Tạp chí toán học AMM của Mỹ
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 September 30, 2008. Additional information, such as gen- eralizations and references, is welcome. The problem number and the solver’s name and address should appear on each solution. An asterisk (*) after the num- ber of a problem or a part of a problem indicates that no solution is currently available. PROBLEMS 11362. Proposed by David Callan, University of Wisconsin, Madison, WI. A bit string arc diagram is an undirected graph in which the vertices are the positions in a single string of bits and the edges are called arcs due to the visual representation in which they are drawn joining positions in the string. To be a good diagram, arcs must occur only between unequal bits, and each bit may be the left endpoint of at most one arc. Thus, the first diagram is good but, for two reasons, the second is not. 1100110 1100110 There are six good diagrams on two bits, four with no arc and two with a single arc. How many good diagrams are there on n bits? 11363. Proposed by Oleh Faynshteyn, Leipzig, Germany. Let m a , m b ,andm c be the lengths of the medians of a triangle T . Similarly, let I a , I b , I c , h a , h b ,andh c be the lengths of the angle bisectors and altitudes of T ,andletR, r,andS be the circumra- dius, inradius, and area of T . Show that I a I b I c + I b I c I a + I c I a I b ≥ 3(2R −r), and m a I b h c + m b I c h a + m c I a h b ≥ 3 5/4 √ S. 11364. Proposed by P ´ al P ´ eter D ´ alyay, Szeged, Hungary. Let p be a prime greater than 3, and t the integer nearest p/6. (a) Show that if p = 6t + 1, then ( p −1)! 2t−1 j =0 (−1) j 1 3 j + 1 + 1 3 j + 2 ≡ 0 (mod p). May 2008] PROBLEMS AND SOLUTIONS 461 (b) Show that if p = 6t − 1, then ( p −1)! 2t−1 j =0 (−1) j 3 j + 1 + 2t−2 j =0 (−1) j 3 j + 2 ≡ 0 (mod p). 11365. Proposed by Aviezri S. Fraenkel, Weizmann Institute of Science, Rehovot, Israel. Let t be a positive integer. Let γ = √ t 2 + 4, α = 1 2 (2 + γ − t),andβ = 1 2 (2 + γ + t). Show that for all positive integers n, nβ=(nα+n(t − 1))α+1 =(nα+n(t − 1) + 1)α−1. 11366. Proposed by Nicolae Anghel, University of North Texas, Denton, TX. Let φ : R → R be a continuously differentiable function such that φ(0) = 0andφ is strictly increasing. For a > 0, let C a denote the space of all continuous functions from [0, a] into R,andfor f ∈ C a ,letI ( f ) = a x=0 ( φ(x) f (x) − xφ( f (x)) ) dx. Show that I has a finite supremum on C a and that there exists an f ∈ C a at which that supremum is attained. 11367. Proposed by Andrew Cusumano, Great Neck, NY. Let x 1 = √ 1 +2, x 2 = 1 + 2 √ 1 + 3, and in general, let x n+1 be the number obtained by replacing the innermost expression (1 + (n + 1)) in the nested square root formula for x n with 1 + (n + 1) √ 1 + (n +2). Show that lim n→∞ x n − x n−1 x n+1 − x n = 2. 11368. Proposed by Wei-Dong Jiang, Weihai Vocational College, Weihai, ShanDong, China. For a triangle of area 1, let a, b,andc be the lengths of its sides. Let s = (a + b + c)/2. Show that the weighted average of (s − a) 2 , (s − b) 2 ,and(s − c) 2 , weighted by the radian measure of the angles opposite a, b,andc respectively, is at least π/ √ 3. SOLUTIONS Generating Functions and Hypergeometric Series and Continued Fractions! 