Problems Source: Mathematics Magazine, Vol. 85, No. 2 (April 2012 ...

Problems Source: Mathematics Magazine, Vol. 85, No. 2 (April 2012), pp. 150-157 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/10.4169/math.mag.85.2.150 . Accessed: 21/10/2014 02:47 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp

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PROBLEMS ´ BERNARDO M. ABREGO, Editor California State University, Northridge

´ Assistant Editors: SILVIA FERNANDEZ-MERCHANT,

California State University, North-

´ ´ A. GOMEZ, ridge; JOSE Facultad de Ciencias, UNAM, México; EUGEN J. IONASCU, Columbus State University; ROGELIO VALDEZ, Facultad de Ciencias, UAEM, México; WILLIAM WATKINS, California State University, Northridge

PROPOSALS To be considered for publication, solutions should be received by September 1, 2012. 1891. Proposed by Raúl A. Simón, Santiago, Chile. Let ABCD be a quadrilateral in the plane. Let A0 , B 0 , C 0 , and D 0 be the centroids of the triangles BCD, ACD, ABD, and ABC, respectively. Prove that quadrilaterals ABCD and A0 B 0 C 0 D 0 are similar with corresponding sides in the ratio 3 : 1. 1892. Proposed by José Luis D´ıaz-Barrero, Applied Mathematics III, Polytechnical University of Catalonia, Barcelona, Spain. Compute the limit √ ! √ n 1 Y n n + (n + 1) k . lim √ √ n→∞ n n n+ k k=1 1893. Proposed by Jerrold W. Grossman and László Lipták, Oakland University, Rochester, MI, and Mike Shaughnessy, Portland State University, Portland, OR. Let n be an integer greater than 1, and let f (a1 , a2 , . . . , an ) =

Y

(ai − a j ).

1≤i< j≤n

What is the greatest common divisor of f (a1 , a2 , . . . , an ) over all choices of distinct integers a1 , a2 , . . . , an ? c Mathematical Association of America Math. Mag. 85 (2012) 150–157. doi:10.4169/math.mag.85.2.150. We invite readers to submit problems believed to be new and appealing to students and teachers of advanced undergraduate mathematics. Proposals must, in general, be accompanied by solutions and by any bibliographical information that will assist the editors and referees. A problem submitted as a Quickie should have an unexpected, succinct solution. Submitted problems should not be under consideration for publication elsewhere. Solutions should be written in a style appropriate for this M AGAZINE. ´ Solutions and new proposals should be mailed to Bernardo M. Abrego, Problems Editor, Department of Mathematics, California State University, Northridge, 18111 Nordhoff St, Northridge, CA 91330-8313, or mailed electronically (ideally as a LATEX or pdf file) to [email protected]. All communications, written or electronic, should include on each page the reader’s name, full address, and an e-mail address and/or FAX number.

150


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1894. Proposed by Michael W. Botsko, Saint Vincent College, Latrobe, PA. Let (X, T) be a topological space, let S be any subset of X , and let iso(S) be the set of isolated points of S. (A point x ∈ S is an isolated point of S if there exists an open set O such that O ∩ S = {x}.) (a) If (X, T) is second countable, prove that iso(S) is countable. (A topological space (X, T) is second countable if it has a countable open base.) (b) If S is a closed set, prove that S \ iso(S) is a closed set. (c) Prove that the conclusion in (a) does not hold if (X, T) is only assumed to be separable. (A topological space (X, T) is separable if it has a countable dense subset.) (d) Prove that the conclusion in (b) does not hold if S is not assumed to be closed. 1895. Proposed by Steven Finch, Harvard University, Cambridge, MA Let ` denote a planar line with slope tan(θ) and x-intercept ξ , where θ and ξ are independent random variables with uniform distributions over the intervals [π/4, 3π/4] and [−1, 1], respectively. Let `1 , `2 , and `3 be independent copies of `. These three lines determine a compact triangle 1 almost surely. Find the probability density function for the maximum angle α in 1. Find the first and second moments of α as well.

Quickies Answers to the Quickies are on page 157. Q1019. Proposed by Michael W. Botsko, Saint Vincent College, Latrobe, PA. Let f : [a, b] → R be a real valued function and 0 = {(x, f (x)) : x ∈ [a, b]} the graph of f . Suppose that 0 is a closed and bounded subset of R2 . Prove or disprove that f is continuous on [a, b]. Q1020. Proposed by Michael W. Ecker, Mathematics Department, Pennsylvania State University, Lehman, PA. Find all values of n for which the sum 1 + 2 + · · · + n is an integer power of 10.

Solutions Variations on the Steiner–Lehmus Theorem

April 2011

1866. Proposed by Sadi Abu-Saymeh and Mowaffaq Hajja, Mathematics Department, Yarmouk University, Irbid, Jordan. Let ABC be a triangle, and L and M interior points on AB and AC, respectively, such that AL = AM. Let P be the intersection of BM and CL. Prove that PB = PC if and only if AB = AC. Solution by I. E. Leonard, J. E. Lewis, A. C. F. Liu, and G. Tokarsky, University of Alberta, Edmonton, Alberta, Canada. Suppose that AB = AC. (See F IGURE (a).) Hence triangle ABC is isosceles. It follows that BL = AB − AL = AC − AM = CM, and so by the SAS Congruency Theorem, 4LBC ∼ = 4MCB. Therefore, ∠PCB = ∠PBC which means that 4BPC is isosceles, and thus BP = PC.


