tan  = a +b ; 2

2

2

2

2

so that

cos  = p a : a +b p Therefore, y = a + b sin(x + ). p b occurs when x +  = 270, so that x = The minimum of , a + p 270 , , and the maximum of a + b occurs when x +  = 90 , so that x = 90 , . If a < 0, then using similar reasoning, we nd the minimum occurs when x = 90 , , and maximum when x = 270 , . Problem 4. Given two positive angles apand b whose sum is 150 , prove that the maximum value of sin a + cos b is 3. 2

2

3

2

2

2

2

2

2

Solution. We have

y = sin a + cos b = sina + sin (90 , b)  = 2 sin a ,2 b + 45 cos a +2 b , 45 p = 2 sin (a , 30 ) cos 30 = 3 sin(a , 30) : p The maximum of 3 occurs when a , 30 = 90 , so that a = 120 , b = 30 . 4

Problem 5. Find the minimum and maximum values of

y = a sin x + b sin x cos x + c cos x; 0  x < 360 , a, b, c 2 R. Solution. Since sin x = (1 , cos 2x), cos x = (1 + cos 2x), and sin x cos x = sin2x, y = a2 (1 , cos 2x) + 2b sin2x + 2c (1 + cos 2x) = a + c + 1 [b sin2x + (c , a) cos 2x] : 2

5

2

2

1 2

1 2

1 2

2

5

2

2

By problem (3), the minimum and maximum values are q

a + c , b + (a , c) 2

respectively.

2

2

q

and

a + c + b + (a , c) 2

2

2

427

J.I.R. McKnight Problems Contest 1979 1. Sand is being poured on the ground forming a pile in the shape of a cone whose altitude is of the radius of its base. The sand is falling at the rate of 20 L/s. How fast is the altitude of the pile increasing when it is 42 cm high? 2. Solve: 2 3

tan,1



x , 1  + tan,  2x , 1  = tan,  23  : x+1 2x + 1 36 1

1

3. A cylinder is inscribed in a cone, the altitude of the cylinder being equal to the radius of the base of the cone. Find the measure of angle , if the ratio of the total surface area of the cylinder to the area of the base of the cone is 3 : 2.

6  R

R

-?

4. Solve the equation jxj + jx , 1j + jx , 2j = 4. 5. Solve for x and y .

2x

y 3x

+

= 6y = 3  2y

+1

6. A xed circle of radius 3 has its centre at (3; 0). A second circle has its centre at the origin, its radius is approaching zero. A line joins the point of intersection of the second circle and the y -axis to the intersection of the two circles. Find the limit of the point of intersection of this line with the x-axis.

428

Mayhem Problems The Mayhem Problems editors are: Richard Hoshino Mayhem High School Problems Editor, Cyrus Hsia Mayhem Advanced Problems Editor, Ravi Vakil Mayhem Challenge Board Problems Editor. Note that all correspondence should be sent to the appropriate editor | see the relevant section. In this issue, you will nd only problems | the next issue will feature only solutions. We warmly welcome proposals for problems and solutions. With the new schedule of eight issues per year, we request that solutions from this issue be submitted by 1 December 1997, for publication in the issue 5 months ahead; that is, issue 4 of 1998. We also request that only students submit solutions (see editorial [1997: 30]), but we will consider particularly elegant or insightful solutions from others. Since this rule is only being implemented now, you will see solutions from many people in the next few months, as we clear out the old problems from Mayhem.

High School Problems Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario, Canada. L3S 3E8 H229. Here's a simple way to remember how many books there are in the Bible. Remember that there are x books in the Old Testament, where x is a two-digit integer. Then multiply the digits of x to get a new integer y, which is the number of books in the New Testament. Adding x and y , you end up with 66, the number of books in the Bible. What are x and y ? H230. Dick and Cy stand on opposite corners (on the squares) of a 4 by 4 chessboard. Dick is telling too many bad jokes, so Cy decides to chase after him. They take turns moving one square at a time, either vertically or horizontally on the board. To catch Dick, Cy must land on the square Dick is on. Prove that: (i) If Dick moves rst, Cy can eventually catch Dick. (ii) If Cy moves rst, Cy can never catch Dick. (Can you generalize this to a 2m  2n chessboard?) H231. Let O be the centre of the unit square ABCD. Pick any point P inside the square other than O. The circumcircle of PAB meets the circumcircle of PCD at points P and Q. The circumcircle of PAD meets the circumcircle of PBC at points P and R. Show that QR = 2  OP .

429

H232. Lucy and Anna play a game where they try to form a ten-digit number. Lucy begins by writing any digit other than zero in the rst place, then Anna selects a di erent digit and writes it down in the second place, and they take turns, adding one digit at a time to the number. In each turn, the digit selected must be di erent from all previous digits chosen, and the number formed by the rst n digits must be divisible by n. For example, 3, 2, 1 can be the rst three moves of a game, since 3 is divisible by 1, 32 is divisible by 2 and 321 is divisible by 3. If a player cannot make a legitimate move, she loses. If the game lasts ten moves, a draw is declared. (i) Show that the game can end up in a draw. (ii) Show that Lucy has a winning strategy and describe it.

Advanced Problems Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada. M1G 1C3 A205. Find all functions f : R ! R such that f (x)  x and f (x + y)  f (x) + f (y) for all x, y 2 R. A206. Let n be a power of 2. Prove that from any set of 2n , 1 positive integers, one can choose a subset of n integers such that their sum is divisible by n. A207. Given triangle ABC , let A00, B00 , and C 0 be points on sides BC , 0 CA, and AB respectively such that 40A 0B 0C  4ABC . Find the locus of the orthocentre of all such triangles A B C . A208. Let p be an odd prime, and let Sk be the sum of the products of the elements f1; 2; : : : ; p , 1g taken k at a time. For example, if p = 5, then S = 1  2  3 + 1  2  4 + 1  3  4 + 2  3  4 = 50. Show that p j Sk for all 2  k  p , 2. 3

Challenge Board Problems Editor: Ravi Vakil, Department of Mathematics, Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544{1000 USA

C74. Prove that the k-dimensional volume of a parallelopiped in Rn

spanned by the vectors ~v , : : : , ~vk is the determinant of the k  k matrix fvi  vj gi;j . 1

430

PROBLEMS Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 "11" or A4 sheets of paper. These may be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later than 1 May 1998. They may also be sent by email to [email protected]. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication. 1 2

Solutions submitted by FAX There has been an increase in the number of solutions sent in by FAX, either to the Editor-in-Chief's departmental FAX machine in St. John's, Newfoundland, or to the Canadian Mathematical Society's FAX machine in Ottawa, Ontario. While we understand the reasons for solvers wishing to use this method, we have found many problems with it. The major one is that hand-written material is frequently transmitted very badly, and at times is almost impossible to read clearly. We have therefore adopted the policy that we will no longer accept submissions sent by FAX. We will, however, continue to accept submissions sent by email or regular mail. We do encourage email. Thank you for your cooperation.

2276.

Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. Quadrilateral ABCD is cyclic with circumcircle ,(0; R). Show that the nine-point (Feuerbach) circles of 4BCD, 4CDA, 4DAB and 4ABC have a point in common, and characterize that point.

431

2277. Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. For n  1, de ne   un = (11;n) ; (22;n) ; : : : ; (nn,,1;1n) ; (n;nn) ;

where the square brackets [ ] and the parentheses ( ) denote the least common multiple and greatest common divisor respectively. For what values of n does the identity un = (n , 1)un, hold? 2278. Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Determine the value of an , which is the number of ordered n{tuples (k ; k ; : : : ; kn ; kn ) of non-negative integers such that 1

2

3

+1

2k + 3k + : : : + nkn + (n + 1)kn 2

3

+1

= n + 1:

2279. Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. With the usual notation for a triangle, prove that sin A cos B cos C = 4sr R 2R , s + (2R + r) : cyclic X

,

3

2

2

2



4

2280.

Proposed by Toshio Seimiya, Kawasaki, Japan. ABC is a triangle with incentre I . Let D be the second intersection of AI with the circumcircle of 4ABC . Let X , Y be the feet of the perpendiculars from I to BD, CD respectively. Suppose that IX + IY = AD. Find \BAC . 2281. Proposed by Toshio Seimiya, Kawasaki, Japan. ABC is a triangle, and D is a point on the side BC produced beyond C , such that AC = CD. Let P be the second intersection of the circumcircle of 4ACD with the circle on diameter BC . Let E be the intersection of BP with AC , and let F be the intersection of CP with AB . Prove that D, E , F , are collinear. 2282. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. A line, `, intersects the sides BC , CA, AB , of 4ABC at D, E , F respectively such that D is the mid-point of EF . Determine the minimum value of jEF j and express its length as elements of 4ABC . 1 2

432

2283. Proposed by Waldemar Pompe, student, University of Warsaw, Poland. You are given triangle ABC with \C = 60 . Suppose that E is an interior point of line segment AC such that CE < BC . Suppose that D is an interior point of line segment BC such that AE = BC , 1 : BD CE Suppose that AD and BE intersect in P , and the circumcircles of AEP and BDP intersect in P and Q. Prove that QE k BC .

2284.

Proposed by Toshio Seimiya, Kawasaki, Japan. ABCD is a rhombus with \A = 60. Suppose that E, F , are points on the sides AB , AD, respectively, and that CE , CF , meet BC at P , Q respectively. Suppose that BE + DF = EF . Prove that BP + DQ = PQ . 2285. Proposed by Richard I. Hess, Rancho Palos Verdes, California, USA. An isosceles right triangle can be 100% covered by two congruent tiles. Design a connected tile so that two of them maximally cover a nonisosceles right triangle. (The two tiles must be identical in size and shape and may be turned over so that one is the mirror image of the other. They must not overlap each other or the border of the triangle.) What coverage is achieved for a 30{60{90 right triangle? 2

2

2

2

2

2

2286.

Proposed by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. It is well known and easy to show that the product of any four consecutive positive integers plus one, is always a perfect square. It is also easy to show that the product of any two consecutive positive integers plus one is never a perfect square. It is possible that the product of three consecutive integers plus one is a perfect square. For example: 234+1=5 and 4  5  6 + 1 = 11 : (a) Find the next largest natural number n such that n(n + 1)(n + 2) + 1 is a perfect square. (b)? Are there any other examples? 2

2

433

SOLUTIONS No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems. 2090. [1995: 307] Proposed by Peter Ivady, Budapest, Hungary. For 0 < x < =2 prove that 

sin x  <  , x : x  +x 2

2

2

2

2

Essentially the same solution was sent in by Murray S. Klamkin, University of Alberta, Edmonton, Alberta; Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA; and Heinz-Jurgen Sei ert, Berlin, Germany. For 0 < x <  , we have 2



1  sin x  = Y x 1 , x (n) n 2

2

2

2

=1

 < 1 , x

2

2

2

x : <  , +x 2

2

2

2

  University of Sarajevo, Sarajevo, Also solved by SEFKET ARSLANAGIC, Bosnia and Herzegovina; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; LUIS V. DIEULEFAIT, IMPA, Rio de Janeiro, Brazil; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, Hong Kong; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer. Two incomplete solutions were received. Most solvers proved a better inequality, and some commented that the interval could be extended to (,;  ). Klamkin and Manes both point out the complimentary inequality (which is not as easy to prove):

1,t sin t 1 + t  t 2

2

for all real t.

See American Mathematical Monthly, 76 (1969), pp. 1153-1154, R. Redheffer, problem 5642. Janous gave the extension to: what is the value of  that gives the best inequality of the type  

sin x

2

x <  , +x 2

2

x which is valid for all x 2 (0; =2)? We leave this as a challenge to our other 2

readers!

2

434

2169. [1996: 274] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.

AB is a xed diameter of circle , (0;R). P is an arbitrary point of its circumference. Q is the projection onto AB of P . Circle , (P PQ) intersects , at C and D. CD intersects PQ at E . F is the midpoint of AQ. FG ? CD, where G 2 CD. Show that: 1. EP = EQ = EG, 2. A, G and P are collinear. 1

2

1

1

Solution by Toshio Seimiya, Kawasaki, Japan. 1. Let H be the second intersection of PQ with , . Since PH ? AB , AB is the perpendicular bisector of PH , so thatPQ = QH and \PAQ = \HAQ. Since PC = PQ = PD, we get \PHC = \PDC = \PCD, so that 4PHC  4PCE , from which we have 1

PH : PC = PC : PE: Thus we have PH  PE = PC = PQ . As PH = 2PQ, we have 2PQ  PE = PQ , so that 2PE = PQ; thus, PE = EQ. Since FG ? CD and PQ ? AB , we have \GFQ = \PEC . As 4PHC  4PCE we get \PEC = \PCH = \PAH = 2\PAQ. Thus \GFQ = 2\PAQ. Since F; E are midpoints of AQ; PQ, we get FE jjAP , so that \PAQ = \EFQ. Thus \GFQ = 2\EFQ, so that \GFE = \EFQ. Hence we have 4GFE  4QFE , so that EG = EQ. Therefore EP = EQ = EG. 2. Since 4GFE  4QFE we have FG = FQ = AF , so that \GAF = \GFQ = \EFQ. Thus we have AGjjFE . Since AP jjFE; A; G and P are collinear. 2

2

2

1 2

Also solved by Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid, Spain; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; GOTTFRIED PERZ, Pestalozzigymnasium, Graz,   Austria; CRISTOBAL SANCHEZ{RUBIO, I.B. Penyagolosa, Castellon,  Spain; and the proposer.

435

2170. [1996: 274] Proposed by Tim Cross, King Edward's School, Birmingham, England. Find, with justi cation, the positive integer which comes next in the sequence 1411; 4463; 4464; 1412; 4466; 4467; 1413; 4469; : : : . [Ed.: the answer is NOT 4470.] Editor's summary based on the solutions and comments submitted by the solvers whose names appear below. Most solvers felt (and the editors agree) that the answer could be anything, as stated in the following comment by Murray Klamkin: Any number can be the next term! A set of numbers is a sequence mathematically if and only if a rule of formation is given. The given set of numbers is not a sequence and so the problem is meaningless. Given any nite set of n numbers, one can always nd an in nite number of formulae which agree with the given n terms, and such that the (n +1)th term is completely arbitrary. However, some solvers did come up with interesting formulae or reasonings to \justify" the answer that they gave. Some examples:   I. (Diminnie) Let gn = n , 3 n , and let 3

+ bn=3c + 1 + (,1)g n g (n) + 76 (,1)g n , 1 : xn = 3 1487 2 , (,1)g n Then x , x , : : : , x agree with the given terms, and x = 4470. In general, if k is any number and if we de ne yn = xn + bn=9c(k , 4470); then y , y , : : : , y agree with the given terms, and y = k. (

1

1

2

2

)

1 2

n

(

)





8

(

)

o

9

8

9

II. (Hurthig and the proposer) Squaring each of the given terms reveals

199021; 19918369; 19927296; 1993744; 19945156; 19954089; 1996569; 19971961;

and the given numbers an (n = 1, 2, : : : , 8), are the least positive integers whose squares begin with the digits 1990, 1991, : : : , 1997. That is, an is the smallest positive integer k such that k begins with the same digits as 1989 + n. This leads to a = 447. III. (Bradley) The given numbers are the integer parts of the square roots of 2

9

1991  10 ; 1992  10 ; 1993  10 ; 1994  10 ; 1995  10 ; 1996  10 ; 1997  10 ; 1998  10 : 3

4

4

3

4

4

3

4

436 Hence the next term would be j

p

p1999  10 k, or 4471.

j

IV. (Hess) Let f (n) = n 10 ,

4

1 3

k

. The the given numbers are

n; f (n) + 2; f (n + 1); n + 1; f (n + 1) + 2; f (n + 2);n + 2; f (n + 2) + 2; with n = 1411. Thus the next term is f (n + 3) = f (1414) = 4471. Other submitted answers include 4479 (Konecny) and 44610 (Ortega and Gutierrez). Solved by HAYO AHLBURG, Benidorm, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.   Y,  KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV KONECN Ferris State University, Big Rapids, Michigan, USA; SOLEDAD ORTEGA and  JAVIER GUTIERREZ, students, University of La Roija, Logro~no, Spain; and the proposer.

2171. [1996: 274] Proposed by Juan-Bosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Let P be an arbitrary point taken on an ellipse with foci F and F , and directrices d , d , respectively. Draw the straight line through P which is parallel to the major axis of the ellipse. This line intersects d and d at points M and N , respectively. Let P 0 be the point where MF intersects NF . 1

1

2

2

1

2

2

1

Prove that the quadrilateral PF P 0 F is cyclic. 1

2

Does the result also hold in the case of a hyperbola? Solution by Richard I. Hess, Rancho Palos Verdes, California, USA, modi ed by the editor. I: The Ellipse: xa + yb = 1. 2

2

2

2

F , F be , a ; 0, ,, a ; 0, respectively, where  > 1, and pLet b =  , a. Let P be (a sin ; b cos ) and M be (a;b cos ). By symmetry, P 0 is on the y {axis. Let P 0 be (0; ,d). 1

2

1

2

437

N

6 q

q

F

P

C

q

F

q

q

d

2

so that d =

M

-

q

2

Thus

q p

P0

1

d

1

d = b cos  ; a= a , a=

b cos  .  ,1 2

Choose C to be the point (0; ) such that CF = CF = CP 0 . Thus 1

2

( + d) = + a ; 2

2

2

2

so that

2 b cos  + b cos  = a = b ;  , 1 ( , 1)   ,1 2

2

giving

2

2

2

2

2

2

2

b , b cos  : = 2 cos  2( , 1) We now show that CP = CP 0 . If this were true, we would have   ; ( , b cos ) + a = sin  = + b cos ,1 2

2

2

2

2

2

or

 b cos  ,2b cos  + b cos  +  ,b 1 sin  = 2b cos , 1 + ( , 1) ; 2

2 2

2

2

2

2

or

2

2

 ,b + b cos, 1 + b cos  + b , 1 =  b, 1 +  b cos ,1 ; 2

2

2

2

2

2

or

2

2

2

2

2

2

2

 ,1 = 1 ;  ,1  ,1 2

2

2

2 2

2

2

438 which is clearly true. Thus C is the centre of the cyclic quadrilateral PF P 0 F . II: The Hyperbola: xa , yb = 1. ,a  , a  pLet,F , F be  ; 0 , ,  ; 0 , respectively, where 0 <  < 1, and b =  a. Let P be (a sinh ; b cosh ) and M be (a;b cosh ). By symmetry, P 0 is on the y {axis. Let P 0 be (0; d). Choose C to be the point (0; ) such that CF = CF = CP 0 . 1

1

1

2

2

2

2

2

2

2

1

6P 0

2

q

N F

q

2

q q

C

M

P

q

q

F

q

1

-

Follow the procedure in case I to show that C is the centre of the cyclic quadrilateral PF P 0 F . Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer (for the ellipse only). Smeenk notes that it is easy to verify that PP 0 is a normal to the ellipse. Bellot Rosado refers to E.A. Maxwell's Elementary Coordinate Geometry, Oxford University Press, 1952, which contains the two following related problems: 1. P , Q are two points on an ellipse with foci S , S 0 , such that PQ is perpendicular to SS 0 . Prove that PS , QS , PS 0 , QS 0 touch a circle, and identify its centre. 2. Tangents TP , TQ are drawn to an ellipse with foci S , S 0 . A line through S parallel to TQ meets S0 T in U , and a line through S0 parallel to TP meets ST in V . Prove that S , S 0 , U , V lie on a circle. 1

2

439

2172. [1996: 274] Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let x; y;z  0 with x + y + z = 1. For xed real numbers a and b, determine the maximum c = c(a; b) such that a + bxyz  c(yz + zx + xy ). Solution by Mihai Cipu, Institute of Mathematics, Romanian Academy, Bucharest, Romania. The answer is  b c(a; b) = min 4a; 3a + 9 ; 

provided that a  0. For x = y = 1=2, z = 0 one obtains c  4a, while by substituting x = y = z = 1=3 it follows that c  3a + b=9. Thus c(a; b)  min(4a; 3a + b=9). Now to nish the proof we shall show that for all real numbers a  0 and b,  b a + bxyz  min 4a; 3a + 9  (xy + yz + zx) 

(1)

for all x; y;z  0, x + y + z = 1. To this end we shall use the fact that for any such triple x; y; z there exists a Euclidean triangle whose sides have lengths 1 , x; 1 , y; 1 , z . The triangle is degenerate if xyz = 0. Let us denote by r, resp. R, the radius of the incircle, resp. circumcircle, of the associated triangle. Using well-known formulae and the hypothesis, one easily nds xyz = r and xy + yz + zx = r + 4Rr: (2) Here r and R have non-negative values subject to the restrictions 16Rr , 5r  1 and R  2r. [Editor's note. The hypothesis x + y + z = 1 means that the associated triangle has semiperimeter s = 1. Thus (2) follows from the known identities X (s , a )(s , a )(s , a ) = r s and (s , a )(s , a ) = r + 4Rr 2

2

2

1

2

2

3

1

2

2

which hold for any triangle with sides a ; a ; a | see for example equations (15) and (16), page 54 of Mitrinovic, Pecaric and Volenec, Recent Advances in Geometric Inequalities. And the restriction 16Rr , 5r  1 is just the known inequality 16Rr ,5r  s ; see (3.6) on page 166 of Recent Advances, or item 5.8 of Bottema et al., Geometric Inequalities.] We note that for x = y = 0, z = 1 one gets a  0 [else there is no solution for c]. Using this fact, in the case b  9a we have 1

2

3

2

2

2

min(4a; 3a + b=9) = 4a

and

a + br  a(1 + 9r )  a(4r + 16Rr); 2

2

2

440 so that (1) holds. In the opposite case b  9a we get

a(1 , 3r , 12Rr)  a(4Rr , 8r )  (4Rr , 8r )b=9 [since 4Rr , 8r = 4r(R , 2r)  0], or equivalently a + br  (3a + b=9)(r + 4Rr); 2

2

2

2

2

2

which is (1) again.   University of Sarajevo, Sarajevo, Also solved by SEFKET ARSLANAGIC, Bosnia and Herzegovina; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MURRAY   Y,  S. KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV KONECN Ferris State University, Big Rapids, Michigan, USA; and the proposer. One incorrect solution was sent in. Most solvers noted that a solution exists only if a  0.

2173. [1996: 275, 1997: 169] Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Let n  2 and x ; : : : ; xn > 0 with x + : : : + xn = 1. Consider the terms 1

1

ln = and

n X k=1

s

(1 + xk ) 1 ,x xk k

rn = Cn

where

n Y k=1

p11+,xxkk

Cn = (pn , 1)n (pn)n=(n + 1)n, : +1

1

1. Show l  r . 2. Prove or disprove: ln  rn for n  3. 2

2

I. Solution to Part 1 by Richard I. Hess, Rancho Palos Verdes, California, USA. For n = 2, we get C = 2=3 and [since x + x = 1] 2

1

2

r r l = (1 + x ) xx + (1 + x ) xx = x + px x+x2x x ;    2 1 + x 1 + x r = 3 px px : 2

2

1

1

2

1

2

1

2

1

1

1

2

2

2

1

2

2

441 Thus

px x (r , l ) = 2 (1 + x + x + x x ) , x , x , 2x x 3 = 13 (1 , 4x x )   = 13 (x + x ) , 4x x = 13 (x , x )  0: Therefore l  r with equality if and only if x = x = 1=2. 1

2

2

1

2

2

1

1

1

2

2

1

2

1

2

1

2

2

2

2

1

2

2

2

1

2

II. Partial solution to Part 2 by the proposer. We show that ln  rn in the case n = 3 (which indeed was the starting point for the whole problem). Putting x = x, x = y , x = z , the desired inequality l  r reads 1

2

3

3

s

3

(1 + x) 1 ,x x + (1 + y) 1 ,y y + (1 + z ) 1 ,z z r

r

p

 3 4 3  p(11+,xx)(1 p1+,zz) p1+,yy)(1 where x; y;z 2 (0; 1) such that x + y + z = 1.