11198 [2006, 79]. Proposed by P. R. Parthasarathy, India Institute of Technology Madras, Chennai, India. Let f (k) = 1 + k if k is odd, f (k) = 1 + k/2ifk is even. Show that 0 i 1 =0 f (i 1 ) 1+i 1 i 2 =0 f (i 2 ) 1+i 2 i 3 =0 f (i 3 ) ··· 1+i n−1 i n =0 f (i n ) = n m=1 m k=1 (−1) m−k m k k n . Composite solution by the proposer and the editors. We begin by forming a generating function from each side. Multiply by (−1) n q n , sum over n ≥ 1,anduse1forthe constant term. It suffices to show that the two resulting generating functions are the same. We start with the one on the left. 462 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 We first convert the power series to a continued fraction. With a = (a 0 , a 1 , ),let p n (a) be the coefficient of q n in the expression 1 1 + a 0 q 1 + a 1 q 1+ . . . = 1 1+ a 0 q 1+ a 1 q 1+ ···= ∞ n=0 (−1) n p n (a)q n . (1) We prove by induction on n that p n (a) is the sum of n j =1 a i j over all n-tuples i 1 , ,i n such that i 1 = 0and0≤ i j ≤ i j −1 + 1for2≤ j ≤ n. This is precisely our generating function when a k = f (k) for k ≥ 0. For n = 0, we have the empty product and coefficient 1. For n ≥ 1, let a = (a 1 , a 2 , ). By the induction hypothesis, ∞ n=0 (−1) n p n (a)q n 1 + a 0 q ∞ k=0 (−1) k p k (a )q k = 1, By examining the coefficient of q n on both sides, it follows that p n (a) = a 0 n−1 =0 p (a) p n−1− (a ). (2) Group the terms in the claimed formula p n (a) according to the maximum index j such that i j = 0. By the induction hypothesis, the sum of the terms for which the last position with subscript 0 in the n-tuple is position l + 1 can be written as p (a)a 0 p n−1− (a ). Hence (2) completes the proof of the formula for p n (a). The sequence from which we form the continued fraction can be expressed as a 2r−2 = r and a 2r−1 = 2r for r ≥ 1. Let s = 1/q. Our next aim is 1 1+ a 0 /s 1+ a 1 /s 1+ ···= 1 2 ∞ k=0 s s + k 1 2 k . (3) On the right is the hypergeometric series 1 2 2 F 1 (s, 1;1 + s; 1 2 ); in standard notation, 2 F 1 (a, b;c;z) = k≥0 a (k) b (k) c (k) z n n! ,wherex (k) is the rising factorial. We seek a continued fraction expansion converging to the same value as the left side. It is convenient to multiply both sides of (3) by 1/s. The continued fraction on the left converges when s has positive real part. A continued fraction converges to the same value as its even part, which is the limit of its even approximants (see L. Lorentzen and H. Waadeland, Continued Fractions with Applications, North-Holland, 1992, p. 84). The even part is 1/s 1 + a 0 /s− a 0 a 1 /s 2 1 + a 1 /s + a 2 /s− a 2 a 3 /s 2 1 + a 3 /s + a 4 /s− ···. With a 2r−2 a 2r−1 = 2r 2 and a 2r−1 + a 2r = 3r + 1, this becomes 1/s 1 + 1/s− 2 · 1 2 /s 2 1 + 4/s− 2 · 2 2 /s 2 1 + 7/s− ··· 2 ·r 2 /s 2 1 +(3r + 1)/s− ···. This continued fraction is equivalent, in the sense of having the same sequence of approximants, to 1 s + 1− 2 · 1 2 s + 4− 2 · 2 2 s + 7− ··· 2 · r 2 s + 3r + 1− ··· . May 