Proposed by Omran Kouba, Higher Institute for Ap- plied Sciences and Technology, Damascus, Syria. LetC and. D be concentric circles with radii r and R, ...
12045. Proposed by Ovidiu Furdui and Alina Sînt˘am˘arian, Technical University of ClujNapoca, Cluj-Napoca, Romania. Prove that the series ! ! ∞ ∞ X X 1 n−1 (−1) n −1 k2 n=1 k=n+1 converges to
π2 16
−
ln 2 2
− 12 .
12046. Proposed by Moubinool Omarjee, Lycée Henri IV, Paris, France. Suppose that R1 f : [0, 1] → R has a continuous and nonnegative third derivative, and suppose 0 f (x) dx = 0. Prove Z 1 Z 1 Z 1 x3 f (x) dx + 6 x f (x) dx ≥ 15 x2 f (x) dx. 10 0
0
0
12047. Proposed by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let C and D be concentric circles with radii r and R, respectively, with r < R. Let A1 A2 · · · An be a convex n-gon with perimeter p inscribed in C. For 1 ≤ k ≤ n, let Bk be the intersection of the ray Ak Ak+1 with the circle D, where An+1 = A1 . Let q be the perimeter of the n-gon B1 B2 · · · Bn . Prove p/q ≤ r/R, and determine when equality holds.
B2
B3
A4
A3
A1
r
R
A2 B4
B1
SOLUTIONS When Three Intervals Seem the Same to an Observer 11915 [2016, 613]. Proposed by Mark E. Kidwell and Mark D. Meyerson, U.S. Naval Academy, Annapolis, MD. Given four points A, B, C, and D in order on a line in Euclidean space, under what conditions will there be a point P off the line such that the angles ∠APB, ∠BPC, and ∠CPD have equal measure? Solution I by Erik I. Verriest, Georgia Institute of Technology, Atlanta, GA. Recall from projective geometry that the cross-ratio (A, B; C, D) =
AC/AD sin ∠APC sin ∠BPD = BC/BD sin ∠BPC sin ∠APD
is invariant under projection. Assume that A, B, C, D are in order on a line and ∠APC = ∠BPC = ∠CPD = α, so 3α lies in the open interval (0, π ). Draw point A′ on PA such that PA′ = PD. Let PB and PC intersect A′ D in points B′ and C′ , respectively. We have (A, B; C, D) = (A′ , B′ ; C′ , D) =
sin2 (2α) 4 − 4 sin2 α . = sin α sin(3α) 3 − 4 sin2 α
This ratio increases from 4/3 at α = 0 to ∞ at α = π /3. Thus (A, B; C, D) > 4/3 is a necessary condition. Conversely, if (A, B; C, D) > 4/3, we have the configuration A′ , B′ , C′ , D described above. Choose A′′ on PA′ so that with the intersection point B′′ = PB′ ∩ A′′ D we have B′′ A′′ /B′′ D = BA/BD. (This defines A′′ and B′′ uniquely.) Also let C′′ = PC′ ∩ A′′ D. Since (B′′ , C′′ ; A′′ , D) = (B, C; A, D), we have C′′ A′′ /C′′ D = CA/CD. Rescaling gives the desired configuration with A = A′′ , B = B′′ , C = C′′ , and D. Thus the condition (A, B; C, D) > 4/3 is sufficient as well.
May 2018]
PROBLEMS AND SOLUTIONS
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