(1)

We now recall the dicult Crux problem 2029 of Jun-hua Huang, solved by Kee-Wai Lau on [1996: 129]:

p wbwc + wcwa + wawb  3F 3;

(2) where wa ; wb; wc; F are the angle bisectors and the area of a triangle. We claim that this inequality is equivalent to inequality (1). Indeed, let us apply the transformation a = y + z , b = z + x, c = x + y where x; y;z > 0, converting any triangle inequality into an algebraic inequality valid for positive numbers. Then it's not dicult to see that x + y + z = s (the semiperimeter of the triangle), whence x = s , a, y = s , b and z = s , c. Furthermore, due to homogeneity we may and do put s = 1, p whence F = s(s , a)(s , b)(s , c) = pxyz. Also, by the known formula s

C=2) = 2ab s(s , c) = 2 ps(s , c)ab wc = 2ab cos( a+b a+b ab a+b (for example, see [1995: 321]) we get (using x + y + z = 1) p p wc = x + y +2 z + z z(y + z)(z + x) = 1 +2 z z(1 , x)(1 , y);

442 and similarly for wa and wb. Hence (2) is equivalent to

pxyp(1 , y)(1 , x)(1 , z) p3pxyz; 4  3 (1 + x)(1 + y) cyclic X

which is equivalent to (1). Since (2) is true, so is (1). For n  4 I do not have any idea of how to settle whether the inequality ln  rn is true. It may be interesting and useful to see a purely algebraic proof of inequality (1).   University of Sarajevo, Part 1 also solved by SEFKET ARSLANAGIC, Sarajevo, Bosnia and Herzegovina; and the proposer. The proposer did Part 2 only in the case n = 3 (given above). One other reader sent in a solution to Part 2 which the editor considers to be faulty. Readers are invited to try nishing o this problem completely, or even just the special case n = 4.

2174. [1996: 275] Proposed by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. Let A be an n  n matrix. Prove that if An = 0 then An = 0. Solution by John C. Tripp, Southeast Missouri State University, Cape Girardeau, Missouri. We consider A as a linear transformation on an n-dimensional vector space. We assume that An = 0. Let x be any element of the vector space. The set of vectors +1

+1

V = fx; Ax; A x; : : : ; Anxg 2

has n + 1 elements, so it is linearly dependent. Let k be the smallest nonnegative integer such that Ak x is a linear combination of the other vectors in V . We have

Ak x = c Ak x + c Ak x + c Ak x +    + cn,k Anx; for some scalars c ; c ; c ; : : : ; cn,k , and Anx = An,k Akx = An,k (c Ak x + c Ak x + c Ak x +    + cn,k An x) = c An x + c An x + c An x +    + cn,k A n,k ) = An (c x + c Ax + c A x +    + cn,k An,k, x) = 0: Since x was arbitrary, we have An = 0. +1

1

1

2

1

1

+1

+1

1

2

+2

3

+3

3

+1

+2

2

+2

2

2

3

3

2

3

+3

+3

2

1

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIHAI CIPU, Romanian Academy, Bucharest, Romania; CON AMORE PROBLEM GROUP, Royal Danish Schoolof Educational Studies, Copenhagen, Denmark; LUZ M. DeALBA, Drake University, Des Moines, Iowa; F.J.

443 FLANIGAN, San Jose State University, San Jose, California, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, Hong Kong; WALDEMAR POMPE, student, University of Warsaw, Poland; E. RAPPOS, University of Cambridge, Eng land; HEINZ-JURGEN SEIFFERT, Berlin, Germany; SKIDMORE COLLEGE PROBLEM GROUP, Saratoga Springs, New York, USA; DIGBY SMITH, Mount Royal College, Calgary, Alberta; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; and the proposer. Most solvers used the Cayley-Hamilton Theorem. Wang comments that this problem is a special case of a more general and well-known result which states that \if A is an n  n complex matrix such that Ak = 0 for some k  1 (that is, A is nilpotent), then An = 0". Indeed, some solvers proved this more general result.

2175. [1996: 275] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. The fraction 61 can be represented as a di erence in the following ways: 1 = 1 , 1; 1 = 1 , 1; 1 = 1 , 1 ; 1 = 1 , 1 : 6 2 3 6 3 6 6 4 12 6 5 30

1 In how many ways can the fraction 2175 be expressed in the form

1 = 1 , 1; 2175 x y

where x and y are positive integers? Solution by D. Kipp Johnson, Valley Catholic High School, Beaverton, Oregon, USA. Notice that

1 1 1 2175 = x , y

so that

y = 2175 , 2175 : x = y 2175 + 2175 y + 2175 Thus x will be an integer if and only if y + 2175 is a factor of 2175 , and x will be positive whenever y is, since then 2175 =(y + 2175) < 2175, and y will be positive whenever y + 2175 > 2175, so we seek factors of 2175 which exceed 2175. But 2175 = 3  5  29 has (2 + 1)(4+ 1)(2+ 1) = 45 2

2

2

2

2

2

4

2

positive factors, one of which is its square root, 2175. Since the factors of 2175 come in pairs whose product is 2175 , exactly half of the other 44 factors exceed 2175, giving 22 solutions in positive integers. The smallest is x = 300;y = 348. This immediately generalizes to the solution in positive 2

2

444 integers of 1=n = 1=x , 1=y . Since  (n ) (the number of divisors of n ) is odd for a perfect square, there will be ( (n ) , 1)=2 solutions to the equation 1=n = 1=x , 1=y. Also solved by HAYO AHLBURG, Benidorm, Spain; SAM BAETHGE, Nordheim, Texas, USA; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; TIM CROSS, King Edward's School, Birmingham, England; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; KEITH EKBLAW, Walla Walla, Washington, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California,  USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VACLAV  Y,  Ferris State University, Big Rapids, Michigan, USA; KEE-WAI KONECN LAU, Hong Kong; J.A. MCCALLUM, Medicine Hat, Alberta; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; LAMARR WIDMER, Messiah College, Grantham, PA, USA; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There were 4 incorrect solutions. Flanigan refers the interested reader to problem #10501 of the American Mathematical Monthly, volume 103, number 2, February 1996, page 171. 2

2

2

 2176. [1996: 275] Proposed by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina. Prove that v u n uY n t

(ak + bk ) 

k=1

v u n uY n t

k=1

ak +

v u n uY n t

k=1

bk

where a ; a ; : : : ; an > 0 and n 2 N. Solution by Sai C. Kwok, San Diego, CA, USA. Using the arithmetic-geometric mean inequality, we have 1

2

n Y

ak

!1

n

 n1

n Y

bk

! n1

 n1

k=1 ak + bk

n X

ak

n X

bk

k=1 ak + bk

and k=1 ak + bk

k=1 ak + bk

The result follows by adding the above two inequalities.

445 Also solved by THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; yMIHAI CIPU, Romanian Academy, Bucharest, Romania, and Concordia University, Montreal, Quebec; ROBERT  GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta; yWALTHER JANOUS, Ursulinengymnasium, Inns  Y,  Ferris State University, Big Rapids, bruck, Austria; VACLAV KONECN Michigan, USA; MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; yCAN ANH MINH, student, University of California Berke ley, Berkeley, CA, USA; ySOLEDAD ORTEGA and JAVIER GUTIERREZ, students, University of La Rioja, Logro~no, Spain; WALDEMAR POMPE, stu dent, University of Warsaw, Poland; JUAN-BOSCO ROMERO MARQUEZ,  Universidad de Valladolid, Valladolid, Spain; yHEINZ-JURGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; GEORGE TSAPAKIDIS, Agrinio, Greece; and the yproposer. A proof for the case n = 4 was submitted by V.N. Murty. ( The symbol y before a solver's name indicates that the solver's solution was virtually the same as the one highlighted above.) Clearly, the condition that b ; b ; : : : ; bn > 0 was inadvertantly left out from the original statement. All solvers assumed, explicitly or implicitly, that this was the case. However, only Hess gave a simple example to show that the inequality need not be true without the aforementioned condition: take n = 2, a = a = 1 and b = b = ,1. Janous pointed out that equality holds if and only if the vectors (a ; a ; : : : ; an ) and (b ; b ; : : : ; bn ) are proportional. Klamkin commented that the given inequality is an immediate special case of Jensen's generalization of Holder's Inequality, and referred readers to D.S. Mitrinovic et. al., Recent Advances in Geometric Inequalities, Kluwer Academic Publishers, 1989, pp. 50{54. Konecny remarked that it is known that if A and B are n  n positive semi-de nite Hermitian matrices, then 1

1

1

2

2

1

1

p

2

2

2

n det(A + B )

 pn det A + pn det B

(see, for example, Inequalities: Theory of Majorization and Its Application by Albert W. Marshall and Ingram Olkin, Academic Press Inc., 1979, p. 475). If we let A and B be the n  n diagonal matrices with diagonal entries ak 's and bk's respectively (i = 1, 2, : : : , n), then the proposed inequality follows immediately. Sei ert pointed out that the proof given above can be found on p. 178 of the book Classical and New Inequalities in Analysis, Kluwer Academic Publishers, 1992.

446

2177. [1996: 317] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABCD is a convex quadrilateral, with P the intersection of its diagonals and M the mid-point of AD. MP meets BC at E . Suppose that BE : EC = (AB) : (CD) . Characterize quadrilateral ABCD. Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria. 2

2

D

Let d(P; CD)    P 

M

d(P; AB)

A

 = \MPA = \EPC;  = \BPE = \DPM:

C E

Then, applying the Sine Rule to the triangles 4APM and 4DPM , we get AM  sin \AMP

B

sin  =

AP

;

, \AMP ) = AM  sin \AMP ; sin  = DM  sin(180 DP DP 

whence

sin  = DP : sin  AP

(3)

Applying the law of sines to the triangles 4CPE , and 4BPE , we get CE  sin \CEP

sin  = ; CP  sin  = BE  sin(180 , \CEP ) = BE  sin \CEP ;

BP

whence

BP

sin  = CE  BP = CD  BP : sin  BE  CP AB  CP 2

2

(4)

From (1), (2) follows that quadrilateral ABCD has the desired property if and only if

AB = AP  BP = [ABP ] = AB  d(P; AB) CD CP  DP [CDP ] CD  d(P; CD) (where [XY Z ] denotes the area of 4XY Z and d(U; V W ) the distance of U from V W ); that is, if AB : CD = d(P; AB) : d(P; CD): 2

2

447 This implies that ABCD is characterized by the fact that in the triangles 4ABP and 4CDP , having \APB = \CPD in common (ABCD is convex!), the ratio of the side opposite to P and the altitude passing through P is the same, which means that 4ABP and 4CDP are directly or inversely similar. In the rst case, we have \BAP = \DCP ; that is, ABCD is a trapezoid with parallel sides AB and CD. In the second case, we have \BAP ( = \BAC ) = \PDC ( = \BDC ), whence ABCD is an inscribed quadrilateral (A and D are at the same side of BC ). Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; CON AMORE PROBLEM GROUP, Royal Danish School of Educational Studies, Copenhagen, Denmark; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; WALDEMAR POMPE, student, University of Warsaw, Poland; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. Four incomplete or incorrect solutions were received.

2178. [1996: 318] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. If A; B; C are the angles of a triangle, prove that sin A sin B sin C  8 sin A cos B cos C + sin B cos C cos A + sin C cos A cos B   3p3 ,cos A + cos B + cos C  : ,

3

3

3

2

2

2

Solution by Florian Herzig, student, Perchtoldsdorf, Austria. I shall prove a stronger version with 6 sin A sin B sin C of the left side. We use the following known identities and inequalities:

sin 2A + sin 2B + sin2C = 4 sin A sin B sin C cos A + cos B + cos C = 1 p , 2 cos A cos B cos C sin A sin B sin C  3 8 3 cos A cos B cos C  18 2

2

2

(1) (2) (3) (4)

where A, B , C are the angles of a triangle. Therefore

sin A cos B cos C + sin B cos C cos A + sin C cos A cos B = sin A(1 , cos A) cos B cos C + sin B (1 , cos B ) cos C cos A + sin C (1 , cos C ) cos A cos B 3

3

3

2

2

2

448

= , cos A cos B cos C (sin A cos A + sin B cos B + sin C cos C ) + sin A cos B cos C + cos A(sin B cos C + sin C cos B) = , 12 cos A cos B cos C (sin2A + sin2B + sin2C ) + sin A(cos B cos C + cos A) = ,2 cos A cos B cos C sin A sin B sin C + sin A sin B sin C = sin A sin B sin C (1 , 2 cos A cos B cos C ) = sin A sin B sin C (cos A + cos B + cos C ): 2

2

2

Combining (2) and (4) yields

cos A + cos B + cos  43 : 2

2

2

By using this inequality and (3) we get

6 sin A sin B sin C  8(sin A cos B cos C + sin B cos C cos A + sin C cos A cos B ) = 8psin A sin B sin C (cos A + cos B + cos C )  3 3(cos A + cos B + cos C ) 3

3

3

2

2

2

2

2

2

as we wanted to show. Equality holds as in (3) and (4) if and only if A = B = C = 60.   University of Sarajevo, Sarajevo, Also solved by SEFKET ARSLANAGIC, Bosnia and Herzegovina; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; MICHAEL  LAMBROU, University of Crete, Crete, Greece; HEINZ-JURGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. Janous, in his solution, used a new identity, which is presented as a problem 2279 in this issue.

Crux Mathematicorum

Founding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell Editors emeriti / Redacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem

Founding Editors / Redacteurs-fondateurs: Patrick Surry & Ravi Vakil Editors emeriti / Redacteurs-emeriti: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

449

THE ACADEMY CORNER No. 15 Bruce Shawyer

All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Memorial University Undergraduate Mathematics Competition

September 25, 1997

Answer as many questions as you can. Complete solutions carry more credit than scattered comments about many problems. 1. Determine whether or not the following system has any real solutions. If so, state how many real solutions exist.

x + x1 = y;

y + y1 = z;

z + z1 = x:

2. The surface area of a closed cylinder is twice the volume. Determine the radius and height of the cylinder given that the radius and height are both integers. 3. Prove that

1 + 41 + 19 + : : : + n12 < 2: 4. Describe the set of points (x; y ) in the plane for which sin(x + y ) = sin x + sin y:

5. In a parallelogram ABCD, the bisector of angle ABC intersects AD at P . If PD = 5, BP = 6 and CP = 6, nd AB . 6. Show that, where k + n  m, n   X

i=0

n m  = m + n: i k+i n+k

Send me your nice solutions!

450

THE OLYMPIAD CORNER No. 186 R.E. Woodrow

All communications about this column should be sent to Professor R.E. Woodrow, Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada. T2N 1N4. This number we give the 24 problems proposed to the jury, but not selected for the 37th International Mathematical Olympiad in July 1996 at Mumbai, India. My thanks go to Ravi Vakil, Canadian Team Leader to the 37th IMO for collecting this and other contest material and forwarding it to me.

PROBLEMS PROPOSED TO THE JURY BUT NOT USED AT THE 37th INTERNATIONAL MATHEMATICAL OLYMPIAD July 1996 | Mumbai, India

that

1. Let a, b and c be positive real numbers such that abc = 1. Prove ab bc ca a5 + b5 + ab + b5 + c5 + bc + c5 + a5 + ca  1:

When does equality hold? 2. Let a1  a2      an be real numbers such that for all integers k > 0,

ak1 + ak2 +    + akn  0: Let p = maxfja1 j; : : : ; jan jg. Prove that p = a1 and that (x , a1 )(x , a2 )    (x , an )  xn , an1 for all x > a1 . 3. Let a > 2 be given, and de ne recursively: ! 2 a n a0 = 1; a1 = a; an+1 = a2 , 2 an : n,1 Show that for all integers k > 0, we have

1 + 1 + 1 +    + 1 < 1 (2 + a , pa2 , 4): a0 a1 a2 ak 2

451

4. Let a ; a ; : : : ; an be non-negative real numbers, not all zero. 1

2

(a) Prove that xn , a1 xn,1 ,  , an,1 x , an = 0 has precisely one positive real root. P P (b) Let A = nj=1 aj , and B = nj=1 jaj and let R be the positive real root of the equation in (a). Prove that AA  RB . 5. Let P (x) be the real polynomial, P (x) = ax3 + bx3 + cx + d. Prove that if jP (x)j  1 for all x such that jxj  1, then

jaj + jbj + jcj + jdj  7:

6. Let n be an even positive integer. Prove that there exists a positive integer k such that

k = f (x)(x + 1)n + g(x)(xn + 1) for some polynomials f (x), g (x) having integer coecients. If k0 denotes the least such k, determine k0 as a function of n. 7. Let f be a function from the set of real numbers R into itself such that for all x 2 R, we have jf (x)j  1 and      13 1 1 f x + 42 + f (x) = f x + 6 + f x + 7 : 

Prove that f is a periodic function (that is, there exists a non-zero real number c, such that f (x + c) = f (x) for all x 2 R). 8. Let the sequence a(n), n = 1; 2; 3; : : : ; be generated as follows: a(1) = 0, and for n > 1,

a(n) = a(bn=2c) + (,1)n(n+1)=2: (Here btc is the greatest integer less than or equal to t.) (a) Determine the maximum and minimum value of a(n) over n  1996 and nd all n  1996 for which these extreme values are attained. (b) How many terms a(n), n  1996, are equal to 0? 9. Let triangle ABC have orthocentre H , and let P be a point on its circumcircle, distinct from A, B , C . Let E be the foot of the altitude BH , let PAQB and PARC be parallelograms, and let AQ meet HR in X . Prove that EX is parallel to AP . 10. Let ABC be an acute-angled triangle with jBC j > jCAj, and let O be the circumcentre, H its orthocentre, and F the foot of its altitude CH . Let the perpendicular to OF at F meet the side CA at P . Prove that \FHP = \BAC .

452

11. Let ABC be equilateral, and let P be a point in its interior. Let the lines AP , BP , CP meet the sides BC , CA, AB in the points A1, B1, C1 respectively. Prove that A1B1  B1C1  C1A1  A1B  B1C  C1A:

12. Let the sides of two rectangles be fa; bg and fc; dg respectively, with a < c  d < b and ab < cd. Prove that the rst rectangle can be placed within the second one if and only if (b2 , a2)2  (bc , ad)2 + (bd , ac)2:

13. Let ABC be an acute-angled triangle with circumcentre O and circumradius R. Let AO meet the circle BOC again in A0 , let BO meet the circle COA again in B 0 and let CO meet the circle AOB again in C 0 . Prove that OA0  OB0  OC 0  8R3 :

When does equality hold? 14. Let ABCD be a convex quadrilateral, and let RA , RB , RC , RD denote the circumradii of the triangles DAB , ABC , BCD, CDA respectively. Prove that RA + RC > RB + RD if and only if \A + \C > \B + \D. 15. On the plane are given a point O and a polygon F (not necessarily convex). Let P denote the perimeter of F , D the sum of the distances from O to the vertices of F , and H the sum 2of the 2distances from O to the lines containing the sides of F . Prove that D , H  P 2 =4. 16. Four integers are marked on a circle. On each step we simultaneously replace each number by the di erence between this number and the next number on the circle, moving in a clockwise direction; that is, the numbers a, b, c, d are replaced by a , b, b , c, c , d, d , a. Is it possible after 1996 such steps to have numbers a, b, c, d such that the numbers jbc , adj, jac , bdj, jab , cdj are primes? 17. A nite sequence of integers a0; a1; : : : ; an is called quadratic if for each i in the set f1; 2; : : : ; ng we have the equality jai , ai,1 j = i2 . (a) Prove that for any two integers b and c, there exists a natural number n and a quadratic sequence with a0 = b and an = c. (b) Find the smallest natural number n for which there exists a quadratic sequence with a0 = 0 and an = 1996. 18. Find all positive integers a and b for which 

a2  +  b2  =  a2 + b2  + ab; b a ab

453 where, as usual, btc refers to the greatest integer which is less than or equal to t. 19. Let N0 refer to the set of non-negative integers. Find a bijective function f from N0 into N0 such that for all m; n 2 N0 ,

f (3mn + m + n) = 4f (m)f (n) + f (m) + f (n):

20. A square (n , 1)  (n , 1) is divided into (n , 1)

unit squares in the usual manner. Each of the n vertices of these squares is to be coloured red or blue. Find the number of di erent colourings such that each unit square has exactly two red vertices. (Two colouring schemes are regarded as di erent if at least one vertex is coloured di erently in the two schemes.) 21. Let k, m, n be integers such that 1 < n  m , 1  k. Determine the maximum size of a subset S of the set f1; 2; 3 : : : ; k , 1; kg such that no n distinct elements of S add up to m. 22. Determine whether or not there exist two disjoint in nite sets A and B of points in the plane satisfying the following conditions: (a) No three points in A [B are collinear, and the distance between any two points in A [ B is at least 1. (b) There is a point of A in any triangle whose vertices are in B, and there is a point of B in any triangle whose vertices are in A. 23. A nite number of beans are placed on an in nite row of squares. A sequence of moves is performed as follows: at each stage a square containing more than one bean is chosen. Two beans are taken from this square; one of them is placed on the square immediately to the left while the other is placed on the square immediately to the right of the chosen square. The sequence terminates if at some point there is at most one bean on each square. Given some initial con guration, show that any legal sequence of moves will terminate after the same number of steps and with the same nal con guration. 24. Let U be a nite set and f , g be bijective functions from U onto itself. Let 2

2

and

S = fw 2 U : f (f (w)) = g(g(w))g

T = fw 2 U : f (g(w)) = g(f (w))g; and suppose that U = S [ T . Prove that, for m 2 U , f (w) 2 S if and only if g (w) 2 S . As always we welcome your nice original solutions which di er from the ocial solutions provided by the proposers and the selection committee.

454 As an example of an Olympiad which may not be as widely circulated, and for which you may not have already seen solutions, we give the four problems of the 4th Class for the Croatian National Mathematical Competition of May 13, 1994 and the three problems of the Croatian Mathematical Olympiad of May 14, 1994. My thanks go to Richard Nowakowski, Canadian Team Leader at the 35th IMO in Istanbul for collecting these problems.

CROATIAN NATIONAL MATHEMATICAL COMPETITION Fourth Class May 13, 1994

1. One member of an in nite arithmetic sequence in the set of natural numbers is a perfect square. Show that there are in nitely many members of this sequence having this property. 2. For a complex number z let w = f (z) = 2 . 3,z

(a) Determine the set fw : z = 2 + iy; y 2 Rg in the complex plane. (b) Show that the function w can be written in the form

w , 1 =  z , 1: w,2 z,2 (c) Let z0 = 12 and the sequence fzn g be de ned recursively by zn = 3 , 2z ; n  1: n,1

Using the property (b) calculate the limit of the sequence fzng. 3. Determine all polynomials P (x) with real coecients such that for some n 2 N we have xP (x , n) = (x , 1)P (x), 8 x 2 R. 4. In the plane ve points P1, P2, P3, P4, P5 are chosen having integer coordinates. Show that there is at least one pair (Pi; Pj ), for i 6= j such that the line PiPj contains a point Q, with integer coordinates, and is strictly between Pi and Pj . .