2008] PROBLEMS AND SOLUTIONS 463 To obtain a continued fraction expansion for 1 2s 2 F 1 (s, 1;s + 1; 1 2 ),weusethecon- tinued fraction expansion for the standard Euler fraction (see p. 308 of the book cited above). With c = c − a and b = b − a, we obtain c · 2 F 1 (a, b;c;z) 2 F 1 (a, b + 1;c + 1;z) = c + (b + 1)z − (c + 1)(b + 1)z c + 1 + (b + 2)z− (c + 2)(b + 2)z c + 2 + (b + 3)z− (c + 3)(b + 3)z c + 3 + (b + 4)z− ··· To obtain the desired formula, we set c = a = s, b = 0, and z = 1/2. Since 2 F 1 (s, 0;s; 1 2 ) = k≥0 s (k) 0 (k) s (k) (1/2) k k! = 1andc + r + (b + r + 1)z = (s + 3r + 1)/2, the previous display reduces to s 2 F 1 (s, 1;s + 1; 1 2 ) = s + (1 − s)/2 − (1)(1)/2 (s + 4)/2− (2)(2)/2 (s + 7)/2− (3)(3)/2 (s + 10)/2− ··· = 1 2 s + 1 − 2 · 1 2 s + 4− 2 · 2 2 s + 7− 2 ·3 2 s + 10− . Taking the reciprocal of the continued fraction yields 1 2s 2 F 1 s, 1;s + 1, 1 2 = 1 s + 1− 2 · 1 2 s + 4− 2 ·2 2 s + 7− 2 · 3 2 s + 10− . We have thus proved (3). It remains only to convert 1 2 ∞ k=0 s s+k 1 2 k into the generating function we obtained from the right side of the problem statement. Written using q instead of s,thesteps (explained below) are 1 2 ∞ k=0 1 1 +kq 1 2 k = 1 2 ∞ n=0 (−1) n q n ∞ k=0 1 2 k k n = 1 + 1 2 ∞ n=1 (−1) n q n ∞ k=0 1 2 k n m=1 S(n, m)k (m) = 1 + ∞ n=1 (−1) n q n n m=1 m!S(n, m) = 1 + ∞ n=1 (−1) n q n n m=1 m k=1 (−1) m−k m k k n . In step 1 we expand (1 + kq) −1 by the geometric series and interchange the order of summation. In step 2 we invoke k n = S(n, m)k (m) ,whereS(n, m) is the Stirling number of the second kind (the number of partitions of n positions into m unordered blocks) and k (m) is the falling factorial. Both sides count the k-ary n-tuples, on the right grouped by the number of distinct entries. In step 3 we use the identity ∞ k=0 k m 2 −k = 2; this is the instance of ∞ k=0 k m x k = x m /(1 − x) m+1 at x = 1/2, where the series converges. Finally, in step 4 we use the inclusion-exclusion formula for the Stirling numbers. No solutions were received. Groups, Rings, Fields, and Power Series 11216 [2006, 366]. Proposed by Ted Chinburg, University of Pennsylvania, Philadel- phia, PA, and Shahriar Shahriari, Pomona College, Claremont, CA. Let K be a field, 464 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 and let G be an ordered Abelian group. The support Supp(a) of a formal sum a = γ a γ t γ with coefficients a γ in K and exponents γ in G is the set {γ ∈ G : a γ = 0}. The generalized power series ring K ((G ≤0 )) is the set of formal sums a for which Supp(a) is a well-ordered subset of the nonpositive elements of G. Addition and mul- tiplication in K ((G ≤0 )) are defined in the same way they are for ordinary power se- ries. Show that K ((G ≤0 )) is Noetherian if and only if either G ={0} or G is order- isomorphic to Z with the usual ordering. (An ordered Abelian group is an Abelian group G with a total order ≤ such that a ≤ b implies a + c ≤ b + c for all a, b, c in G.) Solution by Reid Huntsinger, Lansdale, PA. Let R = K ((G ≤0 )). Assume G ={0}. Given a positive element γ 0 in G,letI (γ 0 ) ={a ∈ R : a γ = 0ifγ>−γ 0 }.Always I (γ 0 ) is an additive subgroup. In fact, I (γ 0 ) is an ideal, since a ∈ R and b ∈ I (γ 0 ) imply (ab) γ = δ≤0 a δ b γ −δ . The sum is zero unless γ − δ ≤−γ 0 for some δ ≤ 0, which requires γ ≤−γ 0 . Thus ab ∈ I (γ 0 ),makingI (γ 0 ) an ideal. On the other hand, if a ∈ R with min Supp(A) = 0thena is invertible in R.