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PROBLEMS AND SOLUTIONS Edited by Gerald A. Edgar, Doug Hensley, Douglas B. West with the collaboration of Paul T. Bateman, Mario Benedicty, Itshak Borosh, Paul Bracken, Ezra A. Brown, Randall Dougherty, Tam ´ as Erd ´ elyi, Zachary Franco, Chris- tian Friesen, Ira M. Gessel, Jerrold Grossman, Frederick W. Luttmann, Vania Mas- cioni, Frank B. Miles, Richard Pfiefer, Cecil C. Rousseau, Leonard Smiley, John Henry Steelman, Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ull- man, Charles Vanden Eynden, and Fuzhen Zhang. Proposed problems and solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before February 28, 2009. Additional information, such as general- izations and references, is welcome. The problem number and the solver’s name and address should appear on each solution. An asterisk (*) after the number of a problem or a part of a problem indicates that no solution is currently available. PROBLEMS 11383. Proposed by Michael Nyblom, RMIT University, Melbourne, Australia. Show that ∞ n=1 cos −1 1 + √ n 2 + 2n √ n 2 + 4n + 3 (n + 1)(n + 2) = π 3 . 11384. Proposed by Moubinool Omarjee, Lyc ´ ee Jean-Lurc¸at, Paris, France. Let p n denote the nth prime. Show that ∞ n=1 (−1) √ n p n converges. 11385. Proposed by Jos ´ eLuisD ´ ıaz-Barrero, Universidad Polit ´ ecnica de Catalu ˜ na, Barcelona, Spain. Let α 0 , α 1 ,andα 2 be the radian measures of the angles of an acute triangle, and for i ≥ 3letα i = α i−3 . Show that 2 i=0 α 2 i α i+1 α i+2 3 +2tan 2 α i 1/4 ≥ 3 √ 3. 11386. Proposed by Greg Markowsky, Somerville, MA. Consider a triangle ABC.Let O be the circumcircle of ABC, r the radius of the incircle, and s the semiperimeter. Let arc(BC) be the arc of O opposite A, and define arc(CA) and arc(AB) similarly. Let O A be the circle tangent to AB and AC and internally tangent to O along arc(BC), and let R A be its radius. Define O B , O C , R B ,andR C similarly. Show that 1 aR A + 1 bR B + 1 cR C = s 2 rabc . October 2008] PROBLEMS AND SOLUTIONS 757 11387. Proposed by Oskar Maria Baksalary, Adam Mickiewicz University, Pozna ´ n, Poland, and G ¨ otz Trenkler, Technische Universit ¨ at Dortmund, Dortmund, Germany. Let C n,n denote the set of n × n complex matrices. Determine the shortest interval [a, b] such that if P and Q in C n,n are nonzero orthogonal projectors, that is, Hermitian idempotent matrices, then all eigenvalues of PQ + QP belong to [a, b]. 11388. Proposed by M. Farrokhi D.G., University of Tsukuba, Tsukuba Ibaraki, Japan. Given a group G,letG 2 denote the set of all squares in G. Show that for each natural number n there exists a finite group G such that the cardinality of G is n times the cardinality of G 2 . 11389. Proposed by Elizabeth R. Chen and Jeffrey C. Lagarias, University of Michi- gan, Ann Arbor, MI. Given a multiset A ={a 1 , ,a n } of n real numbers (not neces- sarily distinct), define the sumset S(A) of A to be {a i + a j : 1 ≤ i < j ≤ n}, a multi- set with n(n − 1)/2 not necessarily distinct elements. For instance, if A ={1, 1, 2, 3}, then S(A) ={2, 3, 3, 4, 4, 5}. (a)Whenn is a power of 2 with n ≥ 2, show that there are two distinct multisets A 1 and A 2 such that S(A 1 ) = S(A 2 ). (b)Whenn