Additional Competition for the Olympiad May 14, 1994

1. Find all ordered triples (a; b;c) of real numbers such that for every three integers x; y; z the following identity holds: jax + by + czj + jbx + cy + azj + jcx + ay + bzj = jxj + jyj + jzj:

455

2. Construct a triangle ABC if the lengths jAOj, jAU j and radius r of the incircle are given, where O is the orthocentre and U the centre of the incircle. 3. Let P be the set of all lines of the plane M . Does there exist a function f : P ! M having the following properties: (a) the function f is an injection: (b) f (p) 2 p, 8 p 2 P ? That should provide some problems for your puzzling pleasure over the next couple of months. Now we return to readers' solutions to problems featured in earlier numbers of the Corner. First, an apology. Somehow, in shifting my les around we misplaced solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain, to two problems that we discussed in the October number of the Corner. His name should be added as a solver of problems 6 and 7 of the Telecom 1993 Australian Mathematical Olympiad in the solutions given [1997: 324{325]. Last number we gave solutions by the readers to the rst ten problems of the \Baltic Way | 92" contest given in the May 1996 number [1996: 157{ 159].

MATHEMATICAL TEAM CONTEST \BALTIC WAY | 92" Vilnius, 1992 | November 5{8

11.

Let Q+ denote the set of positive rational numbers. Show that there exists one and only one function f : Q+ ! Q+ satisfying the following conditions:   (i) If 0 < q < 12 then f (q ) = 1 + f 1,q2q . (ii) If 1 < q  2 then f (q ) = 1 + f (q , 1). (iii) f (q )  f ( q1 ) = 1 for all q 2 Q+. Solution by Michael Selby, University of Windsor, Windsor, Ontario. By a change of variable qe = 1,12q , we have from (i),

qe

!

!

f 1 + 2qe = 1 + f (qe); (0 < qe < 1); or f 1 +1 2 = 1 + f 11 : qe qe Calling t = 1qe and using (iii) we have 1 = 1 + 1 ; 0 < t < 1; t 2 Q+: (1) f (t + 2) f (t) 



456 Then

1 1 1 1 f (t + 4) = f (t + 2 + 2) = 1 + f (t + 2) = 1 + 1 + f (t) = 2 + f (1t) :

Hence, we can evaluate f (t + 2k), k  0, k an integer, if we know f (t). Observe that condition (ii) can be rewritten as f (1 + t) = 1 + f (t), t 2 Q+, 0 < t  1. We can now evaluate f (2k + 1 + q ) as follows: Since

1 1 : 1 = 1 + 1 ; we have = 1 + f (2 + q) f (q) f (2 + 1 + q) f (1 + q) 1 = 1 + 1 . Hence f (3 + q ), 0 < q  1 If 0 < q  1, then f (3 + q) 1 + f (q ) can be evaluated if f (q ) is known. Once f (3 + q ) is known, we obtain 1 = 1 1 ; = 1 + f (5 + q) f (2 + 3 + q) f (3 + q) and

1 1 + 1 + f (2k + 1 + q) f (2k , 1 + q) ; 1  q > 0:

Therefore, we can now evaluate

f (2k + q); f (2k + 1 + q) 0 < q  1; (2) for all k  0, k an integer, if we know f (q ). Furthermore, we can evaluate f (n), n  1. First f (1) = 1 since putting q = 1 in (iii) gives (f (1))2 = 1. Now f (2) = 1 + f (1) = 2 from (ii). We follow recursively, f (3): 1 = 1+ 1 =2 f (3) f (1)

and Similarly

1

1

f (2k + 1) = 1 + f (2k , 1) : 1 1 and f (2) = 2: = 1 + f (2k + 2) f (2k)

Thus any such function is uniquely de ned on the integers. Finally, we can evaluate the function at any q from the values on the positive integers. Let q = ab , where (a; b) = 1.

457 Write a = bq1 + r1 where q1 is a non-negative integer, and 0  r1 < b is an integer. If r1 = 0, f (q ) = f (q1) which is determined. If a  r1 < b, we apply f ( ab ) = f (q1 + rb1 ). This is determined if the value of f ( rb1 ) is known using (2). Now 0 < rb1 < 1. We now compute f ( rb1 ). b = r1q2 + r2, r2 < r1. Continuing, since 0  rk+1 < rk, rj = 0 for some j , and we will have an expression for which f is evaluated at an integer. Hence f exists and is uniquely determined. 12. Let N denote the set of positive integers. Let ' : N ! N be a bijective function and assume that there exists a nite limit nlim !1

'(n) = L: n

What are the possible values of L? Solution by Michael Selby, University of Windsor, Windsor, Ontario. We claim L must be 1. Consider maxf'(1);: : : ; '(n)g = jn . We note that jn  n, since ' is one-to-one. Let in 2 f1; 2; : : : ; ng be such that '(in) = jn . Then

'(in)  1: in

Since

'(n) = L; lim '(in) = L  1: (1) n!1 in n Now consider Sn = fn 2 N : '(n)  ng. Sn must be in nite. First Sn 6= ; for if Sn = ; then '(k) > k for all k and there is no k0 with '(k0 ) = 1. Suppose Sn is nite, with k the largest value in the set. Then '(n) > n for n  k + 1. Consider f1; 2; : : : ; k + 1g. Since '(n) > k + 1 for n  k + 1, the only integers which can be pre-images of f1; 2; : : : ; k + 1g are f1; 2; : : : ; kg. This is not possible, since ' is one-to-one and onto. Therefore Sn = fn 2 N : '(n)  ng is in nite. Choose a sequence, '(nk) = L. However nk 2 Sn with nk ! 1. We now have klim !1 n nlim !1

'(nk)  1: nk

k

Thus From (1) and (2), L = 1.

L  1:

(2)

458

13. Prove that for any positive x ; x ; : : : ; xn, y ; y ; : : : ; yn the in-

equality

1

n X

1  xy

i=1 i i

2

1

2

4n2 i=1 (xi + yi)2

Pn

holds.  Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; by Christopher J. Bradley, Clifton College, Bristol, UK; by Michael Selby, University of Windsor, Windsor, Ontario; by Bob Prielipp, University of Wisconsin{Oshkosh,Wisconsin,USA; by PanosE. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We feature Bradley's solution. Now,

1

4

xy  (x + y)2 since (x + y )2  4xy , as (x , y )2  0. So

n 4 1  X () 2 i=1 (xi + yi) i=1 xi yi Lemma. (a1 + a2 +    + an )(a2 a3    an + a1 a3 a4    an + a1 a2 a4    an +    + a1a2    an,1)  n2a1a2    an . n X

This follows from separate applications of the AM{GM inequality to the two terms on the left. It follows that

n2

a1 + a2 +    + an  Now put and then

1 1 1 a1 + a2 +    + an

ai = (x +1 y )2 ; i i

i = 1; : : : ; n

1 1 1 (x1 + y1)2 + (x2 + y2)2 +    + (xn + yn)2 

Combining this with () shows n X

1  xy

i=1 i i

:

n2 : i=1(xi + yi)2

Pn

4n2 : i=1(xi + yi)2

Pn

14. There is a nite number of towns in a country. They are connected by one direction roads. It is known that, for any two towns, one of them can be reached from the other one. Prove that there is a town such that all the remaining towns can be reached from it.

459 Solutions by Mansur Boase, student, St. Paul's School, London, England; and by Christopher J. Bradley, Clifton College, Bristol, UK. We give the solution by Boase. We prove the result by induction on the number, n, of towns. If n  2 the result is immediate. Label the towns A1 ; A2 ; : : : ; Ak . We shall prove that if the statement holds for all n < k, then it is also true for n = k, so by induction it will be true for all n. We can split up the towns excluding A1 into two sets M and N , M containing those towns which can be reached from A1 and N those which cannot be reached from A1. Thus, every town in N can reach A1 , and there is no route from a town in M to a town in N . If N is empty, then A1 is the desired town. If this is not the case, then, since for any two towns in N , one of them can be reached from the other, and there is no route from outside N into N , the routes in question must pass through towns in N . By the induction hypothesis, since jN j < k, there is a town in N which can reach all other towns in N . It can also reach A1 , and thus all towns in M . Therefore, this is the town which can reach all the other towns in the country, and the result is proved. 15. Noah has 8 species of animals to t into 4 cages of the ark. He plans to put species in each cage. It turns out that, for each species, there are at most 3 other species with which it cannot share the accommodation. Prove that there is a way to assign the animals to their cages so that each species shares with compatible species. Solution by Mansur Boase, student, St. Paul's School, London, England. Let the animals be vertices of a graph. Join two animals by an edge if they are compatible. Now we have a graph with 8 vertices, and each vertex is joined to at least 4 others. So, by Dirac's theorem on Hamiltonian cycles, there must be a Hamiltonian cycle, and if we take consecutive pairs of animals in this cycle, we can put them in the same cage, and we have the required solution. 17. Quadrangle ABCD is inscribed in a circle with radius 1 in such a way that one diagonal, AC , is a diameter of the circle, while the other diagonal, BD, is as long as AB . The diagonals intersect in P . It is known that the length of PC is 25 . How long is the side CD?  Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; and by Christopher J. Bradley, Clifton College, Bristol, UK. We give the solution of Arslanagic.

460

D

C P B

O

A The triangle ABD is isosceles because AB = BD. Let O be the centre of the circumcircle. Then BO ? AD. Because CD ? AD (AC is a diameter), we get CDkBO; that is, 4PCD  4POB , and it follows that

CD = PC ; that is OB PO  PC = 1  25 = 2 : CD = OBPO 3 3 5

18. Show that in a non-obtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle.  Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, Clifton College, Bristol, UK; by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Prielipp's solution. In this solution R; r; s will denote the circumradius, inradius and semiperimeter of a triangle. We shall show that in a non-obtuse triangle the perimeter is always greater than or equal to 2(2R) + 2r. Lemma. If A is an angle of triangle ABC , then cos A is a root of the equation 4R2t3 , 4R(R + r)t2 + (s2 + r2 , 4R2 )t + (2R + r)2 , s2 = 0 Proof. Since a = 2R sin A, and s , a = r cot( A2 ),   s = a + (s , a) = 2R sin A + r cot A2 s

cos A = 2R (1 , cos A)(1 + cos A) + r 11 + , cos A : p

()

461 Thus so

cos A s2 = 4R2(1 , cos A)(1 + cos A) + 4Rr(1 + cos A) + r2 11 + , cos A 4R2(1 , cos A)2(1 + cos A) + 4Rr(1 + cos A)(1 , cos A) +r2 (1 + cos A) , s2(1 , cos A) = 0:

Hence

rR2 cos3 A , 4R(R + r) cos2 A + (s2 + r2 , 4R2 ) cos A +(2R + r)2 , s2 = 0 making cos A a root of the equation (). Corollary 1. If A, B , and C are the angles of triangle ABC , then cos A, cos B , and cos C are the roots of the equation (). Corollary 2. If A, B , and C are the angles of triangle ABC , then 2 R + r)2 : cos A cos B cos C = s , (2 4R2 Corollary 3. If A is the largest angle of triangle ABC , then s > 2R + r if A < 90 s = 2R + r if A = 90 and s < 2R + r if A > 90: Corollary 4. In a non-obtuse triangle the perimeter of the triangle is always greater than or equal to 2(2R) + 2r. Proof. If the triangle is an acute triangle, then s > 2R + r, and 2s > 2(2R) + 2r. If the triangle is a right triangle, then s = 2R + r. Thus 2s = 2(2R) + 2r. Corollary 5. In a non-obtuse triangle the perimeter of the triangle is always greater than twice the diameter of the circumcircle. 19. Let C be a circle in the plane. Let C1 and C2 be nonintersecting circles touching C internally at points A and B respectively. Let t be a common tangent of C1 and C2 , touching them at points D and E respectively, such that both C1 and C2 are on the same side of t. Let F be the point of intersection of AD and BE . Show that F lies on C .

462 Solution by Christopher J. Bradley, Clifton College, Bristol, UK. Let SA be the tangent to C1 and C at A and TB be the tangent to C2 and C at B , as shown. S F t D  A 90 ,  '

C1 C

G O

H

E

'

90 , '

B

C2

Let \SAD =  and \TBE = '. Let O be the centre of C . AO meets C1 again at G and since it is a common radius, AG is a diameter of C1. BO meets C2 again at H and BH is likewise a diameter of C2. We have \DAG = 90 ,  and \DGA = . By the alternate segment theorem \GDE = 90 ,  and since \ADG = 90 (angle in a semicircle) it follows that \FDE = . Similarly \FED = ' and so \DFE = 180 ,  , '. Also \EBH = 90 , '. Considering the angles of the (re-entrant) quadrilateral FAOB we have re ex \AOB = 360 , (90 , ) , (90 , ') , (180 ,  , ') = 2 + 2'. So \AOB = 360 , 2 , 2' = 2\DFE . But O is the centre of circle C , and AB is an arc of C , so F lies on C . (Converse of the angle at the centre

= twice angle at circumference).

20. Let a  b  c be the sides of a right triangle, and let 2p be its perimeter. Show that p(p , c) = (p , a)(p , b) = S (the area of the triangle).  Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; by Mansur Boase, student, St. Paul's School, London, England; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; by Michael Selby, University of Windsor, Windsor, Ontario; by Panos E. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Since the triangle is a right triangle we have c2 = a2 + b2 , p = a+2b+c , and S = ab2 .

463 Then

  a + b + c a + b + c (a + b)2 , c2 p(p , c) = , c = 2 2 4 2 2 2 a + b + 2 ab , c ab = = S; =

4

and

(p , a)(p , b) =



2

a + b + c , a  a + b + c , b 2

2

= c + b2 , a c , b2 + a 2 2 2 2 2 = c , (b , a) = c , (b + a ) + 2ab = ab = S; 4

as required.

4

2

We conclude this number of the Corner with solutions to some of the problems of the 8th Iberoamerican Mathematical Olympiad, September 14{ 15, 1993 (Mexico) which we gave last year [1996: 159{160].

8th IBEROAMERICAN MATHEMATICAL OLYMPIAD September 14{15, 1993 (Mexico) 1. (Argentina) Let x < x <    < xi < xi <    be all the 1

2

+1

palindromic natural numbers, and for each i, let yi = xi+1 , xi . How many distinct prime numbers belong to the set fy1; y2 ; y3; : : : g ? Solutions by Mansur Boase, student, St. Paul's School, London, England; and by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. We give Boase's solution. The rst few palindromic numbers are

1; 2; 3; : : : ; 9; 11; 22; 33; : : : ; 99; 101; 111; : : : : Now 11 , 9 = 2 and 22 , 11 = 11.

We shall show that these are the only two prime values which a yi term can take. If xi and xi+1 have di erent numbers of digits, then xi will be of the form 99 : : : 9 and xi+1 of the form 10 : : : 01, so yi = xk+1 , xi = 2. We can consider, without any loss of generality only yi where xi has more than two digits since yi can only be prime if yi = 2 or 11 for xi with one or two digits from the above list of the rst xi . If xi and xi+1 end in the same digit, then 10 divides yi, so yi cannot be prime. If xi and xi+1 end in di erent digits, say r and s, then s = r + 1,

464

xi is of the form r999 : : : 9r, and xi+1 is of the form (r + 1)0 : : : 0(r + 1). Then yi = xi+1 , xi = (r + 1) , (10 , r) = 11. Thus only two distinct primes belong to the set fy1; y2 ; : : : g.

2. (Mexico) Show that for any convex polygon of unit area, there exists a parallelogram of area 2 which contains the polygon. Solution by Mansur Boase, student, St. Paul's School, London, England. We shall show more generally that there exists a rectangle of area 2 containing the polygon. The result is obviously true for a triangle. To prove this, construct a rectangle on the longest side of the triangle and circumscribing the triangle. If the area of the triangle is 1, then the rectangle will have area 2. A B C If the polygon has more than three vertices, then choose the two vertices of the polygon which are furthest apart. Call them B and C . Draw perpendiculars to the line BC at B and at C to give lines l1 and l2 , respectively. D B l1 C

E l2 All the vertices of the polygon must lie between these two lines. (Otherwise there would be two vertices further apart than jBC j.) Now consider the smallest rectangle which circumscribes the polygon and with one pair of opposite sides lying on l1 and l2 . Suppose this polygon touches the polygon again at D and E . R B D U S E C T Let the vertices of the rectangle be R, S , T and U with D on UR and E on ST . Then it is easy to see that

[RUCB] = 2[BCD]

465 and

[BSTC ] = 2[BEC ] since BC kRU and BC kST . [RSUT ] = 2[CDBE ]  2(area of polygon)= 2 since the polygon is convex. We can nd an even larger rectangle of area 2

containing the polygon. 4. (Spain) Let ABC be an equilateral triangle, and , its incircle. If D and E are points of the sides AB and AC , respectively, such that DE is tangent to ,, show that

AD + AE = 1: DB EC

 Solutions by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; and by Mansur Boase, student, St. Paul's School, London, England. We give Arslanagic's solution. C

a E A

, p

a

I

B a Let AB = AC = BC = a and BD = p; that is, AD = a , p, and CE = q; that is, AE = a , q. The circle , is inscribed in the quadrilateral and we get D

ED + BC = BD + CE

or

ED + a = p + q

or

ED = p + q , a: By the law of cosines for the triangle ADE , it follows that ED2 = AE2 + AD2 , 2AE  AD cos 60; so, from ()

(p + q , a)2 = (a , q )2 + (a , p)2 , (a , q )(a , p)

and from this we obtain Now, we have

a = p3+pqq : AD = a , p = p(2pq+,qp)

()

466 and that is,

AE = a , q = q(2pp+,qq) ; AD + AE = p (2q , p) + q (2p , q) = p + q = 1; DB EC p p + q q p + q p + q

as required. 6. (Argentina) Two non-negative integer numbers, a and b, are \cuates" (friends in Mexican) if the decimal expression of a + b is formed only by 0's and 1's. Let A and B be two in nite sets of non-negative integers, such that B is the set of all the numbers which are \cuates" of all the elements of A, and A is the set of all the numbers which are \cuates" of all the elements of B . Show that in one of the sets A or B there are in nitely many pairs of numbers x, y such that x , y = 1. Solution by Mansur Boase, student, St. Paul's School, London, England. Suppose an integer of A ends with the digit `r'. Then all integers in B must have a last digit the same as for 10 , r or 11 , r in order that they are all \cuates" of A. If B contains elements with last digits the same as for 10 , r and 11 , r, then every element of A must end in the last digit r to be \cuates" with integers of B of both last digits. Thus either set A or set B has all integers ending in the same digit. Without loss, assume that all elements of A end in `r'. Now consider an element of B which is a \cuate" of all the integers of A. Let us say it is of type (i) if it ends with the last digit 10 , r and of type (ii) if it ends with the last digit of 11 , r. If we change the last digit we obtain another number which is a \cuate" of all the elements of A, and hence in B. The di erence between these pairs is 1. It follows therefore that there are equal numbers of each type in B , and as B is in nite, there are in nitely many pairs x; y in B such that x , y = 1. That completes the Corner for this month. Send me your Olympiad contest materials and your nice solutions to problems in the Corner.

467

BOOK REVIEWS Edited by ANDY LIU Learn from the Masters! edited by Frank Swetz, John Fauvel, Otto Bekken, Bengt Johansson, Victor Katz, published by the MAA, Classroom Resource Materials Series, 1995, ISBN# 0-88385-703-0, softcover, 312+ pages, $23.00. Reviewed by Maria de Losada, Bogota, Colombia. A rich and varied collection of thoughts directed primarily toward the use of the history of mathematics for e ective and enriched teaching (and learning), these are papers given at the \Kristiansand (Norway) Conference" of 1988. The areas and topics cover a broad range corresponding to di erent tastes and personal interests, divided in sections on school and higher mathematics. Frank Swetz's Using Problems from the History of Mathematics in Classroom Instruction is a superb example of history transcending anecdotal information and grasping the relationship between problem solving and the gradual construction of meaning (especially clear in the section Illustrating the growth of mathematical pro ciency), an essential component of each individual student's coming to grips with mathematical concepts. The choice of problems and the orientation involving the speci c ways in which they can be used to enrich instruction is excellent. Other papers that focus on mathematical thinking explore algorithms and analogies, modelling and heuristic reasoning, as well as Man-Keung Siu's Mathematical Thinking and History of Mathematics. In the latter section, amidst the very ne selection o ered, the article by Israel Kleiner The Teaching of Abstract Algebra: An Historical Perspective stands out. Kleiner describes his approach as genetic, but notes that he tries \to show how attempts to solve the problems give rise to the abstract theory. This is, of course, the historical sequence of events." He further addresses the question of how \history provides the opportunity to raise a number of general issues about the nature of mathematics". It is most unfortunate that the random sprinkling of photographs of notable mathematicians throughout the book should place that of Emmy Noether opposite the title In Hilbert's Shadow. De nitely of interest to those who lean toward using history as a resource for enriching their teaching and the mathematical thinking of their students, as well as to those whose interest in historical subjects is just beginning to be awakened.

468

Dissecting Rectangular Strips Into Dominoes Frank Chen

student, D. S. McKenzie Junior High School, Edmonton

Kenneth Nearey and Anton Tchernyi

students, Grandview Heights Junior High School, Edmonton

The First Problem A domino is de ned to be a 1  2 or 2  1 rectangle. The rst one is said to be horizontal and the second vertical. In our Mathematics Club, we learned to count the number gn of di erent ways of dissecting a 3  2n strip into dominoes. The sequence fgn g satis es the recurrence relation gn = 4gn,1 , gn,2; (1) for all n  2, with initial conditions g0 = 1 and, as shown in Figure 1, g1 = 3.

Figure 1 The generating function G(x) for the sequence is de ned to be the formal power series

g0 + g1x + g2x2 +    :

It is easy to deduce from (1) that

G(x) = 1 ,14,x +x x2 :

(2)

In examining the dissections of the 3  2n strip, we observed that they fall into two kinds. A dissection of the rst kind can be divided by a vertical line into two substrips without splitting any dominoes. Such a line is called a fault line. A dissection of the second kind, called a fault-free dissection, has no fault lines. Figure 2 shows an example of each, using 3  4 strips.

469

Figure 2 Let fn be the number of fault-free dissections of the 3  2n strip. From Figure 1, f1 = 3. For all n  2, a fault-free dissection cannot start with three horizontal dominoes. It must start o as shown in Figure 3, and continue by adding horizontal dominoes except for a nal vertical one.

Figure 3 It follows that fn = 2 for all n  2, and the sequence satis es a trivial recurrence relation

fn = fn,1 for all n  3, with initial conditions f0 = 1; f1 = 3 and f2 = 2. Let F (x) = f0 + f1x + f2x2 +    be the generating function for the sequence. Then F (x) = ,1 + x + 2(1 + x + x2 +    ) = ,1 + x + 1,2 x . This simpli es

to

F (x) = 1 +12,x ,x x : 2

(3)

Having solved the simpler problem of counting fault-free dissections of the 3  2n strip, we make use of our result to nd an alternative solution to the general problem of nding all dissections of this strip. They can be classi ed according to where the rst fault line is. This is taken to be the right end of the strip if the dissection is fault-free. Then the strip is divided into a 3  2k substrip on the left and a 3  2(n , k) substrip on the right, where 1  k  n. Since the rst substrip is dissected without any fault lines, it can be done in fk ways. The second substrip can be dissected in gn,k ways as we do not care whether there are any more fault lines. Hence

gn = f1gn,1 + f2gn,2 +    + fng0:

(4)

470 From the values of fn, gn = 3gn,1 + 2gn,2 + 2gn,3 +    + 2g0 . If we subtract from this expression gn,1 = 3gn,2 + 2gn,3 +    + 2g0 , we have gn , gn,1 = 3gn,1 , gn,2, which is equivalent to (1). We now derive (2) in another way. It follows from (4) that for all n  1,

2gn = f0gn + f1gn,1 + f2 gn,2 +    + fng0 : (5) Multiplying F (x) and G(x) yields F (x)G(x) = (f0 + f1x + f2x2 +    )(g0 + g1x + g2x2 +    ) = f0g0 + (f0g1 + f1g0 )x + (f0g2 + f1 g1 + f2 g0)x2 +    : In view of (5), this becomes F (x)G(x) = g0 +2g1x +2g2x2 +   = 2G(x) , 1 so that

G(x) = 2 , 1F (x) :

(6)

Substituting (3) into (6) yields (2).