(The inversion algorithm is essentially the customary algorithm for inverting power series with real coefficients with nonzero constant coefficient.) As a result, all proper ideals of R have the form I (γ ) for some positive γ . If γ 0 <γ 1 ,thenI (γ 1 ) ⊂ I (γ 0 ), and vice-versa. Thus the Noetherian property (no in- finite ascending chains of ideals) is equivalent to having no infinite descending chains in G >0 .Inotherwords,R is Noetherian if and only if the positive elements of G are well-ordered. This latter condition means that the positive elements are all powers of the smallest positive element. Thus G is order-isomorphic to Z exactly when R is Noetherian. Also solved by N. Caro (Brazil), R. Chapman (U. K.), D. Fleischman, Szeged Problem Solving Group “Fej ´ en- tal ´ altuka” (Hungary), and the proposers. An Infinite Product Based on a Base 11222 [2006, 459]. Proposed by Jonathan Sondow, New York, NY. Fix an integer B ≥ 2, and let s(n) denote the sum of the base-B digits of n. Prove that ∞ n=0 k odd 0<k<B nB + k nB + k + 1 (−1) s(n) = 1 √ B . Solution for odd B by the GCHQ Problem Solving Group Cheltenham, UK. For odd B we have n ≡ s(n) mod 2. Therefore we can write (−1) n rather than (−1) s(n) in the infinite product. Thus the corresponding infinite series ∞ n=0 (−1) n log k odd 0<k<B nB + k nB + k +1 = ∞ n=0 (−1) n+1 k odd 0<k<B log 1 + 1 nB + k converges by the classical alternating series test. Hence the product converges to a finite positive limit. By induction (for example), it follows for each positive integer M that 2M−1 n=0 k odd 0<k<B nB + k nB + k +1 (−1) n = (2MB)!2 2M (M!) 2 2 2MB (MB)!(MB)!(2M)! May 2008] PROBLEMS AND SOLUTIONS 465 Using Stirling’s Formula N !=N N e −N (2π N ) 1/2 {1 + O(1/N )}, we obtain 2M−1 n=0 k odd 0<k<B nB + k nB + k + 1 (−1) n = B −1/2 {1 + O(1/M)}. Also, for n > 0wehave log k odd 0<k<B nB + k nB + k + 1 (−1) n = k odd 0<k<B log 1 + 1 nB + k < k odd 0<k<B 1 nB + k < 1 n , so 2M n=0 k odd 0<k<B nB + k nB + k + 1 (−1) n = 2M−1 n=0 k odd 0<k<B nB + k nB + k + 1 (−1) n 1 + O(1/M) . Thus for any positive integer N , N n=0 k odd 0<k<B nB + k nB + k + 1 (−1) n = B −1/2 ( 1 + O(1/N ) ) , which yields the desired result with a good error estimate. Editorial comment. The case B = 2 was the subject of M ONTHLY Problem E2692 [85 (1978) 48 and 86 (1979) 394–395]. Other even values of B were treated in a paper by J. O. Shallit [J. Number Theory 21 (1985) 128–134]. The case of even B is somewhat more difficult than the case of odd B,sinceforevenB the exponent (−1) s(n) is a more complicated function of n than (−1) n . In particular the familiar alternating series test for establishing convergence of a series is no longer applicable. For a detailed treatment of even B we refer the reader to Shallit’s paper. Also