is a power of 2 with n ≥ 4, show that if r distinct multisets A 1 , ,A r all have the same sumset, then r ≤ n − 2. (c ∗ )Whenn is a power of 2 with n ≥ 4, can there be as many as 3 distinct multisets with the same sumset? (Distinct multisets are known to have distinct sumsets when n is not a power of 2.) SOLUTIONS Recalling 11159 and Sixty-Sixth Putnam A5 11277 [2007, 259]. Proposed by Prithwijit De, University College Cork, Republic of Ireland. Find π/2 φ=0 π/2 θ=0 log(2 −sin θ cos φ) sin θ 2 − 2sinθ cos φ + sin 2 θ cos 2 φ dθ dφ. Solution by E. Omey, EHSAL, Brussels, Belgium. The answer is (π 2 /16) log 2. The integral to be evaluated is I = π/2 φ=0 π/2 θ=0 f (sin θ cos φ) sin θ dθ dφ, where f (x) = log(2 − x)/(2 − 2x + x 2 ). In an editorial comment attached to the so- lution of Problem 11159 in this M ONTHLY 114 (2007) 167, it was noted that if g is integrable on [0, 1] then π/2 φ=0 π/2 ψ=0 g(cos ψ cos φ) cos ψ dψ dφ = π 2 1 0 g(t) dt. After substituting θ = π/2 − ψ and setting g = f , we see that I = (π/2) 1 0 f (t)dt. Write x = 1 −t. The required computation, 1 0 log(1 + x) 1 + x 2 dx = π log 2 8 , 758 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 appears in various tables and its evaluation was given as Problem A5 on the Sixty-Sixth William Lowell Putnam Competition (see this M ONTHLY 113 (2006) 733-743). Also solved by R. Bagby, D. H. Bailey & J. M. Borwein (U. S. A., Canada), D. Beckwith, R. Chapman (U. K.), H. Chen, K. Dale (Norway), B. E. Davis, G. de Marco (Italy), A. Fok (Hong Kong), O. Furdui, M. L. Glasser, J. Grivaux (France), E. A. Herman, O. Kouba (Syria), G. Lamb, K. McInturff, P. Perfetti (Italy), O. G, Ruehr, H J. Seiffert (Germany), J. G. Simmonds, A. Stadler (Switzerland), V. Stakhovsky, R. Stong, J. Sun, R. Tauraso (Italy), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, University of Sharjah Problem Solving Group (United Arab Emirates), and the proposer. Entire Limit 11278 [2007, 259]. Proposed by Slavko Simic, Mathematical Institute SANU, Bel- grade, Serbia. Let f be a nonconstant entire function with nonnegative Taylor series coefficients. Prove that lim r→∞ f (r)/rf (r) exists and is rational. Solution by John H. Lindsey II, Cambridge, MA. First consider the case that f = a 0 +···+a n x n , a polynomial with a n > 0. We have lim r→∞ f (r) rf (r) = lim r→∞ a 0 +···+a n r n a 1 r +···+a n nr n = lim r→∞ a 0 /r n +···+a n a 1 /r n +···+a n n = 1 n . Now consider the case in which f = ∞ k=0 a k x k with all coefficients nonnegative, and not eventually zero. For n ∈ N with a n > 0, lim sup r→∞ f (r) rf (r) ≤ lim sup r→∞ n−1 k=0 a k r k a n nr n + lim sup r→∞ ∞ k=n a k r k ∞ k=n a k kr k ≤ 0 + lim sup r→∞ ∞ k=n a k r k ∞ k=n a k nr k = 1 n . Since n can be taken arbitrarily large, lim f (r)/rf (r) = 0. Editorial comment. It should be noted that the limit is intended in the sense of a real variable r →+∞, not in the sense of a complex variable r →∞. Also solved by K. F. Andersen (Canada), J. Bak, R. Chapman (U. K.), G. De Marco (Italy), J. Grivaux (France), E. A. Herman, F. Holland (Ireland), E. J. Ionascu, G. Keselman, K W. Lau (China), O. P. Lossers (Nether- lands), A. Nakhash, N. C. Singer, R. Stong, J. Vinuesa (Spain), GCHQ Problem