The Second Problem Let gn be the number of ways of dissecting a 4  n strip into dominoes. Then g0 = 1; g1 = 1 and, as shown in Figure 4, g2 = 5. It is not hard to verify that g3 = 11. We wish to determine the in nite sequence fgn g via recurrence relations and generating functions.

Figure 4 Let fn be the number of fault-free dissections of the 4  n strip. We have f0 = 0; f1 = 1 and from Figure 4, f2 = 4. For odd n  3, the only fault-free dissections are the extensions of the second and third ones in Figure 4, with horizontal dominoes except for a nal vertical one. Hence fn = 2. For even n  4; fn = 3 since we can also include similar extensions of the fourth dissection in Figure 4. As in the solution of the First Problem, we have

gn = f1gn,1 + f2gn,2 +    + fng0:

471 This leads to

gn = gn,1 + 5gn,2 + gn,3 , gn,4 (7) for all n  4, with initial conditions g0 = 1; g1 = 1; g2 = 5 and g3 = 11. Also,

F (x) = 1 + x + 4x2 + 2x3 + 3x4 + 2x5 + 3x6 +    = ,2 , x + 2x2 +3(1 + x2 + x4 + x6 +    ) + 2x(1 + x2 + x4 +    ) 2x = ,2 , x + x2 + 31 + , x2 2 3 4 (8) = 1 + x + 13x, x+2 x , x :

Substituting (8) into (6), which is still valid here, we have 2 G(x) = 1 , x , 15x,2 x, x3 + x4 :

(9)

We now give an alternative solution to the Second Problem, along the line of the solution to the First Problem we learned at the Mathematics Club. We classify the dissections of the 4  n strip into ve types according to how they start. These correspond to those in Figure 4 if we ignore the vertical dominoes in the second column. Call these Types A, B, C, D and E, and let their numbers be an ; bn ; cn ; dn and en , respectively. By symmetry, we have bn = cn so that gn = an + 2bn + dn + en (10) for all n  1. In a Type A dissection, we are left with a 4  (n , 2) strip, which can be dissected in gn,2 ways. Hence an = gn,2 (11) for all n  3, with a1 = 0 and a2 = 1. In a Type B dissection, if we complete the second column with a vertical domino, the remaining 4  (n , 2) strip can be dissected in gn,2 ways. The only alternative is to ll the second column with two horizontal dominoes. The remaining part can be dissected in bn,1 ways, so that bn = gn,2 + bn,1 (12) for all n  3, with b1 = 0 and b2 = 1. In a Type D dissection, the situation is similar except that if we ll the second column with two horizontal dominoes, we must then also ll the

472 third column with two more horizontal dominoes. The remaining part can be dissected in dn,2 ways, so that dn = gn,2 + dn,2 (13) for all n  3, with d1 = 0 and d2 = 1. Finally, in a Type E dissection, after lling the rst column with two vertical dominoes, we are left with a 4  (n , 1) strip which can be dissected in gn,1 ways. Hence en = gn,1 (14) for all n  2, with e1 = 1. Now (7) follows from (10), (11), (12), (13) and (14) since

gn = an + 2bn + dn + en = gn,2 + (2gn,2 + 2bn,1 ) + gn,2 + dn,2 + gn,1 = gn,1 + 4gn,2 + 2(gn,3 + bn,2) + dn,2 = gn,1 + 4gn,2 + 2gn,3 + 2bn,2 + dn,2 = gn,1 + 4gn,2 + 2gn,3 + gn,2 , an,2 , en,2 = gn,1 + 5gn,2 + gn,3 , gn,4: Using (7), (1 , x , 5x2 , x3 + x4 )G(x) simpli es to 1 , x2 . Hence (9) also follows.

Supplementary Problem Let fn denote the number of fault-free dissections of the 4  n strip. Find a recurrence relation for the sequence ffn g with initial conditions.

Acknowledgement This article has been published previously in a special edition of delta-k, Mathematics for Gifted Students II, Vol. 33, 3 1996, a publication of the Mathematics Council of the Alberta Teachers' Association and in AGATE, Vol. 10, 1, 1996, the journal of the Gifted and Talented Education Council of the Alberta Teachers' Association. It is reprinted with permission of the authors and delta-k.

473

THE SKOLIAD CORNER No. 26 R.E. Woodrow

This number we give the 30 problems of the Kangourou Des Mathemati ques, Epreuve Europeenne, which was given Friday March 22, 1996 to about 500,000 students in 16 European countries, and in 8 African countries, without counting French schools around the world. My thanks go to Ravi Vakil, Canadian Team leader to the 37th IMO in Mumbai, India, who collected a great deal of contest material and forwarded it to me. My copy is in French, and we give it in that language. The contestants are given 75 minutes, and no calculators are allowed.

 KANGOUROU DES MATHEMATIQUES March 22, 1996

Time: 75 minutes 1. Les representants de 12 pays ont choisi pour vous ces 30 questions. Chaque question a e te discutee 10 minutes. Quelle a e te la duree totale de la discussion? (3 points) A. 360 min B. 300 min C. 120 min D. 52 min E. 40 min.

2.

Dans la gure ci-contre, l'aire de la region laissee en blanc est (3 points)

6 cm2 . Quelle est l'aire de la region grise?

A. 3 cm2

B. 4 cm2

C. 6 cm2

D. 9 cm2

3. Quel est le plus grand nombre?

A. 1  9  9  6

(3 points)

B. 19  9  6

D. 1  9  96

E. 12 cm2 . C. 1  99  6

E. 19  96.

4. En utilisant une et une seule fois chacun des chi res 1; 2; 3 et 4, je peux e crire di erent nombres. Je peux e crire par exemple 3241. Quelle est le di erence entre le plus grand et le plus petit nombre ainsi fabriques? (3 points) A. 2203 B. 2889 C. 3003 D. 3087 E. 3333.

474

5. Un cercle et un rectangle s'aimaient d'amour tendre, \Mais malheureusement, dit le cercle, que nous croissions ou retrecissions, nous ne pourrons jamais avoir plus de n points communs!" A votre avis, combien vaut n? (3 points) A. 2 B. 4 C. 5 D. 6 E. 8. 6. Le nombre

3 A. 1110

1 10

1 1 + 100 + 1000 est e gale a :

3 B. 1000

111 C. 1000

D.

(3 points) 3 E. 111 .

111 1110

7. La gure ci-contre represente un grand carre d'aire 1 m . Une diag2

onale est partagee en trois segments de m^eme longueur. Le segment median est une diagonale du petit carre gris. Quelle est l'aire de ce petit carre? (3 points)

A. 101 m2

B. 19 m2

C. 61 m2

D. 14 m2

E. 31 m2 .

8. La salle d'un thea^ tre comporte 26 rengees de 24 places chacune. Toutes les places sont numerotees, en commencant par le premier rang. Dans quelle rangee se trouve le siege numerote 375? (3 points) A. 12e me B. 13e me C. 14e me D. 15e me E. 16e me. 9. Parmi les phrases ci-dessous, quelles sont les phrases vraies?(3 points)

(1)La somme de deux nombres negatifs est toujours negative. (2)La somme d'une nombre positif et d'un nombre negatif est toujours positive. (3)La somme d'une nombre negatif et de deux nombres positifs est toujours positive. A. aucune B. la (1) seule C. le (1) et la (2) D. la (2) et la (3) E. toutes les trois.

10. a est un nombre de deux chi res, b est le nombre obtenu en e crivant deux fois, c^ote a c^ote, les deux chi res de a. Quel est le quotient de b par a? (3 points) A. 10 B. 11 C. 99 D. 100 E. 101.

475

11. Un triangle e quilateral et un hexagone sont inscrits dans un m^eme cercle. Si l'on divise l'aire de l'hexagone par l'aire du triangle, quel est le quotient obtenu? (4 points)

A. 1:5

B. 2

C. 3

D. 4

E.  .

12. Un kangourou a dans sa poche 3 chaussettes blanches, 2 chaussettes noires et 5 chaussettes grises. Sans regarder, il veut en prendre une paire. Quel nombre minimum de chaussettes lui faut-il sortir pour e^ tre s^ur qu'il en a bien deux de la m^eme couleur? (4 points) A. 2 B. 3 C. 4 D. 7 E. 10. 13. A la f^ete foraine, une llette a achete cinq echettes. Chaque fois qu'elle touche la cible, elle a deux echettes gratis. Elle a lance en tout 17

echettes. Combien de fois a-t-elle touche la cible. (4 points) A. 4 B. 6 C. 7 D. 12 E. 17. 14. Les bissectrices de deux angles d'un triangle font entre elles un angle de 110. Combien vaut le troisieme angle de ce triangle? (4 points) 110

?

A. 30

B. 40

C. 45

D. 55

E. 70 .

15. Une vieille montre retarde de 8 minutes par vingt-quatre heures. De combien de minutes dois-je l'avancer ce soir a 22 heures si j'ai absolument besoin qu'elle me donne l'heure exacte demain matin a 7 heures? (4 points) A. 1 min 40 s B. 2 min 20 s C. 3 min D. 4 min 30 s E. 6 min.

476

16. La somme, en degres, des angles marques sur la gure ci-contre est e gale a : (4 points) r

r

r

A. 120

B. 150

C. 180

D. 270

E. 360.

17. Un bidon plein de lait pese 34 kg. Le m^eme bidon quand il est a moitie plein pese 17; 5 kg. Combien pese le bidon vide? (4 points) A. 1 kg B. 0; 5 kg C. 1; 5 kg D. 2 kg E. il n'y a pas assez de donnees. 18. Le c^otes d'un triangle mesurent 8 cm, 15 cm et 17 cm. Quelle est son aire? (4 points) A. 40 cm2 B. 60 cm2 C. 68 cm2 D. 80 cm2 E. on ne peut pas la calculer. 19. Entre 6 heures ce matin et 18 heures ce soir, combien de fois les deux aiguilles de ma montre feront-elles un angle droit? (4 points) A. 2 B. 6 C. 12 D. 22 E. 24.

20. Simone a un gros tas de dalles triangulaires. Toutes ces dalles ont une forme identique: ce sont des triangles e quilateraux de 1 dm de c^ote. De combien de dalles Simone aura-t-elle besoin pour daller un grand triangle e quilateral de 2 metres de c^ote? (4 points) A. 200 B. 300 C. 400 D. 600 E. 800. 21. En decoupant un coin d'un cube en bois, on a obtenu le solide ci-contre. Maintenant on decoupe de la m^eme facon les sept autres coins du cube. On a alors un solide qui a quatorze faces (les faces triangulaires ne se touchent pas et ne se recoupent pas). Quel est le nombre s de sommets et le nombre a d'ar^etes du solide obtenu? (5 points) A. s = 24; a = 36 C. s = 10; a = 15

B. s = 36; a = 24 D. s = 24; a = 32 E. s = 36; a = 18.

477

22. On compte le nombre de points d'intersection de quatre droites distinctes. Quel est le nombre qu l'on est s^ur de ne pas trouver? (5 points) A. 0 B. 2 C. 3 D. 5 E. 6. 23. Combien y a-t-il de triangles, dont les c^otes ont pour mesures (en centimetres) des nombres entiers, et dont le perimetre est e gal a 15 cm? (5 points) A. 1 B. 5 C. 7 D. 19 E. 45. 24. Marine et Claire se partagent un c^one glace en le coupant a mihauteur. Marine en a plus que Claire! (5 points) 1 2 1 2

A. 1 fois et demie plus

B. 2 fois plus

D. 7 fois plus

C. 3 fois plus

E. 8 fois plus.

25. Nous sommes sur une ligne de metro circulaire. Vingt-quatre trains s'y deplacent dans la m^eme direction, a intervalles reguliers et roulant tous a la m^eme vitesse. Demain, on doit rajouter des trains a n de diminuer de 20% les intervalles entre deux trains. Combien y aura-t-il de trains supplementaires demain sur la ligne? (5 points) A. 2 B. 3 C. 5 D. 6 E. 12. 26. Dans la gure ci-contre, (AB) est parallele a (CD). De plus AD , DC , CB et AB = AC . Combien vaut l'angle D^ ? (5 points) D ?

C

A A. 108

B. 120

C. 130

D. 150

B E. on ne peut pas la savoir.

478

27. Charles a attribue a tous ses livres un code de trois lettres, en utilisant, l'ordre alphabetique: AAA; AAB; AAC; : : : ; AAZ; ABA; ABB : : : Charles a 2203 livres. Quel est le dernier code utilise par Charles quand il a code toute sa collection? (5 points) A. CFS B. CHT C. DGS D. DFT E. DGU . 28. Cinq personnes sont assises autour d'une table ronde. Chacune arme a son tour: \Mes deux voisins, de droite et de gauche, sont des menteurs". On sait que les menteurs mentent toujours et que quelqu'un qui n'est pas un menteur dit toujours la verite. De plus tout le monde conna^it la verite en ce qui concerne ses deux voisins. Combien y a-t-il de menteurs a cette table? (5 points) A. 2 B. 3 C. 4 D. 5 E. on ne peut pas la savoir. 29. On a coupe quatre dr^oles de parts suivant les diagonales d'un dr^ole de g^ateau plat a quatre c^otes. J'ai mange une part. Mes amis mecontents ont pese les trois restantes et ont trouve 120 g. 200 g. et 300 g. Combien pesait la part que j'ai mangee? (5 points) 120

?

300

200

A. 120 g

B. 180 g

C. 280 g

D. 330 g

E. 500 g.

30. Dans la suite de chi res 122333444455555 : : : , chaque entier est e crit autant de fois que sa valeur. Quel est le 1996e me chi re e crit? (5 points) A. 0 B. 3 C. 4 D. 5 E. 6.

479 Last issue we gave the problems of the Second Round of the Alberta High School Mathematics Competition, of February 11, 1997. The solutions we give were taken from the contest web site

~

http://www.math.ualberta.ca/ ahsmc/sample.htm

where more information about the contest and the solutions may be found. The solutions are selected from contestants' work. My thanks to E. Lewis, University of Alberta, Chair of the contest, for supplying us with materials.

ALBERTA HIGH SCHOOL MATHEMATICS COMPETITION February 11, 1997

Second Round 1. Find all real numbers x satisfying jx , 7j > jx + 2j + jx , 2j. Remark. Note that jaj is called the absolute value of the real number a. It has the same numerical value as a but is never negative. For example, j3:5j = 3:5 while j , 2j = 2. Of course, j0j = 0. Solution by Laura Harms, Lorne Jenken High School, Barrhead, Alberta. If x is in (,1; ,2), the inequality becomes 7 , x > (,2 , x)+(2 , x) which simpli es to x > ,7. Hence all x in (,7; ,2) satisfy the inequality. If x is in [,2; 2], then jx , 2j + jx + 2j = 4, while jx , 7j is never less than 5, so all x in [,2; 2] satisfy the inequality. If x is in (2; 7), then the inequality becomes 7 , x > (x , 2) + (x + 2) which simpli es to x < 73 . Hence all x in (2; 73 ) satisfy the inequality. If x is in [7; 1), then x , 7 < x , 2 < (x , 2) + (x + 2), and the inequality is not satis ed. In summary, x satis es the inequality if and only if ,7 < x < 7=3. 2. Two lines b and c form a 60 angle at the point A, and B1 is a point on b. From B1 , draw a line perpendicular to the line b meeting the line c at the point C1 . From C1 draw a line perpendicular to c meeting the line b at B2 . Continue in this way obtaining points C2 , B3 , C3 , and so on. These points are the vertices of right triangles AB1 C1 ; AB2 C2 ; AB3 C3 ; : : : . If area (AB1C1 ) = 1, nd

area (AB1C1 )+ area (AB2 C2)+ area (AB3 C3)+  + area (AB1997C1997): Solution by Margaret Tong, James Fowler High School, Calgary, Alberta. Clearly, triangles ABn Cn are similar to each other. In a (30; 60 ; 90 ) triangle, the hypotenuse is twice as long as the shorter leg. Let AB1 = x. Then AC1 = 2x and AB2 = 4x. It follows that area(ABn Cn ) = 16  area(ABn,1Cn,1), so that the desired total area is given by T = 1 + 16 + 162 +    + 161996. Multiplying this by 16, we have 16T =

480

16 + 161 2 + 163 +    + 161997. Subtraction yields 15T = 161997 , 1 so that T = ( 15 )(161997 , 1).

3. A and B are two points on the diameter MN of a semicircle. C , D, E and F are points on the semicircle such that \CAM = \EAN = \DBM = \FBN . Prove that CE = DF .

Solution by Byung-Kyu Chun, Harry Ainlay High School, Edmonton, Alberta. Complete the circle. Extend EA to cut it at C 0 , and extend DB to cut it at F 0. By symmetry, AC = AC 0 so that m(AC 0C ) = m(ACC 0). Similarly, m(BF 0F ) = m(BFF 0). Now m(EC 0C ) = 180 , m(CAC 0) = 180 , 2m(CAM ) = 180 , 2m(FBN ) = 180 , m(FBF 0) = m(DF 0F ). Since the arcs CE and DF subtend equal angles at the circle, they have equal measure. It follows that the chords CE and DF are equal. E F D C

A

M C0

D0

B

N

F0 4. (a) Suppose that p is an odd prime number and a and b are positive integers such that p4 divides a2 + b2 and p4 also divides a(a + b)2. Prove that p4 also divides a(a + b). (b) Suppose that p is an odd prime number and a and b are positive integers such that p5 divides a2 + b2 and p5 also divides a(a + b)2 . Show by an example that p5 does not necessarily divide a(a + b). Solution to part (a) by Byung-Kyu Chun, Harry Ainlay High School, Edmonton, Alberta. Note that a(a+b)2 = a(a2 +b2 )+2a2 b. Since p4 divides both a(a+b)2 and a2 + b2 , it must also divide 2a2 b. Since p is an odd prime, p4 divides a2 b. Suppose p2 does not divide a. Then the only powers of p that can possibly divide a2 are p or p2 . Since p4 divides a2 b, it follows that p2 must divide b. Hence p4 divides b2 . However, this contradicts p4 dividing a2 + b2 but not a2 . It follows2that we must have p2 dividing a. Then p4 divides a2 so that it 2 also divides b . Hence p divides b, and it also divides a + b. It follows that

481

p4 divides a(a + b).

Solutionto part (b) given by Byung-Kyu Chun, Harry Ainlay High School, Edmonton, Alberta; and by Jason Ding, Archbishop MacDonald High School, Edmonton, Alberta. We look for a, b and p such that p5 divides a2 + b2 , p2 divides a and b, 3 but p does not divide a + b. Setting a = p2 x and b = p2y , these conditions become p divides x2 + y 2 and p does not divide x + y . We can pick x = 2, y = 1 and p = 5. This gives a = 50 and b = 25 as an example 5. The picture shows seven houses represented by the dots, connected by six roads represented by the lines. Each road is exactly 1 kilometre long. You live in the house marked B . For each positive integer n, how many ways are there for you to run n kilometres if you start at B and you never run along only part of a road and turn around between houses? You have to use the roads, but you may use any road more than once, and you do not have to nish at B . For example, if n = 4, then three of the possibilities are: B to C to F to G to F ; B to A to B to C to B; and B to C to B to A to B.

D

r

A r

B

r

E

C

r

r

F

r

r

G

Solution by Byung-Kyu Chun, Harry Ainlay High School, Edmonton, Alberta. The number of ways to travel n = 1 kilometre is 2, the ways being B to A and B to C . For n = 2, we have 4 ways, B via A to B , B via C to B, B via C to D and B via C to F . Note that the answer is the same had we started at D or F . We guess that the number of ways for any n is exactly 2n . We now prove this by induction, in an unusual manner. We consider separately the cases n = 2k and n = 2k + 1. In the even case, the result is true for k = 0. Suppose that the number of ways for n = 2(k , 1) is exactly 22(k , 1). At this point, we must be at one of B , D, or F . As pointed out before, there are 4 ways to go another 2 kilometres. Hence, for n = 2k, the number of ways is 4  22(k,1) = 22k . For n = 2k +1, we use the established fact that for n = 2k, the number of ways is exactly 22k . Again, after 2k kilometres, we must be at one of B , D, or F . In each2(case, there are two ways to go the extra kilometre, bringing the total to 2  2 k,1) = 22k+1 for n = 2k + 1. That completes the Skoliad Corner for this number. Please send me contest materials for use in the Skoliad as well as any comments or suggestions about what you would like to see featured here.

482

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. All material intended for inclusion in this section should be sent to the Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University, PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic address is still [email protected] The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto). The rest of the sta consists of Richard Hoshino (University of Waterloo), Wai Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices A Note on Convexity A function f is convex on I (I an interval) if f ( x + (1 , )y)  f (x) + (1 , )f (y) 8 x; y 2 I; 2 [0; 1]; and J{convex on I (see [1]) if   f x +2 y  f (x) +2 f (y) 8 x; y 2 I: Similarly, f is concave on I if f ( x + (1 , )y)  f (x) + (1 , )f (y) 8 x; y 2 I; 2 [0; 1];

and J{concave is de ned similarly. Several sources claim that J{convexity is sucient for convexity (see [2]), but we intend to make this more precise. First, f convex clearly implies that f is J{convex. We will prove that: (i) If f is convex on an open interval I , then f is continuous on I , and (ii) If f is J{convex and continuous on an interval I , then f is convex on I .

Proof. (i) Let a 2 I ,  > 0. We will show that on some interval around a, f (x) < f (a) +  and f (x) > f (a) , . Choose any b 2 I . Assume b > a. Then the graph of f (x) on [a; b] lies under the chord joining (a; f (a)) and (b;f (b)) (see Figure 1), which in turn lies under the line y = f (a) +  on some interval to the right of a, including a. Applying a similar argument when b < a, we nd an interval around a on which f (x) < f (a) + .

483 y = f (x)

y = f (x) (c;f (c)) q

q

b; f (b))

(

q

y = f (a) + 

a; f (a))

(

a

Figure 1.

b

y = f (a) ,  c

q

Figure 2.

a; f (a))

(

a b

Now, if f (x)  f (a) for all x 2 I , then we are done, so assume f (b) < f (a) for some b 2 I . Assume b > a. Then f (x) > f (a) for all x < a by convexity, which certainly implies f (x) > f (a) , . Take some c < a, and consider the line joining (c; f (c)) and (a; f (a)). Then the graph of f (x) to the right of a lies above this line, which in turn lies above the line y = f (a) ,  on some interval to the right of a, including a. The case b < a is similar. (ii) Since f is J{convex, it is clear that

f ( x + (1 , )y)  x + (1 , )f (y)

for = 0, 1, 12 , 14 , 34 , and by an induction argument, for any dyadic rational between 0 and 1; that is, a fraction of the form 2mn . Since the dyadic rationals are dense in [0; 1], we can nd a sequence which converges to any given real in [0; 1]. By taking a limit along this sequence, the identity is shown to be true for all 2 [0; 1]. Remark. In (i), I must be open, as seen in the example  2, xn + y n = z n has only a nite number of solutions even p if x, y , and z are allowed to be quite general, for example of the form r + s 2 where r and s are integers. This result, stated appropriately, is a consequence of Siegel's Theorem, and relates in an obvious way to Fermat's Last Theorem.)