solved (at least for odd values of B) by K. L. Bernstein, E. J. Ionascu, K. McInturff, M. A. Prasad (India), A. E. Stadler (Switzerland), A. L. Stenger, R. Tauraso (Italy), and the BSI Problems Group (Germany). The Square Root of a Certain Matrix 11224 [2006, 459]. Proposed by Dietrich Trenkler, University of Osnabr ¨ uck, Os- nabr ¨ uck, Germany, and G ¨ otz Trenkler, University of Dortmund, Dortmund, Germany. Let a and b be linearly independent column vectors in R 3 . Find a formula in terms of a and b for a square matrix A such that A 2 = ba t − ab t .(Here,x t denotes the transpose of x.) Solution by Patrick Corn, University of Georgia, Athens, GA. Let M = ba t − ab t .Let k =a × b (note that k > 0 by linear independence). We compute M 3 = ba t (a · b) 2 − bb t (a · b)(a · a) − ba t (a · a)(b · b) + bb t (a · a)(a · b) − aa t (b · b)(a · b) + ab t (a · a)(b · b) + aa t (a · b)(b · b) − ab t (a · b) 2 = (ba t − ab t ) (a · b) 2 − (a · a)(b · b) , so M 3 + k 2 M = M (a · b) 2 − (a · a)(b · b) +a × b 2 = 0. 466 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 Letting A = (2k 3 ) −1/2 (M 2 − kM),wehave A 2 − M = (2k 3 ) −1 (M 4 − 2kM 3 + k 2 M 2 − 2k 3 M) = (2k 3 ) −1 (M − 2kI)(M 3 + k 2 M) = 0, as desired (here I is the identity matrix). Writing out A in terms of a and b, we obtain A = 1 √ 2a × b 3/2 (ba t − ab t ) 2 −a × b(ba t − ab t ) . Editorial comment. The restriction to vectors in R 3 is unnecessary. Also solved by S. Amghibech (Canada), C.P. Anilkumar (India), M. Bataille (France), S.J., Bernau, D. R. Bridges, R. Chapman, (U.K.) P. Corn, K. Dale (Norway), J.A. Grzesik, E.A. Herman, K.J. Heuvers, R.A. Horn, G.L. Isaacs, G. Keselman, T. Kezlan, J.H. Lindsey II, O.P. Lossers (The Netherlands), P. Magli (Italy), J. McHugh, K. McInturff, M.D. Meyerson, K. Onneweer, L. Pebody, N.C. Singer, A. Stadler (Switzerland), R. Stong, J. Stuart, T. Tam, N. Thornber, E.I. Verriest, W.C. Waterhouse, M. Woltermann, J.B. Zacharias, BSI Problems Group (Germany) GCHQ Problem Solving Group (U.K.), and the proposer. A New Bound for r/R 11245 [2006, 760]. Proposed by Cezar Lupu, University of Bucharest, Bucharest, Ro- mania, and Tudorel Lupu, Decebal High School, Constanza, Romania. Consider an acute triangle with sides of lengths a, b,andc, and with an inradius of r and a circum- radius of R. Show that r R ≤ 2(2a 2 − (b − c) 2 )(2b 2 − (c − a) 2 )(2c 2 − (a − b) 2 ) (a + b)(b + c )(c + a) . Solution by Ronald A. Kopas, Clarion University, Clarion, PA. Side a is the sum of the two segments on either side of the point of tangency with the incircle, so a = r cot B 2 + cot C 2 = r cos(B/2) sin(C/2) + cos(C/2) sin(B/2) sin(B/2) sin(C/2) = r sin(B/2 +C/2) sin(B/2) sin(C/2) = r sin(π/2 − A/2) sin(B/2) sin(C/2) = r cos(A/2) sin(B/2) sin(C/2) , and r = a sin(B/2) sin(C/2)/ cos( A/2). From the law of sines, a = 2R sin A = 4R sin(A/2) cos( A/2),soR = a/ 4sin(A/2) cos(A/2) . Therefore r R = 4sin A 2 sin B 2 sin C 2 . (1) Now we compute a lower bound for E = 2a 2 − (b − c) 2 .Wehave E = 2 a 2 − (b − c) 2 + (b − c) 2 = 2 b 2 + c 2 − 2bc cos A − b 2 − c 2 + 2bc + (b − c) 2 = 4bc ( 1 −cos A ) + (b − c) 2 = (b + c) 2 − (b − c) 2 (1 − cos A) + (b − c) 2 = (b + c) 2 (1 − cos A) + (b − c) 2 cos A = (b + c) 2 2sin 2 A 2 + (b − c) 2 cos A ≥ (b + c) 2 2sin 2 A 2 , noting that cos A ≥ 0 since the triangle is acute. May 2008] PROBLEMS AND SOLUTIONS 467 Therefore √ 2sin( A/2) ≤ 2a 2 − (b − c) 2 /(b + c). Similarly √ 2sin(B/2) ≤ 2b 2 − (c − a) 2 /(c + a) and √ 2sin(C/2) ≤ 2c 2 − (a − b) 2 /(a + b). Multiply the product of these three inequalities by √ 2 and apply (1) to obtain the required result. Editorial comment. In fact the result holds for right and obtuse triangles as well. Four solvers (V. Schindler, R. Stong, R. Tauraso, and the Microsoft Research Problems Group) provided proofs for this stronger result, but their proofs were all computer assisted and much longer than the proof given here. Also solved by A. Alt, S. Amghibech (Canada), O. Bagdasar (Romania), P. P. D ´ alyay (Hungary), P. De (Ire- land),J.Fabrykowski&T.Smotzer,V.V.Garc ´ ıa (Spain), N. Lakshmanan (India), O. P. Lossers (The Nether- lands), J. Minkus, C. R. Pranesachar (India), T L. & V. R ˘ adulescu (Romania), J. Rooin & A. Emami (Iran), V. Schindler (Germany), R. Stong, R. Tauraso (Italy), D. V ˘ acaru (Romania), L. Zhou, Microsoft Research Problems Group, Northwestern University Math Problem Solving Group, and the proposer. Evaluate the Series 11260 [2006, 939]. Proposed by Paolo Perfetti, Mathematics Department, University “Tor Vergata,” Rome, Italy. Find those nonnegative values of α and β for which ∞ n=1 n k=1 α + k log k β + (k + 1) log(k + 1) converges. For those values of α and β,evaluatethesum. Solution by Richard Stong, Rice University, Houston, TX. If α ≥ β,then ∞ n=1 n k=1 α +k log k β + (k + 1) log(k + 1) ≥ ∞ n=1 α α +k log k , which has all positive terms and diverges. So assume β>α≥ 0. Now n k=1 α +k log k β + (k + 1) log(k + 1) = β β − α n k=1 α +k log k β + k log k − n+1 k=1 α +k log k β + k log k , so we have N −1 n=1 n k=1 α +k log k β + (k + 1) log(k + 1) = α β − α − β β − α N k=1 α +k log k β + k log k . Since ∞ k=1 (β − α)/(β + k log k) has all positive terms and diverges, we have ∞ k=1 1 − β − α β + k log k = ∞ k=1 α +k log k β + k log k = 0. Hence for β>α≥ 0, we have convergence: ∞ n=1 n k=1 α + k log k β + (k + 1) log(k + 1) = α β − α . Also solved by O. Furdui, J P. Grivaux (France), O. Kouba (Syria), J. H. Lindsey II, O. P. Lossers (The Netherlands), A. Stadler (Switzerland), Szeged Problem Solving Group “Fej ´ ental ´ altuka” (Hungary), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, and the proposer. 468 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 . should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before September 30, 2008. Additional information, such. that if p = 6t + 1, then ( p −1)! 2t−1 j =0 (−1) j 1 3 j + 1 + 1 3 j + 2 ≡ 0 (mod p). May 2008] PROBLEMS AND SOLUTIONS 461 (b) Show that if p = 6t − 1, then ( p −1)! 2t−1 j =0 (−1) j 3. are the same. We start with the one on the left. 462 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 We first convert the power series to a continued fraction. With a = (a 0 , a 1 , ),let p n (a)