Solving Group (U. K.), Mi- crosoft Research Problems Group, Northwestern University Math Problem Solving Group, and the proposer. Sliding Beads 11279 [2007, 259]. Proposed by Vitaly Stakhovsky, Redwood City, CA. Two test-mass beads are sliding along a vertical circular track under (Newtonian) constant gravity, without friction. The first bead M 1 starts from the highest point on the circle, at some nonzero velocity. Some time later the second bead M 2 starts from the same position at the top of the circle and with the same initial velocity as M 1 had. Prove that there is a circle to which the line through the current positions of M 1 and M 2 is always tangent, and find the center and radius of that circle in terms of the original circle. Solution by the proposer. Assume that the original circle has unit radius and center at the origin O.LetOY point in the upward direction. Describe the position of bead i by its current location P i and the angle ϕ i = P i OY. Assume both particles are moving with ϕ i increasing. Let v i =˙ϕ i be the ith bead’s velocity. The equation of October 2008] PROBLEMS AND SOLUTIONS 759 motion can be written as v 2 i = 1 − μ cos ϕ i for some 0 ≤ μ<1. Note that v 2 1 − v 2 2 = μ(cos ϕ 2 − cos ϕ 1 ) and ˙v i = (μ/2) sin ϕ i . Hence ˙v 1 +˙v 2 v 2 1 − v 2 2 = 1 2 · sin ϕ 1 + sin ϕ 2 cos ϕ 2 − cos ϕ 1 = 1 2 cot ϕ 1 − ϕ 2 2 . Since |P 1 P 2 |=2sin ϕ 1 −ϕ 2 2 we have d dt log |P 1 P 2 |= 1 2 cot ϕ 1 − ϕ 2 2 (v 1 − v 2 ) = ˙v 1 +˙v 2 v 1 + v 2 = d dt log(v 1 + v 2 ). Therefore τ =|P 1 P 2 |/(v 1 + v 2 ) is constant over time. Note that v 2 1 − v 2 2 = μ sin ϕ 1 + ϕ 2 2 |P 1 P 2 |, so we also have μτ = (v 1 − v 2 )/ sin((ϕ 1 + ϕ 2 )/2). Let Q be the point on P 1 P 2 such that |P 1 Q|/|P 2 Q|=v 1 /v 2 .(Q is the instantaneous center of rotation of the beads and will be the point of tangency of the circle.) Let a perpendicular to P 1 P 2 at Q intersect the vertical axis at a point D = (0, d).LetM be the midpoint of P 1 P 2 .NowQ = (v 1 P 2 + v 2 P 1 )/(v 1 + v 2 ) and M = (P 1 + P 2 )/2, hence |MQ|= (v 1 − v 2 )|P 1 P 2 | 2(v 1 + v 2 ) = v 1 − v 2 2 τ = 1 2 μτ 2 sin ϕ 1 + ϕ 2 2 . Hence d = μτ 2 /2, and the point D is constant over time. Also |DP i | 2 = 1 + d 2 − 2d cos ϕ i =|DQ| 2 +|QP i | 2 ,so |DQ| 2 = v 2 1 |DP 2 | 2 − v 2 2 |DP 1 | 2 v 2 1 − v 2 2 = 1 + d 2 − 2d v 2 1 cos ϕ 2 − v 2 2 cos ϕ 1 v 2 1 − v 2 2 = 1 + d 2 − 2 μ d is constant over time. Thus P 1 P 2 is always tangent to the circle with center D and radius 1 + d 2 − 2d/μ. Note: In case P 2 starts when P 1 is at the lowest point, we get r = 0 and hence the points P 1 , P 2 , D are always collinear. Also solved by ´ O. Ciaurri & E. Fern ´ andez & L. Roncal (Brazil), J. Freeman, J. A. Grzesik, J. H. Lindsey II, J. B. Zacharias, and GCHQ Problem Solving Group (U. K.). What We Mean, Not What We Say 11280 [2007, 259]. Proposed by Harris Kwong, SUNY Fredonia, Fredonia, NY. Let f be a positive nondecreasing function on the real line that is twice differentiable and concave down. For any list x of positive real numbers x 1 , ,x n ,letS = n k=1 x k .In terms of f and n,find max x n k=1 ( f (S − x k )) x k /S . Solution by GCHQ Problem Solving Group, Cheltenham, U. K. There is something wrong with the problem