499 Even the application of the lemma to this problem has some subtleties. As before, it suces to show that C123, C234 , and C341 are concurrent. Let Q be the intersection of C234 and C341 (other than p34) as before, and let P be the intersection of C123 and C234 (other than p23). Let C be the cubic curve that is the union of C123 and L4 , A the cubic that is the union of C234 and L1 , and B the cubic that is the union of C341 and L2 . Then C and A intersect at the six points fp12; : : : ; p34 g, and at P . They also intersect at two points at in nity ( 10 ; 0i ) and ( 10 ; ,0i ) (whatever that means!). The cubics C and B intersect at the same nine points (including the two strange ones!), except the P is replaced by Q. By the lemma, P = Q, and we are done. Of course, a lot of further explanation is required to even make sense of all this, but such explanation is beyond the scope of Mayhem. Comments.

1. This problem is a rst of an in nite sequence of theorems, called the Cli ord Theorems. Call the intersection of two lines in general position their Cli ord point. For three lines in general position, there are three Cli ord points of pairs of lines; call the circle through the three points the Cli ord circle of the three lines. Then, according to this problem, given four lines in general position, the four Cli ord circles (of the four triples of lines) are concurrent; call this point the Cli ord point of the four lines. In general, if n is odd, given n general lines, then the n Cli ord points of all (n , 1)-subsets of the n general lines are concyclic, and the resulting circle is called the Cli ord circle of the n general lines. Similarly, if n is even, given n general lines, then the n Cli ord circles of all (n , 1)subsets of the n general lines are concurrent, and the resulting point is called the Cli ord point of the n general lines. The theorems implicit in these de nitions are the Cli ord theorems. This theorem is discussed in Liang-shin Hahn's book Complex Numbers and Geometry (published by the Mathematical Association of America). This is a wonderful, beautiful book, and possibly the best place to learn about how complex numbers can be used to make Euclidean plane geometry simple. Hahn proves all the Cli ord theorems (in Section 2.3) using a simple lemma, which he proves using complex numbers, but which can also be proved with ordinary Euclidean geometry: Lemma. Suppose there are four circles C1 , C2 , C3 , and C4 in a plane. Let C1 and C2 intersect at z1 and w1 , C2 and C3 intersect at z2 and w2 , C3 and C4 intersect at z3 and w3 , C4 and C1 intersect at z4 and w4 . Then the points z1 , z2 , z3 , z4 are concyclic if and only if w1 , w2 , w3 , w4 are concyclic. If anyone has a slick proof of the Cli ord theorems (or even the next case), we would be interested in seeing it. 2. What needs to be done to make the rst (geometric) proof rigorous? 3. Can you make sense of the weird points at in nity ( 10 ; 0i ) and ( 01 ; ,0i ) described in the sketch of the second solution? In your way of making sense of them, can you see why they lie on every circle? 4. Without understanding the intricacies of the second sketch, can you use the sketch's lemma, with a little hand waving, to prove Hahn's lemma above? Can you use it to prove any other well-known results in Euclidean geometry? (Pappus' theorem seems like a good possibility.) If so, we would love to hear from you!

500

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution, together with references and other insights which are likely to be of help to the editor. When a submission is submitted without a solution, the proposer must include sucient information on why a solution is likely. An asterisk (?) after a number indicates that a problem was submitted without a solution. In particular, original problems are solicited. However, other interesting problems may also be acceptable provided that they are not too well known, and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted without the originator's permission. To facilitate their consideration, please send your proposals and solutions on signed and separate standard 8 12 "11" or A4 sheets of paper. These may be typewritten or neatly hand-written, and should be mailed to the Editor-in-Chief, to arrive no later than 1 June 1998. They may also be sent by email to [email protected]. (It would be appreciated if email proposals and solutions were written in LATEX). Graphics les should be in epic format, or encapsulated postscript. Solutions received after the above date will also be considered if there is sucient time before the date of publication. Where to send your solutions and proposals

There has been an increase in the number of solutions and proposals sent to the Canadian Mathematical Society's Head Oce in Ottawa, Ontario. Please note the instructions above and send them directly to the Editor-inChief.

Solutions submitted by FAX

There has been an increase in the number of solutions sent in by FAX, either to the Editor-in-Chief's departmental FAX machine in St. John's, Newfoundland, or to the Canadian Mathematical Society's FAX machine in Ottawa, Ontario. While we understand the reasons for solvers wishing to use this method, we have found many problems with it. The major one is that hand-written material is frequently transmitted very badly, and at times is almost impossible to read clearly. We have therefore adopted the policy that we will no longer accept submissions sent by FAX. We will, however, continue to accept submissions sent by email or regular mail. We do encourage email. Thank you for your cooperation.

501

2287.

Proposed by Victor Oxman, University of Haifa, Haifa, Israel. Let G denote the point of intersection of the medians, and I denote the point of intersection of the internal angle bisectors of a triangle. Using only an unmarked straightedge, construct H , the point of intersection of the altitudes. 2288. Proposed by Victor Oxman, University of Haifa, Haifa, Israel. In the plane are a circle (without centre) and ve points A, B , C , D, E, on it such that arc AB = arc BC and arc CD = arc DE. Using only an unmarked straightedge, construct the mid-point of arc AE . 2289?. Proposed by Clark Kimberling, Evansville, IN, USA. Use any sequence, fck g, of 0's and 1's to de ne a repetition-resistant sequence s = fsk g inductively as follows: 1. s1 = c1 , s2 = 1 , s1 ; 2. for n  2, let L = maxfi  1 : (sm,i+2 ; : : : ; sm ; sm+1) = (sn,i+2 ; : : : ; sn ; 0) for some m < ng; L0 = maxfi  1 : (sm,i+2 ; : : : ; sm ; sm+1) = (sn,i+2 ; : : : ; sn ; 1) for some m < ng: (so that L is the maximal length of the tail-sequence of (s1; s2 ; : : : ; sn ; 0) that already occurs in (s1 ; s2 ; : : : ; sn ), and similarly for L0 ), and 8 < 0 if L < L0 ; sn+1 = : 1 if L > L00 ; cn if L = L :  For example, if ci = 0 for all i, then

s = (0; 1; 0; 0; 0; 1; 1; 0; 1; 0; 1; 1; 1; 0; 0; 1; 0; 0;  1; 1; 1; 1; 0; 1; 1; 0; 0; 0; 0; 0; 1; 0; 1; 0; : : : ) Prove or disprove that s contains every binary word. 2290. Proposed by Panos E. Tsaoussoglou, Athens, Greece. For x; y;z  0, prove that

(x + y )(y + z)(z + x) 2  xyz (2x + y + z )(2y + z + x)(2z + x + y ):

,



2291.

Proposed by K.R.S. Sastry, Dodballapur, India. Let a, b, c denote the side lengths of a Pythagorean triangle. Suppose that each side length is the sum of two positive integer squares. Prove that 360jabc.

502

2292.

Proposed by K.R.S. Sastry, Dodballapur, India. A convex quadrilateral Q has integer values for its angles, measured in degrees, and the size of one angle is equal to the product of the sizes of the other three. Show that Q is either a parallelogram or an isosceles trapezium. 2293. Proposed by Claus Mazanti Sorensen, student, Aarhus University, Aarhus, Denmark. A sequence, fxn g, of positive integers has the properties: 1. for all n > 1, we have xn,1 < nxn ; 2. for arbitrarily large n, we have x1 x2 : : : xn,1 < nxn ; 3. there are only nitely many n dividing x1 x2 : : : xn,1 . Prove that

1 X

(,1)k is irrational. k=1 xk k!

2294. Proposed by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. For the annual Sino-Japanese \Go" tournament, each country sends a team of seven players, Ci 's and Ji 's, respectively. All players of each country are of di erent ranks (strengths), so that C1 < C2 < : : : < C7

and

J1 < J2 < : : : < J7:

Each match is determined by one game only, with no tie. The winner then takes on the next higher ranked player of the opponent country. The tournament continues until all the seven players of one country are eliminated, and the other country is then declared the winner. (For those who are not familiar with the ancient Chinese \Chess" game of \Go", a better and perhaps more descriptive translation would be \the surrounding chess".) (a) What is the total number of possible sequences of outcomes if each country sends in n players? (b)? What is the answer to the question in part (a) if there are three countries participating with n players each, and the rule of the tournament is modi ed as follows: The rst match is between the weakest players of two countries (determined by lot), and the winner of each match then plays the weakest player of the third country who has not been eliminated (if there are any left). The tournament continues until all the players of two countries are eliminated.

503

2295.

Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. Find three positive integers a, b, c, in arithmetic progression (with positive common di erence), such that a + b, b + c, c + a, are all perfect squares. 2296. Proposed by Vedula N. Murty, Andhra University, Visakhapatnam, India. x > 2x2 for 0 < x < 1. Show that sin2 2

2

1+x

Hence or otherwise, deduce that 
0] so f1 decreases for x 2 [1; 1). Thus the sequence F (a) is strictly decreasing if

f10 (x) =



and only if



2

2 +2 n2 ,log n+1 , (n + 1)2 log nn+1 n a>  2 2 (n + 1) log nn+2 , n ,log n+1 +1 n

(2)

for all integers n  1. Next we show that the right-hand side of (2) is a decreasing function of n. It is more convenient to study the right-hand side as a function of a continuous variable x on [1; 1). Thus we are to show 

2



2

(x + 1)2 log xx+2 , (x + 2)2 log xx+3 +1 +2  2  2 x +3 x +2 (x + 2) log x+2 , (x + 1) log x+1



2

2 x2 ,log x+1 , (x + 1)2 log xx+2 x +1 < :  2 ,  x +1 2 x +2 (x + 1) log x+1 , x log x

We work on a more general case of an inequality, for appropriate functions f : [1; 1) ! R, of the form

(x + 1)f (x + 1) , (x + 2)f (x + 2) < xf (x) , (x + 1)f (x + 1) : f (x + 2) , f (x + 1) f (x + 1) , f (x)

(3)

Here putting f equal to the function f1 in (1) gives the desired inequality. We shall need the following easy lemma. Lemma. If f : [b; 1) ! R is a strictly monotonic, twice di erentiable function with f (x) > 0 for all x in its domain, and further satis es f (x)f 00 (x) < 2 (f 0(x))2 there, then inequality (3) holds. Proof. Indeed, observe that the function g (x) = 1=f (x) satis es 0(x))2 , f (x)f 00 (x) 2 ( f 00 g (x) = 3

(f (x))

> 0;

506 so g is strictly convex. Applying convexity to x, x + 2 and 21 (x + (x + 2)) = x + 1 we see that

1 1  1 + 1 ; < f (x + 1) 2 f (x) f (x + 2)

that is, 



2f (x)f (x + 2) < f (x) + f (x + 2) f (x + 1):

(4)

If f is strictly monotonic, then the quantity

f (x + 1) , f (x) ,f (x + 2) , f (x + 1)

,

is strictly positive. Multiplying (3) by this quantity and simplifying, we get (4), which is therefore equivalent to (3). 2 Returning to the proof, we will show that the function f1 given by (1), which we have already shown is strictly monotonic, satis es the rest of the conditions of the lemma. Clearly f1 (x) > 0 on [1; 1), and we need to show f1f100 < 2(f10 )2. Here 2  f10 (x) = log x +x 1 , x +2 1 log x +x 1

and

2 2 x+1 f100(x) = x(x + 1)2 , x(x + 1)2 log x :

Thus upon cancellation we need to show

1 < (x + 1) log x + 1 , 2 1 , log x + x x 

2

(5)

on [1; 1). This unfortunately is a little tedious, as the sides of (5) are almost equal for large x, so we will delay its proof until the end. Assuming (5) and returning to inequality (2), if we call its right-hand side an , then we have shown that (an ) is strictly decreasing. Hence the condition a > an for all integers n  1 is equivalent to

1:5) , (log 2) a > a1 = 4(log (log 2)2 , 2(log 1:5)2  1:168188898: Conclusion: F (a) is strictly decreasing if and only if a > a1 = 1:16818889 : : : . Similarly, F (a) is strictly increasing if and only if a < an for all integers n  1. But as (an ) decreases and is bounded below (by zero), this is equivalent to a  lim an . We show that lim an = 1. This can be done by 2

2

507 l'H^opital's Rule taking x ! 1 on the continuous analogue of an ; or, arguing asymptotically, we have

1 log n + n



2

and hence



 2  2 log 1 + n1 = n1 , 2n1 2 + 3n1 3 + O(1=n4)   = n12 1 , n1 + 1211n2 + O(1=n3)

=





1 1 , n1 + 1211n2 + O(1=n3) , 1 , n+1 + 12(n11+1)2 + O(1=n3)   an = ,1  1 1 1 n+1 , (n+1)2 + O(1=n3) , n , n2 + O(1=n3) ,1 2 + O(1=n3) = (n,+1) 1 + O(1=n3) ,! 1; (n+1)2

,

as claimed. 1 To complete the proof we must show (5). Set w = x+1 for x  1 so x that w  1=2 and x+1 = 1 , w. Now (5) becomes

 2 1 + log(1 , w) < w1 log(1 , w) + 2 ; and hence we have to show on 0 < w  1=2 that [log(1 , w)]2 + (4w , w2 ) log(1 , w) + 3w2 > 0: (6) Now we need the following estimates of log(1 , w): for 0 < w  1=2

we have

,w , w2 , w3 , w4 , w5 , 1342w < log(1 , w) < ,w , w2 , w3 , w4 , w5 , w6 : 2

3

4

5

6

2

3

4

5

6

(7)

The second inequality is simply a truncation of the Taylor series of log(1,w). The rst simply says

13w6 > w6 + w7 + w8 +    ; 42 6 7 8 this is true enough, for 0 < w  1=2 implies 0 < 1,ww  1, and we have

w6 + w7 + w8 +    < w6 + w7 (1 + w + w2 +    ) 6 7 8 6 7   6 7 6 6 6 w w 1 = 6 + 7 1 , w  w6 + w7 = 1342w :

508 Substituting inequalities (7) into the left-hand side of (6), we get that it is enough to show for 0 < w  1=2 that 2 3 4 5 6 2 w w w w w w+ + + + +



2

3

4

5

6

2 3 4 5 6 +(4w,w2) ,w , w2 , w3 , w4 , w5 , 1342w +3w2 > 0:



The lengthy but routine calculation gives

1 4 1 5 19 6 71 7 12 w + 6 w + 90 w , 210 w + (positive terms) > 0: But this last inequality is certainly true since, for 0 < w  1=2, we have 1 w4 , 71 w7  1 w4 , 71 w4  1 3 = 69w4 > 0: 12 210 12 210 2 1680

[Editor's note: some minor adjustments seemed to be needed to Lambrou's asymptotic proof that lim an = 1, and also to one of his coef cients toward the end of the solution. These have been made in the above writeup.] Also solved by CON AMORE PROBLEM GROUP, The Royal Danish School of Educational Studies, Copenhagen, Denmark; RICHARD I. HESS,   Y,  Ferris State Rancho Palos Verdes, California, USA; and VACLAV KONECN University, Big Rapids, Michigan, USA. Two other readers sent in incorrect or incomplete solutions. A couple of readers mentioned that F (a) is eventually decreasing (for large enough n) whenever a > 1, as can be seen from the above solution. One reader pointed out the similar problem 442 in the College Mathematics Journal, solution in Vol. 23 (1992) pp. 71{72, in which the exponent p is n + a instead of n(n + a).

2180. [1996: 318] Proposed by Juan-Bosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Prove that if a > 0; x > y > z > 0; n  0 (natural), then 1. ax (yz )n(y , z ) + ay (xz )n(z , x) + az (xy )n(x , y )  0; 2. ax cosh x(y , z ) + ay cosh y (z , x) + az cosh z (x , y )  0. Solution by Joe Howard, New Mexico Highlands University, Las Vegas, NM, USA. A characterization for a function f to be convex is that, for x > y > z > 0,

(y , z)f (x) + (z , x)f (y ) + (x , y)f (z )  0:

509 [See, for example, D.S. Mitrinovic, Analytic Inequalities, Springer, Berlin, 1970, p. 16.] Since result 1 is equivalent to x

y

z

(y , z) xan + (z , x) yan + (x , y ) zan  0;

at

it is sucient to show that f (t) = n is a convex function of t for n  0. t The case n = 0 is easy, so suppose that n > 0, and without loss of generality, suppose that a = e. Then

et ,t2 , 2t + n(n + 1) : tn+2 The discriminant of the quadratic equation t2 , 2t + n(n + 1) is ,4(n2 + n , 1) < 0. Hence f 00 (t) > 0, and so f is convex. For result 2, we must show that f (t) = at cosh t is convex. Again, assume without loss of generality, that a = e. Thus f 00 (t) = 2et (cosh t + sinh t) = 2e2t > 0: Thus f is convex, and the inequality follows. f 00(t) =

  University of Sarajevo, SaraAlso solved by SEFKET ARSLANAGIC, jevo, Bosnia and Herzegovina; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta ( rst part only);  MICHAEL LAMBROU, University of Crete, Crete; HEINZ-JURGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. Klamkin read the second part as cosh((x(y , z )), etc., which, on re ection, is a reasonable interpretation, and gives a trivial result. We wonder how many other readers did this?  2181. [1996: 318] Proposed by Sefket Arslanagi c , Berlin, Germany.

Prove that the product of eight consecutive positive integers cannot be the fourth power of any positive integer. Solution by Joe Howard, New Mexico Highlands University, Las Vegas, NM, USA. [Slightly modi ed by the editor.] Without loss of generality, let

P = (n , 3)(n , 2)(n , 1)n(n + 1)(n + 2)(n + 3)(n + 4):

510 Then

P = n8 + 4n7 , 14n6 , 56n5 + 49n4 + 196n3 , 36n2 , 144n:

If P is a 4th power, then it would be of the form

F = ,n2 + an + b4 ; where a, b are constants. But then b = 0 since P has a zero constant term. Thus F = n4 (n + a)4 , and clearly a 6= 0. But this implies that the coecients of n3, n2 and n in F are all zero. However, P has these coecients non-zero. Hence P cannot be of the form F .

It has been pointed out by many readers that this problem has appeared before. Most readers referred to the American Mathematical Monthly, 1936, p. 310 for the solution to #3703 (posed by Victor Thebault in 1934, p. 522). Another reference was made to Honsberger's monograph Mathematical Morsels, where it appears on p. 156 as \A Perfect 4th Power". Several readers also made reference to the general problem of proving that the product of (two or more) consecutive integers is never a square, which was established in 1975 by Erdos } and Selfridge [1]. Because the solutionby Cautis was quite di erent from any of these published solutions, we have decided to publish it here. The interested reader is directed to these other sources for a di erent solution. Comments and/or solutions were submitted also by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADAM BROWN, Scarborough, Ontario; GORAN CONAR, student, Varazdin, Croatia; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon; MURRAY S.   Y,  KLAMKIN, University of Alberta, Edmonton, Alberta; VACLAV KONECN Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Greece; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; HARRY SEDINGER, St. Bon aventure University, St. Bonaventure, NY, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH, Mount Royal College, Calgary, Alberta; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; KENNETH M. WILKE, Topeka, Kansas, USA; PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the proposer. Sei ert remarks that A. Guibert proved the result in 1862. This is stated by L.E. Dickson in his History of the Theory of Numbers, Vol. II, 1952, pp. 679-680.

511 Janous notes the following deep theorem by Erdos } and Selfridge: The product of two or more consecutive positive integers is never a power of a positive integer; that is, the Diophantine equation

(n + 1)(n + 2) : : : (n + k) = xk has no integer solutions with k; l  2 and n  0. 1.

Reference P. Erdos } and J.L. Selfridge, The product of consecutive integers is never a power, Illinois J. Math. 19 (1975), 292{301.

2182. [1996: 318] Proposed by Robert Geretschlager, Bundesrealgymnasium, Graz, Austria. Many CRUX readers are familiar with the card game \Crazy Eights", of which there are many variations. We de ne the game of \Solo Crazy Eights" in the following manner: We are given a standard deck of 52 cards, and are dealt k of these at random, 1  k  52. We then attempt to arrange these k cards according to three rules: 1. Any card can be chosen as the rst card of a sequence; 2. A card can be succeeded by any card of the same suit, or the same number, or by any eight; 3. Anytime in the sequence that an eight appears, any suit can be \called", and the succeeding card must be either of the called suit, or another eight. (This means that, in e ect, any card can follow an eight). The game is won if all dealt cards can be ordered into a sequence according to rules 1{3. If no such sequence is possible, the game is lost. What is the largest value of k for which it is possible to lose the game? Solution by Michael Lambrou, University of Crete, Greece. We show that if 34 or fewer cards are dealt, then it is possible to lose the game, but for any 35 cards or more, the game always ends successfully. Thus k = 34. For example, the 34 or fewer cards may be chosen as follows; (a) ace of hearts, (b) no eights, (c) any subset of the 33 cards consisting of any suit which is not a heart and any rank which is neither an ace nor an eight. In this situation the ace cannot be linked to any other card (as there are no other aces nor hearts nor eights); thus we have a losing hand. Let us now show that for 35 cards or more, the game can end successfully. Note that if eights were present, we could exchange them with any undealt cards, arrange the cards, and re-exchange the eights (if there are insucient undealt cards, we could simply place the eights at the end). Thus

512 it is sucient to show that we can obtain success on 35 or more cards which do not include eights. The algorithm below is based on two simple observations: 1. Any set of ranks each of which appears 3 times among the dealt cards (that is, is represented by 3 suits) can be arranged in a sequence following the rules of the game. This is so because any two such ranks have at least two suits in common so, running through all suits of a xed rank, we can link it to the next rank utilising a common suit which we can go through similarly leaving last a suit that links it with a further rank in the set. (Clearly a common suit between the second and the third rank, which is di erent from the suit that linked the rst to the second rank, always exists). This can be repeated until we exhaust the set. 2. Any rank which appears twice among the dealt cards can be linked to a rank which appears three times, since at least one suit is common to both. Let Ai , (i = 0; 1; 2; 3; 4), denote the set of ranks that appear i times (that is, represented with i suits) among the dealt cards, and let ni = jAi j, the number of elements in Ai . Since we are excluding eights, we have and we also have

n0 + n1 + n2 + n3 + n4 = 12

n1 + 2n2 + 3n3 + 4n4  35: We will show that n1 + n2  n4 + 1 and that if n2 = 0 this can be improved to n1  n4 . We argue by contradiction. Suppose that n4 + 1 < n1 + n2 . Then

35  n1 + 2n2 + 3n3 + 4n4 = n1 + 2n2 + 3(12 , n0 , n1 , n2 , n4) + 4n4 = 36 , 3n0 , 2n1 , n2 + n4  36 , 2n1 , n2 + n4 < 36 , 2n1 , n2 + (n1 + n2 , 1) = 35 , n1; which is impossible since n1  0. If further we have n2 = 0, and we assume that n4 < n1 we would get

35  n1 + 2n2 + 3n3 + 4n4 = n1 + 0 + 3(12 , n0 , n1 , 0 , n4 ) + 4n4 = 36 , 3n0 , 2n1 + n4  36 , 2n1 + n4 < 36 , 2n1 + n1 = 36 , n1;

513 which forces n1 < 1, but then 0  n4 < n1 is impossible. We are now in position to describe an algorithm arranging all 35 cards according to rules 1{3. Suppose rst that n2 = 0. Then n1  n4 and we may write A1 = fb1 ; b2; : : : ; bn1 g and A4 = fc1 ; c2 ; : : : ; cn4 g. We start with a card, say of rank b1 , in A1 . We link progressively this card with each card of A1 having the same suit. After this suit has been exhausted within A1, we link with a card, say of rank c1 , of A4 of the same suit (as the elements of A4 have all four suits dealt, this is always possible). We then run through all four cards of rank c1 , in some order ending with a suit for which there still exist cards in A1 . (This is always possible unless A1 is exhausted.) Then exhaust the cards of A1 with the same suit and jump back to A4 (this is always possible as n1  n4). Repeat this process until A1 is exhausted, then go through the rest of A4 ending with a suit that is the same as some suit for some rank in A3 . Finally jump to A3 and complete the process, which is possible from observation 1 above. On the other hand, if n2 6= 0, and thus n1 + n2  n4 + 1, we modify the algorithm from above as follows. After we exhaust A1 and return to A4 , we then go to A2 , select a rank, exhaust the (two) suits of this rank, and jump back to A4 . Continue back and forth between A4 and A2 , running each rank through all its suits, being sure to order the suits for each rank from A4 so that it matches a suit for a rank still remaining in A2 . If A2 is exhausted rst we complete the rest of A4 and jump to A3 as described for the rst case. If A4 is exhausted rst we go from A2 to A3 which can be done by observation 2 above, as long as we have taken care to end up for the last rank of A2 with a suit which matches the suit for at least one rank from A3 . (This may take a little bit of planning for the ordering of the last rank of A4 as well!) Also solved by the proposer. There were two incomplete solutions. Geretschlager also asks about a generalization to replace 52 = 4  12+4 by n  k + j . He observes that for n = 4 (that is, 4k + j ), an analogous argument to the one presented would yield a maximum number of 3k , 2. He then asks for general conditionssuch that the resulting maximum number is (n , 1)(k , 1)+1, or to nd out what other numbers (if any) could turn up.