statement: apart from constant functions, there are no positive concave functions on the real line, let alone twice differentiable nondecreasing ones. 760 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 If n = 1 the answer is f (0) whatever properties f has. For general n, we show that if f is a positive nondecreasing function on the real line (not necessarily concave or differentiable, or even continuous), then sup x n k=1 f (S − x k ) x k /S = lim y→+∞ f (y). (1) We claim first that the limit exists. Indeed, if f is bounded above, its least upper bound is its limit; if f is not bounded above, the limit is +∞. Let this limit be ϕ. Next, if ϕ is finite, then n k=1 ( f (S − x k )) x k /S ≤ n k=1 ϕ x k /S = ϕ. Finally, if x 1 =···=x n = z, then n k=1 f (S − x k ) x k /S = f ((n − 1)z) 1/n = f ((n − 1)z) → ϕ as z →∞. This completes the proof of (1). The maximum is attained if and only if the limit is. There are of course functions on the nonnegative half of the real line with the full range of properties given in the problem statement. The result given here applies to them. Editorial comment. It seems that in the intended problem f is defined on the positive real line, and n and S are given values. Then maximize over all lists x satisfying the conditions. Answer: f (1 − 1/n)S . Also solved by R. A. Agnew, K. F. Andersen (Canada), J P. Grivaux (France), E. A. Herman, J. H. Lindsey II, V. Stakhovsky, R. Stong, M. Tetiva (Romania), and Microsoft Research Problems Group. A Triangular Result 11285 [2007, 358]. Proposed by Yakub Aliyev, Qafqaz University and Baku State Uni- versity, Baku, Azerbaijan. Let six points be chosen in cyclic order on the sides of triangle ABC: A 1 and A 2 on BC, B 1 and B 2 on CA,andC 1 and C 2 on AB.LetK denote the intersection of A 1 B 2 and C 1 A 2 , L the intersection of B 1 C 2 and A 1 B 2 ,and M the intersection of C 1 A 2 and B 1 C 2 .LetT , U,andV be the intersections of A 1 B 2 and B 1 A 2 , B 1 C 2 and B 2 C 1 ,andC 1 A 2 and C 2 A 1 , respectively. Prove that lines AK, BL,andCM are concurrent if and only if points T , U,andV are collinear. Solution I (one direction) by Apostolis Denis, Varvakeio High School, Athens, Greece. We prove the first direction of the implication. Suppose AK, BL, CM are concur- rent. Write u.v for the intersection of lines u and v (extended if necessary). Let X = A 1 C 2 .B 1 A 2 , Y = B 1 A 2 .C 1 B 2 ,andZ = C 1 B 2 .A 1 C 2 . Given triangles ABC and abc, Desargues’ Theorem with its dual states that Aa, Bb, Cc are concurrent if and only if AB.ab, BC.bc, CA.ca are collinear. We consider parallel lines to intersect at infinity and allow for collinearity and concurrency at infinity (see Editorial comment below). Application of Desargues’ Theorem to the following three pairs of triangles gives the claimed implication: (i) [ABC and KLM]: AK, BL, CM are concurrent if and only if AB.KL, BC.LM, CA.MK are collinear. Since C 1 lies on AB,andB 2 lies on KL,wehave AB.KL = C 1 B.B 2 L. Similarly, BC.LM = BA 2 .B 1 L,andCA.MK = B 1 B 2 .C 1 A 2 . (ii) [C 1 BA 2 and B 2 LB 1 ]: C 1 B.B 2 L, BA 2 .LB 1 ,andA 2 C 1 .B 1 B 2 are collinear if and only if C 1 B 2 , BL, A 2 B 1 are concurrent. Note: Since C 1 B 2 .A 2 B 1 = Y , this