2183. [1996: 319] Proposed by Vaclav Konecny, Ferris State University, Big Rapids, Michigan, USA. Suppose that A, B , C are the angles of a triangle and that k, l, m  1. Show that: 0 < sink A sinl B sinm B   S2  k2   2 l m2 k l m 2 , , 2 , 2  k l m S (Sk + P ) (Sl + P ) (Sm + P ) ; where S = k + l + m and P = klm.

514 Editor's and Proposer's comments. This problem has already been posed; see 908 [1984: 19] Proposed by Murray S. Klamkin, University of Alberta, Edmonton, Alberta. Determine the maximum value of

P  sin A  sin B  sin C; where A, B , C are the angles of a triangle and , , are given positive numbers. A solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria, is given in [1985: 93]. He sent in a solution this time pointing out that he had done so before! Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MICHAEL  LAMBROU, University of Crete, Crete, Greece; HEINZ-JURGEN SEIFFERT, Berlin, Germany; and the proposer.

2184. [1996: 319] Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Let n be a positive integer and let an denote the sum bn= X2c

(,1)k

n , k: k



k=0 Prove that the sequence fan : n  0g is periodic.

Composite solution from Michael Lambrou, University of Crete, Crete, Greece and Kee-Wai Lau, Hong Kong. Since

n , k = k



we have

an = = =



n , 1 , k + n , 1 , k; k k,1

bn= X2c

    bn= X2c (,1)k n , k1 , k + (,1)k n ,k ,1 ,1 k k=0 k=1

bn= X2c

(,1)k

k=0 b(n,X 1)=2c

n , 1 , k , b(n,X1)=2c(,1)kn , 2 , k k k k=0



(,1)k

k=0 = an,1 , an,2:

n , 1 , k , b(n,X2)=2c(,1)kn , 2 , k k k k=0



515 It follows that an+3 = an+2 , an+1 = (an+1 , an ) , an+1 = ,an , and so an+6 = ,an+3 = an, showing that the sequence fan : n  0g is periodic with period 6. In fact, the sequence takes consecutively the values 1; 1; 0; ,1; ,1; 0, inde nitely. Also solved by PAUL BRACKEN, CRM, Universite de Montreal, Quebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;   Y,  Ferris D. KIPP JOHNSON, Beaverton, Oregon, USA; VACLAV KONECN  State University, Big Rapids, Michigan, USA; HEINZ-JURGEN SEIFFERT, Berlin, Germany; and the proposer. Bradley and Hess commented that the corresponding sum without the factor (,1)k would yield the sequence of Fibonacci numbers. [Ed.: This can be found in many elementary books on Combinatorics, for example, Exercise 21 on page 87 of Basic Techniques of Combinatorial Theory by Daniel I.A. Cohen.] Flanigan pointed out that this problem appears, with technical di erences in the de nition of an , in the book Concrete Mathematics by Graham, Knuth and Patashuik, (1989), 177{179. Lau pointed out that with the recurrence relation an = an,1 , an,2 and the initial values a0 = a1 = 1, one can prove easily, by induction, that   1 sin  n  ; an = cos n + p 3 3 3

n = 0; 1; 2; : : : :

This fact was also derived by the proposer.

2185. [1996: 319] Proposed by Bill Sands, University of Calgary, Calgary, Alberta. Notice that 22 + 42 + 62 + 82 + 102 = 4  5 + 5  6 + 6  7 + 7  8 + 8  9; that is, the sum of the rst n (in this case 5) even positive squares is equal

to the sum of some n consecutive products of consecutive pairs of positive integers. Find another value of n for which this happens. (NOTE: this problem was suggested by a nal exam that I marked recently.) I. Solution by Tim Cross, King Edward's School, Birmingham, England. We require to nd positive integers n; k for which

22 +42 +    +(2n)2 = k(k +1)+(k +1)(k +2)+   +(k + n , 1)(k + n);

516 which is equivalent to

4

n X

r=1

r2 =

n X

(k + r , 1)(k + r) = k(k , 1)

r=1

n X r=1

1 + (2k , 1)

n X r=1

r+

3  n6 (n + 1)(2n + 1) = k(k , 1)n + (2k , 1) n2 (n + 1); 1 (n + 1)(2n + 1) + 1 (n + 1) = k(k , 1) + k(n + 1); 2 2

and nally

k2 + nk , (n + 1)2 = 0:

n X r=1

r2 ;

(1)

We thus look for positive integer solutions

p n + 4(n + 1) , n + k= 2

2

2

and we require the discriminant  = 5n2 +8n +4 to be a perfect square, say  = 2 for some positive integer . This condition leads to a Pell equation (5n + 4)2 , 5 2 = ,4. Examining the more general form x2 , 5y 2 = ,4, we nd solutions

(x; y ) = (1; 1); (4; 2); (11; 5); (29; 13); : : : : The solution x = 5n + 4 = 29 gives n = 5, the example given.

Pell equations have solution-pairs which satisfy similar second-order recurrence relations. In this case, xk and yk both satisfy uk = 3uk,1 , uk,2 ; k  3; (2) with (x1; y1) = (1; 1) and (x2; y2) = (4; 2). (Notice that the sequence fykg = 1; 2; 5; 13; : : : is that of alternate Fibonacci numbers.) If we take the sequence fxk g = 1; 4; 11; 29; : : : we see that terms are alternately  1 mod 5 and  4 mod 5. Since we need x  4 mod 5, we put vk = x2k and can then derive from (2) the sequence vk = 7vk,1 , vk,2; k  3; with v1 = 4 and v2 = 29: [Editor's note. For example, from (2) we get

x2k = 3x2k,1 , x2k,2 = 3(3x2k,2 , x2k,3 ) , x2k,2 = 7x2k,2 , (3x2k,3 , x2k,2 ) = 7x2k,2 , x2k,4 and the recurrence for the v 's follows.] Then, since nk = (vk , 4)=5, we can deduce the sequence fnk g de ned by nk = 7nk,1 , nk,2 + 4; k  3; with n1 = 0 and n2 = 5:

517 This gives the sequence fnkg = 0 (trivially); 5; 39; 272; 1869; 12815; : : : of suitable values of n. II. Solution by John Oman and Bob Prielipp, University of Wisconsin{ Oshkosh, Wisconsin, USA. [Editor's note: Oman and Prielipp rst derived equation (1), which however they wrote in the form n2 , (k , 2)n , (k2 , 1) = 0: (3) Here and below, their notation has been changed to agree with Solution I.] Considering equation (3) as a quadratic in n, a necessary condition for it to have integer solutions is for the discriminant k(5k , 4) to be a perfect square. Thus 5k , 4 = x2 and k = y 2 for some positive integers x and y . [Editor's note. Since gcd(k; 5k , 4) = 1; 2 or 4, the only alternative is 5k , 4 = 2x2 and k = 2y 2, which implies 5y 2 , 2 = x2 and thus x22  32 mod 5, impossible. Note that we thus get the same equation x = 5y , 4 as in Solution I.] The following sequences for km and nm solve (3) and provide additional solutions to the problem:

km = ym 2

where

and

and

2 nm = km , 2 + 2km(5km , 4) = ym , 2 2+ xmym ;

p

p5 !m p5 !m p p 3 + 3 , xm = (2 + 5) 2 + (2 , 5) 2 p

ym = 5 +52 5

p! p! p! 3+ 5 m+ 5,2 5 3, 5 m: 2 5 2

!

The rst few values generated by these formulas are

m x y 0 1 2 3

4 11 29 76

k

n

2 4 5 5 25 39 13 169 272 34 1156 1869

[Editor's note. Oman and Prielipp then noted that xm = L2m+3 and ym = F2m+3;

518 the (2m + 3)rd Lucas and Fibonacci numbers, respectively (where, as usual, F1 = F2 = 1 and L1 = 1, L2 = 3, both sequences then generated by the familiar Fibonacci recurrence). This follows because

p

p

3 5 = 1 5 2 2 and so and

!2

and

p5 ! p 1  2 5 = ; 2 3

p5 !2m+3 1 , p5 !2m+3 1 + xm = + = L2m+3 2 2 p5 !2m+3 1 1 , p5 !2m+3 1 + 1 , p5 2 = F2m+3: ym = p5 2

Thus km = F22m+3 and (after some manipulations) nm = F2m+3 F2m+4 , 1.] Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADAM BROWN, Scarborough, Ontario; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; KEITH EKBLAW, Walla Walla, Washington, USA; JEFFREY K. FLOYD, Newnan, Georgia, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns bruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA; VACLAV   KONECNY, Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; J.A. MCCALLUM, Medicine Hat, Alberta; ROBERT P. SEALY, Mount Allison  University, Sackville, New Brunswick; HEINZ-JURGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. One incorrect solution was sent in. Besides Oman and Prielipp, and (to a smaller extent) Cross, no other readers seem to have noticed the presence of the Fibonacci numbers in the solution to this problem. One more occurrence, which Oman and Prielipp don't mention, is that the largest number k + n on the right-hand side of the given equation is F22m+4 , which ts nicely with the fact that the smallest number on the right-hand side is k = F22m+3 . In fact, only Brown commented on the fact that these numbers are squares! So for example, the equation arising from the next-smallest solution n = 39 is

22 + 42 +    + 782 = 25  26 + 26  27 +    + 63  64;

519 where 25 = F52 and 64 = F62 (and 78 = 2F5F6 , 2, as follows from Solution II). In general, the required equation reads

22 + 42 +    + (2F2t,1F2t , 2)2 = F22t,1 (F22t,1 + 1) + (F22t,1 + 1)(F22t,1 + 2) +    + (F22t , 1)F22t for any integer t  2.

2186. [1996: 319] Proposed by Vedula N. Murty, Andhra University, Visakhapatnam, India. Let a, b, c respectively denote the lengths of the sides BC , CA, AB of triangle ABC . Let G denote the centroid, let I denote the incentre, let R denote the circumradius, r denote the inradius, and let s denote the semiperimeter. Prove that GI 2 = 9(a +1b + c) (a , b)(a , c)(b + c , a) 



+ (b , c)(b , a)(c + a , b) + (c , a)(c , b)(a + b , c) : Deduce the (known) result

GI 2 = 19 s2 + 5r2 , 16Rr : 



Solution by Kee-Wai Lau, Hong Kong. Let A be the origin, AB = u and AC = v. It is well known that

AG = 31 (u + v) and AI = a + bb + c u + a + cb + c v: Hence

   b 1 c 1 GI = a + b + c , 3 u + a + b + c , 3 v: 

520 Since u  u = c2 ; v  v = b2 and u  v = bc cos A = 12 (b2 + c2 , a2 ), so

GI = 2

= = =

b , 1 2 c2 +  c , 1 2 b2 a+ b + c 3   a + b + c  3 + a + bb + c , 13 a + cb + c , 31 (b2 + c2 , a2 )  1 2 2 2 2 9(a + b + c)2 (a + c , 2b) c + (a + b ,2c) b + (a + c , 2b)(a + b , 2c)(b2 + c2 , a2)  1 3 3 3 2 2 9(a + b + c) , a , b , c + 2a b + 2ab + 2b2c + 2bc2 + 2c2a + 2ca2 , 9abc  1 9(a + b + c) (a , b)(a , c)(b + c , a)  + (b , c)(b , a)(c + a , b) + (c , a)(c , b)(a + b + c) 

as required. Since

s

s = a + 2b + c ; r = (s , a)(s ,s b)(s , c) and R = abc 4rs ; we have 1 ,s2 + 5r2 , 16Rr 9  2 )(a + c , b)(a + b , c) , 8abc  = 19 (a + b4 + c) + 5(b + c , a4( a + b + c) a+b+c  3 + 5(b + c , a)(a + c , b)(a + b , c) , 32abc  1 ( a + b + c ) = 9(a + b + c) 4  1 3 3 3 2 = 9(a + b + c) , a , b , c + 2a b + 2ab2 

+ 2b2 c + 2bc2 + 2c2 a + 2ca2 , 9abc

= GI 2 ; as required. Bellot Rosado notes that a variation of this problem was proposed by Cezar Cosnita , solved by T.C. Esty, with the second form of GI given by D.L. MacKay [Problem E415, American Mathematical Monthly (1940), solution p. 712]. Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,

521 Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece; BOB PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.

2187. [1996: 320] Proposed by Syd Bulman-Fleming and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. It is easy to show that the maximum number of bishops that can be placed on an 8  8 chessboard, so that no two of them attack each other, is 14. (a) Prove or disprove that in any con guration of 14 non-attacking bishops, all the bishops must be on the boundary of the board. (b) Describe all of the con gurations with 14 non-attacking bishops. Solution by Kee-Wai Lau, Hong Kong. (a) We prove the result. Clearly we need consider only the black bishops. The diagram below shows the black squares of a chessboard rearranged so that the bishops will now move vertically or horizontally. [The diagram is labelled the same as the standard labelling of the squares of a chessboard, with the rows (ranks) numbered 1 to 8 and the columns ( les) labelled a to h. | Ed.]

d8 b8 c7 a7 b6 a5

f8 e7 d6 c5 b4 a3

h8 g7 f6 e5 d4 c3 b2 a1

h6 g5 f4 e3 d2 c1

h4 g3 h2 f 2 g1 e1

In any column and any row there is at most one bishop. [Thus there are at most 7 black bishops and similarly at most 7 white bishops, for the given maximum of 14. Moreover, to attain this maximum, there must be a black bishop in each column in the above diagram. | Ed.] Denote the bishop in the kth column by Bk . If B1 is on a7 then B7 must be on h2, and if B1 is on b8 then B7 must be on g 1. Next, B2 must be on a5 or d8 and correspondingly B6 must be on h4 or e1. Next B3 must be on a3 or f 8 and correspondingly B5 must be on h6 or c1. Finally B4 must be on a1 or h8. This shows that the non-attacking bishops must be on the boundary of the board. (b) From part (a) we see that there are 24 = 16 ways to locate the black bishops. Similarly there are 16 ways to locate the white bishops. Thus there are 256 con gurations with 14 non-attacking bishops.

522 Also solved (usually the same way) by SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; and the proposers. Part (a) was also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria.

2188. [1996: 320] Proposed by Victor Oxman, University of Haifa, Haifa, Israel. Suppose that a, b, c are the sides of a triangle with semi-perimeter s and area 4. Prove that 1+1+1< s: a b c 4

USA.

I. Solution Paul Yiu, Florida Atlantic University, Boca Raton, Florida,

Let C be the angle opposite to side c of the triangle. Since 1 1 b : Here, equality holds if and only if 1 2 2 24 the angle opposite to c is a right angle. Similarly, 1b  2c4 , and 1c  2a4 . Since equality cannot hold simultaneously, we have

4 = ab sin C  ab, we have a 

1 + 1 + 1 < b+c+a = s : a b c 24 4

II. Solutionby Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. I will prove the stronger inequality:

p

1 1 1 3 s a + b + c  2  4: Let a = x + y; b = y + z; c = z + x, where x; y;z > 0. Then  1 + 1 + 1 2 =  1 + 1 + 1 2 a b c x+y y+z z+x   2  2p1xy + 2p1yz + 2p1zx  2 1 1 1 1 = 4 pxy + pyz + pzx   3 1 1 1  4 xy + yz + zx (by Cauchy-Schwarz) q

q

y z= p xx yy zz xyz = x xyz xy + yz + zx . p s Thus a + b + c   4 : The equality holds only when a = b = c.

Now 4s =

+ + ( + + )

1

1

1

3 2

+ +

1

1

1

523 Also solved by HAYO AHLBURG, Benidorm, Spain; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; CLAUDIO ARCONCHER,   University of Sarajevo, Sarajevo, Jundia, Brazil; SEFKET ARSLANAGIC, Bosnia and Herzegovina; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADAM BROWN, Scarborough, Ontario; SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; GORAN CONAR, student, Gymnasium Varazdin, Varazdin, Croatia; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta;   Y,  Ferris State University, Big Rapids, Michigan, USA; VACLAV KONECN STEFAN and ALEXANDER LAMBROU, students, Crete, Greece; MICHAEL LAMBROU, University of Crete, Crete, Greece; CAN AN+H MINH, University of California, Berkeley, California; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria (2 solutions); BOB PRIELIPP, University of Wisconsin{  Oshkosh, Wisconsin, USA; JUAN-BOSCO ROMERO MARQUEZ, Universidad  de Valladolid, Valladolid, Spain; HEINZ-JURGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; GEORGE  UBIS MATII NEZ, student, I.B. TSAPAKIDIS, Agrinio, Greece; ADRIAN Sagasta, Logro~no, Spain; MELETIS VASILIOU, Elefsis, Greece; and the proposer.

2189. [1996: 361] Proposed by Toshio Seimiya, Kawasaki, Japan.

The incircle of a triangle ABC touches BC at D. Let P and Q be variable points on sides AB and AC respectively such that PQ is tangent to the incircle. Prove that the area of triangle DPQ is a constant multiple of BP  CQ. Solution by Florian Herzig, student, Perchtoldsdorf, Austria. Let E; F; T be the points of contact of the incircle with CA; AB and PQ; de ne x := AP , y := AQ as well as z := PQ. Then

x + y + z = x + y + PT + QT = (x + PF ) + (y + QE ) = AF + AE = 2(s , a)

(1)

where s is the semi-perimeter of 4ABC . Applying the cosine rule to 4APQ yields 2 2 2 z2 = x2 + y2 , 2xy cos A = x2 + y2 , 2xy  b +2cbc, a :

524 From equation (1) we get

2 2 2 4(s , a)2 , 4(s , a)(x + y )+ x2 +2xy + y2 = x2 + y2 , xy  b + cbc , a

which is equivalent to

 2 2 2 xy  b + 2bcbc+ c , a + 4(s , a)2 = 4(s , a)(x + y) xy ,(b + c)2 , a2  + 4bc(s , a)2 = 4bc(s , a)(x + y) 4xys(s , a) + 4bc(s , a)2 = 4bc(s , a)(x + y ) xys + bc(s , a) = bc(x + y): (2) Now let R be the circumradius of 4ABC and use the sine law [in the second

equality] and (2) [in the third] to get

 sin B [DPQ] = [ABC ] , AP  AQ2  sin A , BP  BD 2 CD  CQ  sin C , 2   1 = 4R abc , xya , (c , x)(s , b)b , (b , y )(s , c)c h = 41R abc , axy , bc(s , b) , bc(s , c) + bx(s , b)  i + cy (s , c) + xys + bc(s , a) , bc(x + y )  = 41R abc , axy , abc + bx(s , b , c) + cy(s , b , c)  + xys + bc(s , a)   = 41R xy (s , a) , bx(s , a) , cy(s , a) + bc(s , a) = (s4,Ra) (c , x)(b , y) = (s4,Ra)  BP  CQ:

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym  Y,  Ferris State University, Big nasium, Innsbruck, Austria; VACLAV KONECN Rapids, Michigan, USA; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer. Several solvers mentioned that this problem generalizes the proposer's earlier 1862 [1993: 203], [1994: 172-173].

525

2190. [1996: 361] Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria. Determine the range of sin2 A + sin2 B + sin2 C

A

B

C

where A, B , C are the angles of a triangle. Solution by Kee-Wai Lau, Hong Kong. Denote the function of the problem by f (A;B; C ). We show that

0 < f (A;B; C )  427 :

The rst inequality follows from the de nitions. By considering the degenerate triangle A =  , B = 0, C = 0, we see that it is also sharp. We now prove the second inequality. For 0 < x  2 , let g (x) = sinx2 x . We have

dg = sin x(2x cos x , sin x) and dx x2 2 d g = h(x) ; where h(x) = 1 , cos(2x) + 2x2 cos(2x) , 2x sin(2x) : dx2 x3 Since dh = ,4x2 sin(2x)  0 and h(0) = 0, we have h(x)  0. It follows dx 2 d g2  0, so that g is concave. Hence, if 4ABC is acute angled, then that dx     A + B + C  = 27 : f (A;B; C )  3g = 3 g 3 3 4 Equality holds if and only if A = B = C = 3 . Now, suppose that one of the angles of 4ABC is obtuse. We may assume that \A > 2 , \B + \C < 2 . 2 , B +C  2 2 sin 2 sin B sin C 2 2 Hence f (A;B; C ) < +  B + C 2   + 2 B+2 C [by the concavity of g (x) together with sinA A < A1 < 2 for A > 2 ]. It is easy to show that tan x  2x for 0  x  4 . Hence dgdx(x)  0 for 0  x  4 . It follows that 2 , 2 sin f (A;B; C ) <  + 2  4 = 6 < 427 : 4

Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; and the proposer.

526 Three unsatisfactory solutions were received. Two of these depended on an inequality that was passed o as \known". Such claims must be accompanied by a reference. Otherwise, how are the editors supposed to know that the claim is true? More importantly, the purpose of the solution should be to explain why the result is true; consequently any reference should be accessible to CRUX with MAYHEM readers. A key step in the third rejected submission was claimed to be \obvious". The only thing obvious to this editor was that such claims do not belong in mathematical arguments.  2191. [1996: 361] Proposed by Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina. Find all positive integers n, that satisfy the inequality

1 < sin    < p1 : 3 n 2 2

Editor's composite solution based on the ones submitted by the solvers whose names appear below. Since sin  = 0, n = 1 is clearly not a solution. ,  Since sin n is decreasing for n  2 and since

 q p  1 sin 8 = 2 2 , 2 > p1 2 2 , we have sin n > 2p1 2 for 2  n  8.  the other hand, since sin x < x for x > 0, we have, for all n  10, , On sin n < n  10 < 13 . 

We now show that

1 < sin    < p1 3 9 2 2

(1)

and so n = 9 is the only solution. [Ed: Though (1) can be veri ed numerically by using a calculator, as many solvers did, the proposer's original intention was an \analytic" proof like the one presented below.] Let  = 9 and let r = sin . Then from

p3

  3 3 2 = sin 3 = sin(3) = 3 sin  , 4 sin  = 3r , 4r ; p we see that r is a positive root of the polynomial f (x) = 4x3 , 3x + 23 . 