concurrency is equivalent to BL = LY. Similarly, AK = KX and CM = MZ so KX, LY, MZ are concurrent. October 2008] PROBLEMS AND SOLUTIONS 761 (iii) [KLM and XY Z]: KX, LY, MZ are concurrent if and only if T = KL.XY, U = LM.YZ, V = MK.ZX are collinear. Solution II by The GCHQ Problem Solving Group, Cheltenham, United Kingdom. Pas- cal’s Theorem states that the intersection points of pairs of opposite sides of a hexagon are collinear if and only if the vertices of the hexagon lie on a conic. With the notation of Solution I: (a) Applying Desargues’ Theorem to ABC and KLM,wehave:AK, BL, CM are concurrent if and only if AB.KL, BC.LM, CA.MK are collinear. (b) { AB.KL, BC.LM, CA.MK } = { C 1 C 2 .A 1 B 2 , A 1 A 2 .B 1 C 2 , B 1 B 2 .C 1 A 2 } ,soby Pascal’s Theorem these are collinear if and only if C 1 , C 2 , A 1 , A 2 , B 1 , B 2 lie on a conic. (c) { T, U, V } = { A 1 B 2 .B 1 A 2 , B 1 C 2 .B 2 C 1 , C 1 A 2 .C 2 A 1 } , so again by Pascal’s The- orem, these are collinear if and only if A 1 , B 2 , B 1 , C 2 , C 1 , A 2 lie on a conic. Thus, AK, BL, CM are concurrent if and only if T , U, V are collinear. Editorial comment. The statement of the problem is true projectively, but for affine (or Euclidean) geometry, it degenerates when X, Y ,orZ fails to exist, or if KLM is a degenerate triangle. Also solved by R. Chapman (U. K.), P. D ´ alyay (Hungary), M. Englefield (Australia), W. Fosheng (China), O. P. Lossers (Netherlands), M. Tetiva (Romania), L. Zhou, and the proposer. Continuous Blackjack 11287 [2007, 359]. Proposed by Stephen J. Herschkorn, Highland Park, NJ. Players 1 through n play “continuous blackjack.” At his turn, Player k considers a random number X k drawn from the uniform distribution on [0, 1]. He may either accept X k as his score or draw a second number Y k from the same distribution, in which case his score is X k + Y k if X k + Y k < 1 and 0 otherwise. The highest score wins. Give a rule for when player k should draw a second number, in terms of k, n, the result of X k ,and the highest score attained so far. Solution by BSI Problems Group, Bonn, Germany, and the editors. We renumber the players P 0 , P 1 , ,P n−1 in reverse order, so that P k has k players yet to draw after his turn. It turns out that the optimal strategy for P k is the same whether the remaining players play as a team or selfishly. Let Z k be P k ’s score; that is, Z k = X k if P k accepts his first number, Z k = X k + Y k if he draws a second number Y k and X k + Y k < 1, and Z k = 0ifhedrawsasecond number and X k +Y k ≥ 1. For k = 0, ,n −1, let M k = max j≥k Z j ,andletM n = 0. We assume selfish play. That is, P k should take a second number if and only if doing so improves his probability of winning, assuming P k−1 through P 0 do the same. We ignore ties, which have zero probability. Note that P k ’s situation depends only on the number k of players yet to draw and on M k+1 .HowM k+1 was produced, and how many players came earlier to produce it, are irrelevant. Let f k (z) denote the probability that if players P j for j ≥ k have already played and the best score so far is M k = z, then for all j < k, Z j < z—that is, all later players achieve scores lower than z,soz is the winning score. We will prove that for all k: 1. f k is a continuous, nondecreasing function on [0, 1] and f k (z)>0forz > 0. 2. There is a unique number g k ∈[0, 1) such that f k (g k ) = 1 g k f k (t) dt,andP k should take a second number if and only if X k < max(M k+1 , g k ). We first observe that statement 2 follows from statement 1. To see why, suppose statement 1 holds. Clearly 1 z f k (t) dt is a strictly decreasing continuous function of z, 762 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 1 0 f k (t) dt ≥ f k (0),and 1 1 f k (t) dt = 0 < f k (1), so by the intermediate value theo- rem there is a unique number g k ∈[0, 1) such that f k (g k ) = 1 g k f k (t) dt.Furthermore, if z < g k then 1 z f k (t) dt > f k (z),andifz > g k then 1 z f k (t) dt < f k (z). To deter- mine how player P k should play, suppose that all players P j for j > k have already played. Clearly if X k < M k+1 then player P k must take a second number to have any hope of winning. If X k > M k+1 , then player P k ’s probability of winning is f k (X k ) if he doesn’t take a second number, and 1 X k f k (t) dt if he does. The latter probability is larger than the former if and only if X k < g k ,soP k should take a second number if and only if X k < max(M k+1 , g k ). To prove statement 1, we proceed by induction on k. In the case k = 0, if all of the players have played and M 0 = z,thenz is definitely the winning score, so f 0 (z) = 1. Now assume statement 1 (and therefore also statement 2) holds for k. Suppose that players P j for j ≥ k + 1 have already played and M k+1 = z. There are two ways for player P k to end up with a score Z k < z: either X k + Y k < z,orX k + Y k > 1. The probability of the first event is z 0 (z − x) dx = z 2 /2, and the probability of the second is max(z,g k ) 0 xdx= (max(z, g k )) 2 /2. Thus, the probability that Z k < z is: h k (z) = 1 2 (g 2 k + z 2 ), if z ≤ g k ; z 2 , if z > g k . If Z k < z then M k = z, and therefore the probability that Z j < z for all j < k is f k (z).Thus,wehave f k+1 (z) = h k (z) f k (z), and combining our formula for h k (z) with statement 1 for k, we see that statement 1 holds for k + 1 as well. Finally, we wish to determine the numbers g k . We first observe that for all k, g k+1 > g k . To see why, suppose that g k+1 ≤ g k . Then, on putting l = k + 1, we have f l (g l ) = h k (g l ) f k (g l ) ≤ h k (g l ) f k (g k ) = h k (g l ) 1 g k f k (t) dt ≤ h k (g l ) 1 g l f k (t) dt < 1 g l h k (t) f k (t) dt = 1 g l f l (t) dt = f l (g l ), a contradiction. It follows that g k+1 > g k , as claimed. Next, we observe that for z ≥ g k , f k (z) = z 2k . This follows easily by induction from the facts that f 0 (z) = 1, f k+1 (z) = h k (z) f k (z),andg k+1 > g k , and the formula for h k (z). Finally, we use this formula for f k (z) to get an equation that we can solve to find g k : g 2k k = f k (g k ) = 1 g k f (t) dt = 1 g k t 2k dt = 1 2k + 1 (1 − g 2k+1 k ). In summary, the best strategy is for P k to take a second number if and only if X k < max(M k+1 , g k ),whereg k is the unique nonnegative solution to z 2k = (1 − z 2k+1 )/(2k + 1). Editorial comment. If the remaining players are not selfish, but cooperate to try to stop P k , this is still P