527 Note that

p3

f (,1) = ,1 + 2 < 0; p3   23 1 f 3 = , 27 + 2 > 0; and

p3

f (0) = 2 > 0; p3   1 , 5 f 2p2 = 4p2 + 2 < 0;

p3

f (1) = 1 + 2 > 0:

Since f is a continuous function, we conclude that it has three real   roots, one in each of the three intervals: (,1; 0), 13 ; 2p1 2 and 2p1 2 ; 1 . ,  But sin 9 < 9  0:349 < 2p1 2 , and so (1) follows. Solved by GERALD ALLEN, CHARLES DIMINNIE, TREY SMITH and ROGER ZARNOWSKI (jointly) Angelo State University, San Angelo, TX, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; F.J. FLANIGAN, San Jose State University, San Jose, California, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA;   Y,  Ferris State University, Big Rapids, Michigan, USA; VACLAV KONECN  MICHAEL LAMBROU, University of Crete, Crete, Greece; HEINZ-JURGEN SEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands;  and JAVIER GUTIERREZ,  DIEGO SOTES students, University of Rioja, Logro~no, Spain; and DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA. There were also nine incomplete or (partially) incorrect solutions. Though these solutions all give the correct answer n = 9, they contain various errors. Most of these errors pertain to the analysis of the three roots of f (x) which is actually more subtle than it appears to be. Some solvers erroneously claimed that one of the roots is greater than one. Using MAPLE, we nd easily that the three roots are: ,0:9848077, 0:3420201, and 0:6427876 , 19 . Only, one ( joint ) solver stated that the three roots are, in fact, cos 18 ,  ,  sin 9 , and sin, 79 . However, they made the mysterious and wrong statement that sin 79 < 13 .   ,  Some other solvers, after checking that f 13 and f 2p1 2 have oppo,  site signs, jump to the conclusionthat sin 9 must be the root in the interval   1 p1 3 ; 2 2 . Clearly, to make this argument valid, one has to show that the other positive root is not in this interval.

528

2192. [1996: 362] Proposed by Theodore Chronis, student, Aristotle University of Thessaloniki, Greece. Let fan g be a sequence de ned as follows:   an+1 + an,1 = aa2 an; 1 a a Show that if 2  2, then n  n. a1 a1

n  1:

Solution by Can Anh Minh, University of California, Berkeley. By the triangle inequality we have

a2 a 2 jan+1j + jan,1 j  jan+1 + an,1 j = a1 an = a1 jan j  2janj Thus we have jan+1 j , jan j  jan j , jan,1 j, and therefore janj , jan,1j  jan,1 j , jan,2j      ja2j , ja1j  ja1j

Thus we have

jan j , ja j = 1

n X k=2

(jak j , jak,1 j)  (n , 1)ja1 j

It follows that jan j  nja1 j, or equivalently

an  n: a1

Also solved by GERALD ALLEN, CHARLES DIMINNIE, TREY SMITH, and ROGER ZARNOWSKI (jointly), Angelo State University, San Angelo, Texas; CHETAN T. BALWE, Pune, India; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursul  Y,  Ferris State Uniinengymnasium, Innsbruck, Austria; VACLAV KONECN versity, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; DIGBY SMITH, Mount Royal College, Calgary, Alberta; and the proposer.

529

2193. [1996: 362] Proposed by Luis V. Dieulefait, IMPA, Rio de Janeiro, Brazil. (a) Prove that every positive integer is the di erence of two relatively prime composite positive integers. (b) Prove that there exists a positive integer n0 such that every positive integer greater than n0 is the sum of two relatively prime composite positive integers. Solution to (a) by Michael Lambrou, University of Crete, Crete, Greece, modi ed slightly by the editor. Note that for all natural numbers k, we have: 2k + 1 = (k + 1)2 , k2 (1) and

2k = (2k + 1)(8k + 1) , (4k + 1)2 (2) Since gcd(k; k + 1) = 1, (1) gives a required representation for all odd integers greater than 3. But 1 = 9 , 8 and 3 = 25 , 22, and so a required representation exists for all odd natural numbers. On the other hand, straightforward computations show that

,(8k + 4)(2k + 1)(8k + 1) + (8k + 5)(4k + 1) = 1 2

and so ,



gcd (2k + 1)(8k + 1); (4k + 1)2 = 1: Thus (2) gives a required representation for all even natural numbers. Solution to (b) by the proposer, modi ed by the editor. Let (n) denote Euler's totient function and let  (n) denote the primecounting function. [Ed: that is,  (n) is the number of primes p such that p  n.] It is known (see Hardy and Wright, An Introduction to the Theory of Numbers) that

liminf n!1

(n) n

log log n

= e, > 12 ;

where is Euler's constant. Hence, for suciently large n, we have

(n) > 2 log2log n :

(1)

(n) < 65 logn n

(2)

On the other hand, by Tschebyche 's theorem, we have

530 if n is large enough. From (1) and (2), we obtain

(n) > lim 5 log n = 1; n!1 12 log log n which implies, for large enough n, that (n)  2 ,(n) + 1 : (3) Note that, if n = a + b, then (a; n) = 1 is equivalent to (b;n) = 1, and to (a; b) = 1. Consider the (n) ordered decompositions of n : n = k + (n , k), where 1  k  n, such that gcd(k;n) = 1. If we strike out those pairs in which the rst or second summand is a prime or 1, then we are deleting at most 2( (n) + 1) pairs, and so from (3), we conclude that there is at least one decomposition n = a + b where a and b are relatively prime composite nlim !1  (n)

natural numbers, and our proof is complete. Solved (both parts) by RICHARD I. HESS, Rancho Palos Verdes, California, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; and the proposer. Part (a) only was solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria; and WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria. Regarding (a), Janous actually obtained a stronger result by showing that there are, in fact, in nitely many required representations. Regarding (b), Hess remarked that a computer search seems to indicate that n0  210. Using arguments similar to those given in the proof above, Lambrou proved a stronger result, namely: For any given integer k  1, there exists a positive integer mk such that every integer greater than mk can be written as the sum of two relatively prime composite natural numbers in at least mk di erent ways. It is not dicult to see that this result also follows easily from the proposer's proof presented above.

2194. [1996: 362] Proposed by Christopher J. Bradley, Clifton College, Bristol, UK. Prove or disprove that it is possible to nd a triangle ABC and a transversal NML with N lying between A and B , M lying between A and C , and L lying on BC produced, such that BC , CA, AB, NB, MC , NM , ML, and CL are all of integer length, and NMCB is a cyclic inscriptable quadrilateral. Editor's comment. There is a terminology problem here. It appears (from his solution) that the proposer had intended the word inscriptable to mean that a circle can be inscribed in the quadrangle NMCB . Although several of the solvers

531 assumed this to be the case, all reference books that I consulted disagree. The Oxford English Dictionary and Nathan Altshiller Court (in his 1925 College Geometry) both say that an inscriptible quadrangle (note the spelling) is one that can be inscribed in a circle; today one more commonly calls such quadrangles (that can be inscribed in circles) cyclic, concyclic, or inscribable | take your choice. On the other hand, one must keep in mind that English (the language in which, for example, in ammable means ammable and unravel means ravel) is so unpredictable that the proposer might well have references to back up his terminology. For a quadrangle that contains an inscribed circle, one would say circumscribing or avoid the issue and just say \a quadrangle with an inscribed circle". The matter really becomes muddy here: Court calls the latter quadrangle circumscriptible, while the OED agrees in one place (under inscriptible), but contradicts itself in another, where a quadrangle is said to be circumscribable if it \may be circumscribed by a circle". The moral of this story: be circumspect (or maybe, circumspectable). As it turns out, our featured solutions all produce cyclic quadrangles having inscribed circles!

I. Solution by Michael Lambrou, University of Crete, Crete, Greece (somewhat edited). We show that any triangle ABC with rational sides and \A 6= \C has a transversal LMN such that NMCB satis es both circle conditions, and has all relevant lengths rational. Hence, multiplying by an appropriate number, we can obtain a similar triangle with the required properties. Let a triangle ABC with rational sides be given. By renaming, we may assume that B , which may be acute, right or obtuse, is larger than C . Note that B cannot equal C , as the transversal | in order to produce a cyclic quadrangle | would then be parallel to BC . Consider as transversal LMN such that \AMN = B (making NMCB cyclic), and LM is tangent to the incircle of triangle ABC

(4) (5)

(forcing NMCB to have an inscribed circle). We must show that the lengths NB, MC , NM , ML and CL are all rational. We have that the triangles ANM and ACB are similar. Let  = AM=AB be the similarity ratio. We use the fact that the sides of triangle ABC are rational to show that  2 Q.

(BC + AB + AC ) = MN + AM + AN = MN + (AC , MC ) + (AB , NB ) = [(MN + BC ) , (NB + MC )] + AB + AC , BC = 0 + AB + AC , BC 2 Q : But BC + AB + AC 2 Q, so that  2 Q as well.

532 Next, by Menelaus'sTheorem applied to triangle ABC with transversal

LMN , it follows that

BL  CM  AN = 1 2 Q ; CL AM BN so that BL=CL 2 Q. Since BC=CL = (BL , CL)=CL, we have LC 2 Q and BL 2 Q. Similarly, by Menelaus's Theorem applied to triangle MLC with transversal AB , we see that LN 2 Q, and the desired result follows.

II. Families of solutionsby Michael Lambrou, University of Crete, Crete, Greece and (independently) by Richard I. Hess, Rancho Palos Verdes, California, USA. Let AMN and LBN be congruent right triangles with right angles at M and B. Denote the legs by p ( = AMp = LB), q ( = MN = BN ), and the hypotenuse by r ( = NL = NA = p2 + q 2). The solution is achieved by choosing p, q , r so that p divides r + q . Hess provided the examples in column 2, and Lambrou, column 3. Hess Lambrou p = AM = LB 2k + 1 2, q = MN = BN 2k(k + 1) , r = NL = NA (k + 1)2 + k2  , where 2 + 2 =  2, 2 r + q = AB = LM (2k + 1) ( +  ), q(q + r)=p = MC 2k(k + 1)(2k + 1) ( +  ),

= BC r(q + r)=p = AC (2k +, 1) ( +  )  = LC (k + 1)2 + k2 = 2 + 2 +  . In each case, NMCB has a circumscribed circle because of the right angles at B and M , while it has an inscribed circle because of its symmetry (so that NM + CB = NB + CM ).

Lambrou also sent in a family of asymmetric solutions for positive integers m > n:

AB AC BC AN

= = = =

nm2(m , n)(n2 + 1); n2m(m , n)(m2 + 1); mn(m , n)(m + n)(mn , 1); and 1 1 mn AC; AM = mn AB:

We leave it to the reader to verify the details. (Remember to check the various triangle inequalities.) Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.

533

2195. [1996: 362] Proposed by Bill Sands, University of Calgary, Calgary, Alberta. A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw. What is the probability that the three-element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any, is repeated? For example, the sequences 453 and 383 are acceptable, while the sequences 327 and 388 are not. (NOTE: this problem was suggested by a nal exam that I marked recently.) I. Solution by David Hankin, Hunter College Campus Schools, New York, NY, USA. Let uk be the number of acceptable sequences chosen from balls numbered 1 to 2k, and let ak be the number of these acceptable sequences that contain 1. We rst show that, for k  1, uk , ak = uk,1 ; where u0 = 0. Note that uk , ak is the number of acceptable sequences that do not contain 1. The number of these sequences is the same as the number of acceptable sequences that can be chosen from balls numbered 3 to 2k. Clearly, this is equal to uk,1 . To nd ak , note that sequences that contain 1 must have one, two or three 1's. There is one sequence with three 1's. For sequences with two 1's, there are 2k , 1 ways to choose the third element and 3 arrangements of the three elements; so there are 3(2k , 1) such sequences. Similarly, there are 6,2k2,1 = 6(2k , 1)(k , 1) sequences with one 1. Thus

ak = 1+3(2k , 1)+6(2k , 1)(k , 1) = 12k2 , 12k +4 = 4[k3 , (k , 1)3]: Now (since u0 = 0) un =

n X

(uk , uk,1 ) =

k=1

n X

k=1

ak = 4

n X

[k3 , (k , 1)3 ] = 4n3:

k=1

Thus the requested probability is

4n3 = 1 : (2n)3 2

II. Solution by Michael Lambrou, University of Crete, Crete, Greece. We show that there is a one-to-one onto pairing that pairs each favourable triplet abc (which for convenience we will denote (a; b; c)) with an unfavourable one. Once this is done, the independence of events shows that

534 the sought probability is 1=2, as the favourable events exactly match the unfavourable ones. We denote the smallest element among (a; b; c) by 2s , 1, where 1  s  n. The pairing will be described for the case when 2s , 1 occurs in the rst position (and is perhaps repeated in one or both of the other positions). The other two possibilities are dealt with in a similar fashion, by cyclic change of order. Note that some care must be taken not to double count the triplets that belong to more than one situation. Here is how we do our pairing.  For xed s and p with 2s , 1 < p  2n , 1, map, in any one-to-one onto fashion, the 2n , 2s triplets of the form (2s , 1; p; q) where 2s , 1 < q  2n and q 6= p to the 2n , 2s triplets of the form (2s; p + 1; v ) where 2s < v  2n; for 2s , 1 < q  2n , 1, map (2s , 1; 2s , 1; q ) to (2s; 2s; q + 1);  for 2s , 1 < q  2n , 1, map (2s , 1; 2n;q ) to (2s , 1; q;q );  map (2s , 1; 2s , 1; 2n) to (2n; 2n; 2s , 1); and nally  map (2s , 1; 2s , 1; 2s , 1) to (2s; 2s; 2s). It is easy to see that this map is not ambiguous. It is also easy to check that all favourable and all unfavourable triplets have been taken into account once and once only. This completes the argument. Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,  UK; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA; W. MOSER, McGill University, Montreal, Quebec; ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick; D. N. SHETH, Sir Parashuram College, Pune, India; DIGBY SMITH, Mount Royal College, Calgary, Alberta; and the proposer. Two incorrect solutions were sent in. Lambrou was the only solver to nd a \combinatorial" proof that the probability is exactly 1=2. He also sent in two other solutions, one of which is similar to solution I. Bradley's solution is also similar to solution I. 

535

2196. [1996: 362] Proposed by Juan-Bosco Romero Marquez, Universidad de Valladolid, Valladolid, Spain. Find all solutions of the diophantine equation 2(x + y ) + xy = x2 + y 2;

with x > 0, y > 0. Solution by Sam Baethge, Nordheim, Texas, USA. Let y = rx with r rational. The given equation becomes a quadratic in r:

r2x2 , r(x2 + 2x) + (x2 , 2x) = 0 p Then r = x + 2  ,3x2 + 12x + 4 =2x. The discriminant can be written as 16 , 3(x , 2)2 and must be the square of an integer. Since x is a positive integer the only solutions are (x; r) = (4; 1), (4; 21 ), or (2; 2). This produces (x; y ) = (4; 4), (4; 2), or (2; 4). 

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; MANSUR BOASE, student, St. Paul's School, London, England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, Ontario; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; GORAN CONAR, student, Varazdin, Croatia; PAUL-OLIVIER DEHAYE, Brussels, Belgium; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; C. FESTRAETS-HAMOIR, Brussels, Belgium; F.J. FLANIGAN, San Jose  State University, San Jose, California, USA; ROBERT GERETSCHLAGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; DAVID HANKIN, Hunter College Campus Schools, New York, NY, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; IGNOTUS, Villeta, Colombia; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA; KEE-WAI LAU, Hong Kong; J.A. MCCALLUM, Medicine Hat, Alberta; CAN ANH MINH, University of California, Berkeley; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, Newfoundland; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; BOB PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; ROBERT P. SEALY, Mount Allison University, Sackville, New  Brunswick; HEINZ-JURGEN SEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH, Mount Royal College, Calgary, Alberta; DAVID R. STONE, Georgia Southern University, Statesboro, Geor UBIS MATII NEZ, Logro~no, Spain; DAVID C. VELLA, Skidgia, USA; ADRIAN more College, Saratoga Springs, New York; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There were seven incorrect solutionsand one incomplete solution. Vella (in one of his two submitted solutions) used a change of variables to turn the given equation into the equation of an ellipse with centre on the

536

x{axis and major axis aligned with the x{axis, and then used a geometric

approach to solve the problem. The proposer also suggests a generalization: Solve the diophantine equation:

z(x + y) + xy = x2 + y2

2197. [1996: 363] Proposed by Joaqun Gomez  Rey, IES Luis Bu~nuel, Alcorcon, Madrid, Spain. Let n be a positive integer. Evaluate the sum: 1 X

,2k

k

: 2k+1 k=n (k + 1)2 Solution by David Doster, Choate Rosemary Hall, Wallingford, Connecticut, USA. 1 2n. This follows directly from The value of the sum is n

4

1 X

,2k

nX ,1

,2k

n

k =1 2k+1 ( k + 1)2 k=0

and

(1)

  1 = 1 , 4n 2nn : 2k+1 ( k + 1)2 k=0

k

(2)

To prove (1), note that it is easy to show that ,2k



1 2



k+1 (k + 1)22k+1 = , k + 1 (,1)  1  3  5    (2k , 1) . Therefore they both simplify to 2k+1(k + 1)! k

1 X

,2k

k

k=0 (k + 1)2

k

2 +1

1 X



1



1 X

1

(,1)k+1 k +2 1 = , (,1)k k2 k=1 k=0 !   1 1 X p = , ,1 + (,1)k k2 = 1 , 1 + (,1) = 1: = ,

k=0

537 To prove (2), we use induction. After checking at n = 1, we use the inductive assumption to get n X

,2k

2n 1 2n + n = 1 , 2k+1 n n 2n+1 ( k + 1)2 4 ( n + 1)2 k=0      1 = 1 , 41n 2nn 1 , 2(n 1+ 1) = 1 , 41n 2nn 2(2nn + + 1)    n + 2) = 1 , 1 2n + 2 ; = 1 , 4n1+1 2nn (2n +(n1)(2 + 1)2 4n+1 n + 1

k

,



which proves the identity. Also solved by PAUL BRACKEN, Universite de Montreal, Montreal, Quebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;  HEINZ-JURGEN SEIFFERT, Berlin, Germany; STAN WAGON, Macalester College, St. Paul, Minnesota, USA; and the proposer. Janous ndsn the ,sum oby considering the generating function of the se2k 1 quence fCk g = k+1 k of Catalan numbers and in doing so, obtains, as a by-product, the recurrence relation nX ,1 Cn = n +1 1 4n , 2 Ck  4n,1,k : k=0 !

He believes that this is a new identity, and wonders whether there is a purely combinatorial proof and/or interpretation of it. Wagon obtained the sum by using MATHEMATICA, and remarked that more sophisticated symbolic algebra software ( work of Petkovsek, Wilf and Zeilberger) can provide certi cates that the closed-form formula obtained is actually valid. He also commented that: \Thus sums such as this are really best left to computers, : : : ". This editor is interested in knowing how many of our readers would agree with him.

Crux Mathematicorum

Founding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell Editors emeriti / Redacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem

Founding Editors / Redacteurs-fondateurs: Patrick Surry & Ravi Vakil Editors emeriti / Redacteurs-emeriti: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

538

YEAR END FINALE

Again, a year has own by! This year saw the merger of CRUX and MAYHEM come to fruition. The logistical problems were greater than I expected, and we were slow at getting the rst few issues to press. But now, it seems to ow smoothly. Thanks to all for their patience and assistance. The online version of CRUX with MAYHEM continues to attract attention. Thanks are due to LOKI JORGENSON, NATHALIE SINCLAIR, and the rest of the team at SFU who are responsible for this. There are many people that I wish to thank most sincerely for particular contributions. First and foremost is BILL SANDS. Bill is of such value to me and to the continuance of CRUX with MAYHEM. As well, I thank most sincerely, CATHY BAKER, ROLAND EDDY, CHRIS FISHER, BILL SANDS, JIM TOTTEN, and EDWARD WANG, for their regular yeoman service in assessing which solutions should be highlighted; DENIS HANSON, D. FARENICK, C. FISHER, A. LIU, R. MCINTOSH, J. MACLAREN, D. RUOFF, M. TSATSOMEROS, H. WESTON, for ensuring that we have quality arti cles; ANDY LIU, ROBERT GERETSCHLAGER, MURRAY S. KLAMKIN, MARI A FALK   GOTTFRIED PERZ, JIM TOTTEN, for ensuring that de LOSADA, JOSZEF PELIKAN, we have quality book reviews, ROBERT WOODROW (and JOANNE LONGWORTH), who carries the heavy load of two corners, one somewhat new and the other of long standing, and RICHARD GUY for sage advice whenever necessary. The editors of the MAYHEM section, NAOKI SATO, CYRUS HSIA, ADRIAN CHAN, RICHARD HOSHINO, RAVI VAKIL and WAI LING YEE, all do a sterling job. I also thank two of our regulars who assist the editorial board with proof reading; THEODORE CHRONIS and WALDEMAR POMPE. The quality of these people are vital parts of what makes CRUX with MAYHEM what it is. Thank you one and all. As well, I would like to give special thanks to our retiring Associate Editor, CLAYTON HALFYARD, for keeping me from printing too many typographical errors; and my colleagues, PETER BOOTH, RICHARD CHARRON, EDGAR GOODAIRE, ERIC JESPERS, MIKE PARMENTER, DONALD RIDEOUT, NABIL SHALABY, in the Department of Mathematics and Statistics at Memorial University for their occasional sage advice. I have also been helped by some Memorial University students, DON HENDER, COLIN HILLIER, PAUL MARSHALL, SHANNON SULLIVAN, as well as a WISE Summer student, ALYSON FORD. The staff of the Department of Mathematics and Statistics at Memorial University deserve special mention for their excellent work and support: ROS ENGLISH, MENIE FRENCH, WANDA HEATH, and LEONCE MORRISSEY; as well as the computer and networking expertise of RANDY BOUZANE. Also the assistance of ELLEN WILSON at Mount Allison University is much appreciated. Not to mention GRAHAM WRIGHT, Managing Editor, would be a travesty. Graham has kept so much on the right track. He is a pleasure to work with. The CMS's TEX Editor, MICHAEL DOOB has been very helpful in ensuring that the printed master copies are up to the standard required for the U of T Press who continue to print a ne product. Finally, I would like to express real and heartfelt thanks to the Heads of my Department, BRUCE WATSON and HERBERT GASKILL, and to ALAN LAW, Dean of Science of Memorial University, and WILLIE DAVIDSON, Acting Dean of Science of Memorial University, without whose support and understanding, I would not be able to do the job of Editor-in-Chief. Last but not least, I send my thanks to you, the readers of CRUX. Without you, CRUX would not be what it is. Keep those contributions and letters coming in. I do enjoy knowing you all.