k ’s best strategy. For j < k,playerP j now accepts any score X j that is better than M j . The result is that h j (z) = z 2 and f j (z) = z 2 j for j < k. Thus P k should still take a second number if and only if X k < max(M k+1 , g k ). October 2008] PROBLEMS AND SOLUTIONS 763 Also solved by D. R. Bridges, J. Chachulska & W. Matysiak (Poland), R. Chapman (U. K.), J. Freeman, J. H. Lindsey II, G. Pastor (Mexico), T. Rucker, B. Schmuland (Canada), R. Staum, T. Tam, E. A. Weinstein, GCHQ Problem Solving Group (U. K.), and the proposer. Formulas Involving the Angles of a Triangle 11289 [2007, 359]. Proposed by Oleh Faynshteyn, Leipzig, Germany. Let ABC be a triangle with sides a, b,andc, all different, and corresponding angles α, β,andγ . Show that (a) (a + b) cot(β + 1 2 γ)+ (b + c) cot(γ + 1 2 α) +(a + c) cot(α + 1 2 β) = 0. (b) (a − b) tan(β + 1 2 γ)+ (b − c) tan(γ + 1 2 α) + (c − a) tan(α + 1 2 β) = 4(R + r), where R is the circumradius of the triangle and r the inradius. Solution by Marian Tetiva, Birlad, Romania. Since α + β + γ = π ,wehaveβ + 1 2 γ = π 2 − α−β 2 ,socot(β + 1 2 γ) = tan α−β 2 and tan(β + 1 2 γ) = cot α−β 2 .Fromthelaw of sines and trigonometric identities, we get a +b = 2R(sin α + sin β) = 4R sin α + β 2 cos α −β 2 , a −b = 2R(sin α − sin β) = 4R sin α − β 2 cos α +β 2 . Similar equations are obtained by cyclically permuting a, b, c and α, β, γ . We will write “ ” for a sum with three terms obtained by cyclic permutation from the one term given. For part (a): (a + b) cot β + 1 2 γ = 4R sin α +β 2 cos α − β 2 tan α − β 2 = 4R sin α +β 2 sin α − β 2 = 4R sin 2 α 2 − sin 2 β 2 = 0. For part (b): (a − b) tan β + 1 2 γ = 4R sin α −β 2 cos α +β 2 cot α −β 2 = 4R cos α +β 2 cos α − β 2 = 4R cos α = 4(R +r). For the last step we used the formula cos α = 1 + 4sin α 2 sin β 2 sin γ 2 = 1 + r R . Also solved by Z. Ahmed (India), S. Amghibech (Canada), M. Bataille (France), R. Chapman (U.K.), P. P. D ´ alyay (Hungary), P. De (Ireland), A. Demis (Greece), A. Fok (China), W. Fosheng (China), V. V. Garc ´ ıa (Spain), M. Goldenberg & M. Kaplan, O. Kouba (Syria), H. Kwong, K W. Lau (China), O. P. Lossers (Nether- lands), Y. Mikata, E. Mouroukos (Greece), R. D. Nelson (U.K.), J. H. Nieto (Venezuela), P. E. Nuesch (Switzer- land), C. R. Pranesachar (India), J. Rooin & A. Alikhani (Iran), V. Schindler (Germany), L. Sega, H J. Seiffert (Germany), M. Shattuck, R. A. Simon (Chile), A. Stadler (Switzerland), D. Vacaru (Romania), M. Vowe (Switzerland), L. Zhou, GCHQ Problem Solving Group (U.K.), Microsoft Research Problems Group, North- western University Math Problem Solving Group, and the proposer. 764 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 . Vanden Eynden, and Fuzhen Zhang. Proposed problems and solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address. Define O B , O C , R B ,andR C similarly. Show that 1 aR A + 1 bR B + 1 cR C = s 2 rabc . October 2008] PROBLEMS AND SOLUTIONS 757 11387. Proposed by Oskar Maria Baksalary, Adam Mickiewicz University,. computation, 1 0 log(1 + x) 1 + x 2 dx = π log 2 8 , 758 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 appears in various tables and its evaluation was given as Problem A5 on the Sixty-Sixth William