539

INDEX TO VOLUME 23, 1997 Crux Articles A Probabilistic Approach to Determinants with Integer Entries Theodore Chronis .................................................................... 23 Folding the Regular Heptagon Robert Geretschlager ................................................................. 81 Heronian Triangles with Associated Inradii in Arithmetic Progression Paul Yiu ............................................................................. 146 Folding the Regular Nonagon Robert Geretschlager ............................................................... 210 A Fermat-Fibonacci Collaboration K.R.S. Sastry ........................................................................ 274 Packing Boxes with N -tetracubes Andris Cibulis ....................................................................... 336 More Unitary Divisor Problems K.R.S. Sastry ........................................................................ 407 Dissecting Rectangular Strips Into Dominoes Frank Chen, Kenneth Nearey and Anton Tchernyi ................................. 468 The Academy Corner Bruce Shawyer February No. 8 .......................................................................... 3 March No. 9 ............................................................................ 65 April No. 10 ........................................................................... 129 May No. 11 Bernoulli Trials 1997 Christopher Small ................................. 193 September No. 12 Bernoulli Trials 1997 { Hints and Answers Christopher Small .... 257 October No. 13 ....................................................................... 321 November No. 14 ..................................................................... 385 December No. 15 ..................................................................... 449 The Olympiad Corner R.E. Woodrow February No. 179 ........................................................................ 6 March No. 180 ......................................................................... 66 April No. 181 .......................................................................... 131 May No. 182 .......................................................................... 196 September No. 183 ................................................................... 260 October No. 184 ...................................................................... 322 November No. 185 .................................................................... 388 December No. 186 .................................................................... 450 The Skoliad Corner R.E. Woodrow February No. 19 ........................................................................ 25 March No. 20 ........................................................................... 89 April No. 21 ........................................................................... 150 May No. 22 ........................................................................... 218 September No. 23 ..................................................................... 278 October No. 24 ....................................................................... 343 November No. 25 ..................................................................... 411 December No.26 ...................................................................... 473 Miscellaneous Welcome / Bienvenue ................................................................... 1 Editorial ................................................................................ 30 In memoriam | Dr. Leon Banko Clayton Dodge ................................... 145

540 Book Reviews Andy Liu Shaking Hands in Corner Brook and other Math Problems by Peter Booth, Bruce Shawyer and John Grant McLoughlin (Editors) Reviewed by Robert Geretschlager and Gottfried Perz .............................. 20 Leningrad Mathematical Olympiads 1987{1991 by Dmitry Fomin and Alexey Kirichenko Reviewed by Joszef  Pelika n .......................................................... 78 The Lighter Side of Mathematics by Richard K. Guy and Robert E. Woodrow (Editors) Reviewed by Murray S. Klamkin ................................................... 143 Cien Problemas de Matem a ticas:  Combinatoria, Algebra, Geometra, One Hundred Mathematics Problems: Combinatorics, Algebra, Geometry by Francisco Bellot Rosado and Mara Ascensio n L opez  Chamorro Reviewed by Mara Falk de Losada ................................................. 209 The Puzzle Arcade by Jerry Slocum Quantum Quandaries by Timothy Weber (Editor) Reviewed by Andy Liu .............................................................. 272 A Mathematical Mozaic by Ravi Vakil Reviewed by Jim Totten ............................................................ 334 144 Problems of the Austrian-Polish Mathematics Competition 1978{1993 by Marcin Emil Kuczma 40th Polish Mathematics Olympiad 1989/90 by Marcin Emil Kuczma Reviewed by Andy Liu .............................................................. 405 Learn from the Masters! edited by Frank Swetz, John Fauvel, Otto Bekken, Bengt Johansson and Victor Katz Reviewed by Maria de Losada ..................................................... 467 Problems February 2137, 2201{2212 ............................................................. 45 March 2214{2225 ..................................................................... 109 April 2173, 2226{2237 ................................................................. 166 May 2238{2250 ....................................................................... 242 September 2251{2262 ................................................................. 299 October 2265{2275 .................................................................... 363 November 2276{2286 ................................................................. 431 December 2297{2300 .................................................................. 500 Solutions February 2101{2112 .................................................................... 49 March 2113{2124 ..................................................................... 112 April 1940, 2125{2128, 2130{2132, 2134{2140 ....................................... 170 May 1940, 2133, 2141{2144 ........................................................... 246 September 2145{2157 ................................................................. 302 October 1940, 2158{2168 ............................................................. 367 November 2090, 2169{2178 ........................................................... 434 December 2179{2197 ................................................................. 504

541 Mathematical Mayhem February ................................................................................ 30 March .................................................................................. 94 April ................................................................................... 152 May ................................................................................... 224 September ............................................................................. 283 October ................................................................................ 348 November ............................................................................. 418 December ............................................................................. 482 Mayhem Articles A Journey to the Pole | Part I Miguel Carrion  A lvarez ............................................................... 36 Matrix Exponentials | An Introduction Donny Cheung ....................................................................... 94 A Journey to the Pole | Part II Miguel Carrion  A lvarez .............................................................. 152 A Pattern in Permutations John Linnell ......................................................................... 158 What is the next term? Naoki Sato .......................................................................... 226 Euler's and DeMoivre's Theorem Soroosh Yazdani ..................................................................... 231 Decimal Expansion of Fractions Cyrus Hsia .......................................................................... 285 An Ambivalent Sum Naoki Sato .......................................................................... 290 Unique Forms Naoki Sato .......................................................................... 351 Solving the Quartic Cyrus Hsia .......................................................................... 420 Pizzas and Large Numbers Shawn Godin ........................................................................ 422 Minima and Maxima of Trigonometric Expressions Nicholae Gusita ..................................................................... 426 The Equation of the Tangent to the nth Circle Krishna Srinivasan ................................................................. 485 Combinatorial Games Adrian Chan ........................................................................ 487 Shreds and Slices February ................................................................................ 30 May ................................................................................... 224 September ............................................................................. 283 October ................................................................................ 348 November ............................................................................. 482 Positive Matrices and Positive Eigenvalues ............................................ 31 Newton's Relations .................................................................... 32 Mathematically Correct Sayings ....................................................... 34 Fibonacci Residues .................................................................... 224 Playful Palindromes ................................................................... 283 The Best of the WEB .................................................................. 284 Invariants of Inscribed Regular n-gons ............................................... 348 Four Large Spheres and a Small Sphere .............................................. 349 Factorial Fanaticism ................................................................... 418 A Note on Convexity .................................................................. 482

542 Mayhem Problems February ................................................................................ 42 March ................................................................................. 100 April ................................................................................... 163 May ................................................................................... 235 September ............................................................................. 296 October ................................................................................ 356 November ............................................................................. 429 December ............................................................................. 494 High School Problems February: H217, H218, H219, H220 ................................................... 43 April: H221, H222, H223, H224 ...................................................... 164 September: H225, H226, H227, H228 ................................................ 297 November: H229, H230, H231, H232 ................................................ 429 High School Solutions March: H205, H206, H207 ............................................................ 101 May: H208, H209, H210 .............................................................. 236 October: H211, H212, H213 .......................................................... 357 December: H214, H215, H216 ........................................................ 494 Advanced Problems February: A193, A194, A195, A196 ..................................................... 44 April: A197, A198, A199, A200 ........................................................ 165 September: A201, A202, A203, A204 ................................................. 298 November: A205, A206, A207, A208 .................................................. 430 Advanced Solutions March: A185, A186, A187 ............................................................. 102 May: A184, A188, A189 ............................................................... 239 October: A190, A191, A192 ........................................................... 358 December: A191 ...................................................................... 496 Challenge Board Problems February: C70, C71, C72 ............................................................... 44 April: C70, C71, C72 .................................................................. 165 September: C73 ....................................................................... 298 November: C74 ....................................................................... 430 Challenge Board Solutions March: C64, C86, C69 ................................................................. 106 October: C70, C72 ..................................................................... 360 December: C70, C71 .................................................................. 496 Contests J.I.R. McKnight Problems Contest 1978 .............................................. 293 1984 Swedish Mathematics Olympiad ................................................ 294 1983 Swedish Mathematics Olympiad Solutions ..................................... 296 1996 Balkan Mathematical Olympiad ................................................. 355 J.I.R. McKnight Problems Contest 1979 .............................................. 428 J.I.R. McKnight Problems Contest 1980 .............................................. 493 Miscellaneous Contest Dates .......................................................................... 35 IMO Report Richard Hoshino .......................................................... 41 IMO CORRESPONDENCE PROGRAM E.J. Barbeau ................................. 161 IMO Report Adrian Chan ............................................................. 350 Book Reviews Donny Cheung ......................................................... 491

543 Proposers, solvers and commentators in the PROBLEMS and SOLUTIONS sections for 1997 are: Hayo Ahlburg: 2170, 2175, 2188 Gerald Allen: 2191, 2192 Miguel Amengual Covas: 2102, 2104, 2112, 2114, 2151, 2162, 2166, 2188, 2189, 2196

Claudio Arconcher: 2114, 2120, 2124, 2142, 2188 Federico Ardila: 2102, 2103, 2106, 2107, 2108, 2111, 2112, 2113, 2114, 2115, 2117, 2119, 2120, 2121, 2124  Sefket Arslanagic: 2090, 2103, 2113, 2114, 2117, 2121, 2124,

2128, 2132, 2133, 2137, 2150, 2153, 2157, 2163, 2167, 2172, 2173, 2176, 2178, 2180, 2181, 2180, 2181, 2188, 2191 Charles Ashbacher: 2118, 2119, 2158 Sam Baethge: 2114, 2118, 2122, 2124, 2125, 2126, 2143, 2144, 2154, 2166, 2175, 2196 Chetan T. Balwe: 2192 Niels Bejlegaard: 2128, 2130, 2136, 2137, 2142, 2150 Francisco Bellot Rosado: 2102, 2103, 2114, 2128, 2130, 2137, 2148, 2149, 2153, 2156, 2162, 2164, 2169, 2171, 2181, 2186, 2188, 2189 Mansur Boase: 2111, 2112, 2115, 2118, 2119, 2122, 2196 Carl Bosley: 2102, 2102, 2105, 2111, 2114, 2115, 2124 Paul Bracken: 2184, 2197 Christopher J. Bradley: 2102, 2103, 2105, 2107, 2108, 2109, 2112, 2114, 2117, 2118, 2120, 2121, 2122, 2124, 2125, 2126, 2127, 2128, 2130, 2133, 2137, 2140, 2141, 2142, 2143, 2144, 2145, 2147, 2148, 2149, 2150, 2155, 2164, 2165, 2166, 2167, 2170, 2171, 2174, 2175, 2177, 2178, 2181, 2184, 2185, 2186, 2188, 2189, 2191, 2192, 2194, 2195, 2196, 2197

Jeremy T. Bradley: Adam Brown: 2181, 2195, 2188 Sydney Bulman{Fleming: 2123, 2187 Miguel Angel Cabez on  Ochoa: 2102, 2112, 2114, 2114, 2144, 2147, 2150, 2151, 2154, 2156, 2157, 2158, 2160, 2162 Joseph Callaghan: 2126, 2128 Sabin Cautis: 2177, 2181, 2188, 2192, 2196 Krzysztof Chelminski: 2121 Ji Chen: 1940, 2101 Han Ping Davin Chor: 2102, 2103, 2108, 2113, 2125, 2126, 2127, 2137 Theodore Chronis: 2090, 2102, 2105, 2108, 2112, 2113, 2140, 2143, 2144, 2145, 2147, 2150, 2153, 2157, 2158, 2159, 2161, 2163, 2174, 2175, 2176, 2180, 2181, 2185, 2188, 2191, 2192, 2196 Jan Ciach: 2168 Mihai Cipu: 2154, 2157, 2158, 2159, 2163, 2167, 2172, 2174, 2176 Goran Conar: 2181, 2188, 2196 Tim Cross: 2103, 2108, 2112, 2114, 2118, 2121, 2122, 2124, 2125, 2126, 2143, 2144, 2150, 2170, 2175, 2185 Luz M. DeAlba: 2174 Paul-Olivier Dehaye: 2196 Georgi Demizev: 2151, 2153, 2154, 2155, 2156, 2157, 2158, 2159, 2161, 2162, 2163 Luis V. Dieulefait: 2090, 2115, 2193 Charles R. Diminnie: 2111, 2118, 2122, 2124, 2125, 2126, 2132, 2140, 2143, 2144, 2147, 2150, 2157, 2163, 2170, 2175, 2185, 2191, 2192 C. Dixon: 2166 David Doster: 2102, 2103, 2124, 2132, 2140, 2145, 2150, 2166, 2184, 2192, 2196, 2197 Jordi Dou: 2110, 2114, 2117, 2124 Keith Ekblaw: 2111, 2147, 2152, 2175, 2185 Hans Engelhaupt: 2102, 2103, 2107, 2114, 2118, 2119, 2122, 2124, 2125, 2126, 2131, 2133, 2137, 2140, 2143, 2144, 2150, 2151, 2154, 2155, 2157, 2158, 2162, 2166, 2167, 2174, 2175 Russell Euler: 2150, 2163 Noel Evans: 2122 C. Festraets-Hamoir: 2196 F.J. Flanigan: 2104, 2132, 2140, 2145, 2147, 2150, 2153, 2174, 2175, 2181, 2184, 2191, 2196

Je rey K. Floyd: 2112, 2125, 2185 Ian June L. Garces: 2125, 2126, 2143, 2147 Toby Gee: 2115 Robert Geretschlager: 2145, 2158, 2176, 2182, 2195, 2196 Kurt Girstmair: 2129 Shawn Godin: 2111, 2112, 2118, 2122, 2123, 2125, 2126, 2140,

2143, 2144, 2155, 2157, 2158, 2163, 2187, 2191, 2195, 2196 Joaqun Gomez  Rey: 2135, 2184, 2197 Javier Gutierrez: 2157, 2170, 2176, 2191 David Hankin: 2114, 2118, 2143, 2144, 2147, 2150, 2195, 2196 G.P. Henderson: 2129, 2136 Fabian Martin Herce: 2163 Florian Herzig: 2101, 2102, 2103, 2105, 2107, 2108, 2112, 2113, 2114, 2118, 2119, 2121, 2122, 2124, 2125, 2127, 2130, 2131, 2132, 2133, 2135, 2137, 2140, 2143, 2144, 2147, 2148, 2150, 2151, 2152, 2153, 2156, 2157, 2158, 2159, 2160, 2162, 2163, 2165, 2166, 2167, 2169, 2172, 2174, 2175, 2176, 2178, 2180, 2181, 2183, 2184, 2185, 2186, 2187, 2188, 2189, 2190, 2192, 2193, 2194, 2195, 2196, 2197 Richard I. Hess: 2109, 2111, 2112, 2113, 2114, 2115, 2118, 2119, 2122, 2123, 2124, 2125, 2126, 2128, 2129, 2130, 2131, 2132, 2133, 2135, 2143, 2144, 2145, 2147, 2150, 2151, 2152, 2154, 2155, 2156, 2157, 2158, 2159, 2160, 2161, 2162, 2163, 2164, 2165, 2166, 2167, 2168, 2169, 2170, 2171, 2172, 2173, 2174, 2175, 2176, 2177, 2178, 2179, 2180, 2181, 2183, 2184, 2185, 2186, 2187, 2188, 2189, 2190, 2191, 2192, 2193, 2194, 2195, 2197 John G. Heuver: 2102, 2107, 2120, 2133, 2167, 2176, 2188 Joe Howard: 2163, 2180, 2181 Cyrus Hsia: 2102, 2107, 2111, 2112, 2192, 2196 Peter Hurthig: 2090, 2114, 2116, 2123, 2124, 2126, 2170 Ignotus: 2196 Peter Ivady: 2090 Walther Janous: 2090, 2101, 2102, 2103, 2104, 2105, 2106, 2107, 2108, 2112, 2113, 2114, 2115, 2116, 2118, 2120, 2121, 2122, 2123, 2124, 2125, 2126, 2130, 2132, 2133, 2135, 2137, 2140, 2141, 2143, 2144, 2145, 2147, 2148, 2150, 2151, 2152, 2153, 2154, 2155, 2156, 2157, 2158, 2159, 2160, 2161, 2162, 2163, 2166, 2167, 2169, 2170, 2171, 2172, 2173, 2174, 2175, 2176, 2178, 2179, 2180, 2181, 2183, 2184, 2185, 2186, 2188, 2189, 2190, 2191, 2192, 2193, 2195, 2196, 2197 D. Kipp Johnson: 2143, 2144, 2151, 2154, 2155, 2157, 2158, 2159, 2161, 2163, 2175, 2181, 2184, 2185, 2191, 2195, 2196 Michael Josephy: 2115, 2123 Yang Kechang: 2106, 2116 Amit Khetan: 2144, 2150 Murray S. Klamkin: 2105, 2108, 2112, 2113, 2116, 2155, 2157, 2159, 2161, 2162, 2162, 2163, 2167, 2168, 2170, 2172, 2178, 2180, 2181, 2183, 2188 Vaclav Konecny: 2101, 2102, 2106, 2107, 2109, 2112, 2114, 2116, 2117, 2118, 2124, 2129, 2133, 2137, 2140, 2143, 2144, 2145, 2147, 2150, 2166, 2167, 2170, 2172, 2175, 2176, 2179, 2181, 2183, 2184, 2185, 2188, 2189, 2191, 2192 Marcin E. Kuczma: 1940, 2113, 2120 Mitko Kunchev: 2102, 2103, 2113, 2114, 2124, 2130, 2145, 2148, 2150, 2151, 2153, 2154, 2155, 2156, 2157, 2158, 2159, 2161, 2162, 2163, 2176 Sai Chong Kwok: 2176 Alexander Lambrou: 2188 Michael Lambrou: 2177, 2178, 2179, 2180, 2181, 2183, 2184, 2185, 2186, 2187, 2188, 2191, 2192, 2193, 2194, 2195, 2197 Stefan Lambrou: 2188 Luke Lamothe: 2112 Kee-Wai Lau: 1940, 2090, 2101, 2103, 2106, 2108, 2109, 2113, 2116, 2117, 2123, 2124, 2132, 2145, 2147, 2150, 2167, 2174, 2175, 2184, 2185, 2186, 2187, 2190, 2192, 2196 Thomas Leong: 2134, 2147, 2152, 2157, 2159

544 Kathleen E. Lewis: 2112, 2122, 2125, 2126 Mara Ascensi on  Lopez  Chamorro: 2151, 2160

2103, 2114, 2149,

B. M???y: 2143, 2144 David E. Manes: 2090, 2111, 2112, 2113, 2118, 2119, 2122, 2123,

2125, 2126, 2131, 2132, 2143, 2144, 2145, 2147, 2150, 2157, 2161 Beatriz Margolis: 2140, 2150 Giovanni Mazzarello: 2125, 2126, 2143, 2147 Tara McCabe: 2122 J.A. McCallum: 2150, 2175, 2185, 2186 John Grant McLoughlin: 2112, 2122, 2125 Vasiliou Meletis: 2103 Norvald Midttun: 2132 Can Anh Minh: 2157, 2176, 2188, 2192, 2196 W. Moser: 2195 Vedula N. Murty: 1940, 2108, 2113, 2167, 2186, 2188 John Oman: 2167, 2185 Soledad Ortega: 2157, 2170, 2176 Victor Oxman: 2091, 2109, 2128, 2142, 2188 Michael Parmenter: 2123, 2147, 2181, 2196 Yolanda Pellejer: 2157 P. Penning: 2102, 2103, 2107, 2109, 2112, 2114, 2118, 2119, 2120, 2122, 2123, 2124, 2125, 2126, 2127, 2130, 2133, 2136, 2137, 2158 Gottfried Perz: 2102, 2123, 2125, 2126, 2137, 2143, 2150, 2156, 2165, 2167, 2169, 2177, 2188, 2196 Carl Pomerance: 2134 Waldemar Pompe: 2102, 2103, 2114, 2117, 2127, 2134, 2139, 2177 Luiz A. Ponce: 2151 Bob Prielipp: 2167, 2195, 2186, 2188, 2196 Cory Pye: 2112, 2118, 2122, 2125 Stanley Rabinowitz: 2129 E. Rappos: 2174 L. Rice: 2140, 2145, 2148, 2150 Juan-Bosco Romero Marquez: 2108, 2149, 2161, 2167, 2171, 2176, 2180, 2188, 2196 Kristian Sabo: 2145, 2150 Jawad Sadek: 2163 Mara Mercedes Sanchez Benito: 2064, 2065, 2067, 2068, 2069, 2071, 2077 Crist obal  S a nchez{Rubio: 2150, 2151, 2164, 2169 Bill Sands: 2125, 2126, 2185, 2195 K.R.S. Sastry: 2104, 2133, 2140, 2154, 2166 Joel Schlosberg: 2107, 2108, 2112, 2115, 2118, 2119, 2122, 2125, 2134, 2143, 2145 Robert P. Sealy: 2104, 2111, 2112, 2123, 2125, 2126, 2132, 2157, 2163, 2175, 2185, 2195, 2196

Harry Sedinger: 2140, 2181 Heinz-Jurgen Sei ert: 2101, 2102, 2105, 2108,

2112, 2113, 2118, 2122, 2125, 2128, 2132, 2135, 2140, 2143, 2144, 2145, 2147, 2150, 2154, 2156, 2157, 2158, 2159, 2161, 2163, 2167, 2174, 2176, 2178, 2180, 2181, 2183, 2184, 2185, 2188, 2191, 2196 Toshio Seimiya: 2102, 2103, 2114, 2117, 2120, 2124, 2127, 2128, 2130, 2133, 2137, 2141, 2142, 2146, 2148, 2149, 2151, 2154, 2156, 2160, 2162, 2164, 2165, 2166, 2167, 2169, 2171, 2177, 2188, 2189 Zun Shan: 2104 D. N. Sheth: 2195 Catherine Shevlin: 2124 Shailesh Shirali, Rishi Valley School, India: 2117, 2148, 2149 D.J. Smeenk: 2102, 2103, 2107, 2114, 2117, 2120, 2124, 2127, 2130, 2133, 2137, 2141, 2148, 2149, 2150, 2151, 2156, 2160, 2162, 2164, 2166, 2167, 2169, 2171, 2181, 2188, 2189, 2191, 2194, 2196 Digby Smith: 2112, 2116, 2118, 2140, 2143, 2144, 2145, 2147, 2150, 2153, 2155, 2163, 2174, 2181, 2185, 2192, 2195, 2196 Trey Smith: 2191, 2192 Lawrence Somer: 2147 Diego Sotes: 2191 David R. Stone: 2118, 2119, 2122, 2123, 2125, 2126, 2132, 2134, 2143, 2144, 2147, 2150, 2175, 2181, 2185, 2192, 2195, 2196 John C. Tripp: 2174 Panos E. Tsaoussoglou: 2102, 2105, 2108, 2113, 2116, 2117, 2118, 2122, 2124, 2143, 2144, 2145, 2147, 2150, 2158, 2166, 2167, 2175, 2176, 2177, 2178, 2180, 2186, 2188 George Tsapakidis: 2102, 2107, 2117, 2176, 2188 Adrian Ubis Matiinez: 2188, 2196 Meletis Vasiliou: 2102, 2107, 2114, 2117, 2120, 2124, 2188 David C. Vella: 2196 Nishka Vijay: 2150 Edward T.H. Wang: 2123, 2125, 2132, 2145, 2150, 2163, 2167, 2174, 2181, 2185, 2187, 2196 Hoe Teck Wee: 2101, 2103, 2105, 2111, 2119, 2131, 2147, 2156, 2165 Chris Wildhagen: 2090, 2102, 2111, 2112, 2113 Kenneth M. Wilke: 2125, 2143, 2147, 2175, 2181, 2196 Lamarr Widmer: 2175 Aram A. Yagubyants: 2137, 2148 Paul Yiu: 2118, 2130, 2137, 2181, 2188 Roger Zarnowski: 2191, 2192

Con Amore Problem Group: 2174, 2177, 2179

2164, 2165, 2166, 2167, 2168,

Skidmore College Problem Group: 2140, 2150, 2174

Proposers, solvers and commentators in the MAYHEM PROBLEMS and SOLUTIONS sections for 1997 are:  Miguel Carrion Alvarez :

H205, H206, H207, H211, H212, H213, H214, A184, A188, C69 Donny Cheung: A184 J. Chris Fisher: C70 Nicolas Guay: A184 Waldemar Pompe: A191 Bob Prielipp: H207, H208, H211 Naoki Sato: C64

Matt Szczesney: C73 Vin de Silva: C68 Ravi Vakil: A184 Sam Vandervelde: C68 Edward T.H. Wang: A184, A185, A189 Eric Wepsic: C68 Samuel Wong: H205, H207 Wai Ling Yee: H209, A195, A197, A188, A189, A190, A192