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MATHEMATICS MAGAZINE

Conversely, suppose that AL = AM, PB = PC but, without loss of generality, AB > AC. (See F IGURE (b).) Let D be the point in the ray AC such that AB = AD. Let Q be the intersection of BM and DL. By the sufficiency part BQ = DQ. Because C is between M and D, it follows that P is between Q and M. Hence, by the triangle inequality, PB = PQ + QB = PQ + QD > PD. An exterior angle of a triangle is greater than either of the opposite interior angles, and triangles ALM and ABD are similar isosceles triangles. Thus ∠DCP > ∠ALC > ∠ALM = ∠ADB > ∠CDP. In a triangle, the larger angle is opposite to the longer side, thus ∠DCP > ∠CDP implies that PD > PC, and therefore PB > PC. Editor’s Note. As pointed out by Eugene A. Herman the statement of the problem needs to assume that L 6 = B or M 6 = C, otherwise P is not uniquely defined. We modified the statement to avoid this ambiguity. We received many solutions which used a similar argument or end with an application of Ceva’s Theorem instead. Many solutions were based on relatively hard calculations. This problem is closely related to the classical problem (Steiner–Lehmus Theorem) of showing that two angle bisectors in a triangle are equal if and only if the triangle is isosceles. Also solved by Dionne Bailey, Michel Battaille (France), Elton Bojaxhiu (Germany) and Enkel Hysnelaj (Australia), Bruce S. Burdick, Robert Calcaterra, CMC 328, Tim Cross (United Kingdom), David Doster, Robert L. Doucette, Thomas Gettys, Ahmad Habil (Syria), Eugene A. Herman, Jeong Hwan Kim (Korea), L. R. King, Omran Kouba (Syria), Victor Y. Kutsenok, Elias Lampakis (Greece), Kee-Wai Lau (China), Longxiang Li and Luyuan Yu (China), Graham Lord, Masao Mabuchi (Japan), Kathleen Marsh, Joel Schlosberg, Vasile G. Teodorovici (Canada), Stuart V. Witt, Michael Woltermann, and the proposers. There were eight incomplete or incorrect submissions.

A Mean Value Theorem for n values

April 2011

´ 1867. Proposed by Angel Plaza and César Rodr´ıguez, Department of Mathematics, Universidad de Las Palmas de Gran Canaria, Las Palmas, Spain. R1 Let f : [0, 1] → R be a continuous function such that 0 f (t) dt = 1 and n a positive integer. Show that 1. there are distinct c1 , c2 , . . . , cn in (0, 1) such that f (c1 ) + f (c2 ) + · · · + f (cn ) = n, 2. there are distinct c1 , c2 , . . . , cn in (0, 1) such that 1 1 1 + + ··· + = n. f (c1 ) f (c2 ) f (cn )


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I. Solution by Tania Moreno García, University of Holguín, Cuba, and Jose Pablo Suárez, University of Las Palmas de Gran Canaria, Spain. By the Mean Value Theorem, for each 1 ≤ k ≤ n, there exists ck ∈ ((k − 1)/n, k/n) such that Z k/n Z k/n 1 f (t) dt = n f (t) dt = f (ck ). k/n − (k − 1)/n (k−1)/n (k−1)/n Adding these identities gives n=n

1

Z 0

f (t) dt = n

n Z X

k/n (k−1)/n

k=1

f (t) dt =

n X

f (ck ).

k=1

This proves the first part. For the second part, note that there are numbers R0 = a0 < a1 < . . . < an−1 < an = Ra ak 1 such that 0 k f (t)dt = k/n for 1 ≤ k ≤ n. Thus ak−1 f (t)dt = 1/n for 1 ≤ k ≤ n. Applying the Mean Value Theorem to each of these integrals shows that there are ck ∈ (ak−1 , ak ) such that Z ak 1 1 = f (t) dt = f (ck ); n(ak − ak−1 ) ak − ak−1 ak−1 that is 1/ f (ck ) = n(ak − ak−1 ). Adding these identities gives n X k=1

n

X 1 =n (ak − ak−1 ) = n. f (ck ) k=1

II. Solution by Arnold Adelberg and Eugene A. Herman, Department of Mathematics, Grinnell College, Grinnell, IA. We prove a more general result: Suppose g is a continuous, real-valued function on an open interval I , y is in g(I ), and n is a positive integer. If either g is constant on I or y is not an extreme value of g, then there are distinct c1 , c2 , . . . , cn in I such that g(c1 ) + g(c2 ) + · · · + g(cn ) = ny.

(1)

In other words, any non-extremal value of a continuous non-constant function is the average of n nearby values, for any n. If g is constant, then equation (1) holds for any c1 , c2 , . . . , cn in I . Otherwise, since g(I ) is an interval and y is not an endpoint, there exists δ > P 0 such that (y−δ, y+δ) ⊆ g(I ). Thus, if a1 , a2 , .P . . , an are distinct real numbers with nk=1 ak = 0 and |ak | < δ for 1 ≤ k ≤ n, then nk=1 (y + ak ) = ny and y + ak ∈ g(I ) for 1 ≤ k ≤ n. Since y + ak = g(ck ) for distinct ck ∈ I , the result follows. This result implies the statement in the problem as follows. By the Mean Value R1 Theorem, there exists a c ∈ (0, 1) such that 1 = 0 f (x) d x = f (c). By continuity, if f is not constant, then there are a, b ∈ (0, 1) such that f (a) < 1 < f (b). Also by continuity we may assume that f > 0 on (a, b). For the first part in the problem, let g = f , I = (0, 1), and y = 1; for the second part, let g = 1/ f , I = (a, b), and y = 1. Also solved by Michael Andreoli; Dionne Bailey, Elsie Campbell, and Charles Diminnie; Michel Bataille (part 1 only, France); Elton Bojaxhiu (Germany) and Enkel Hysnelaj (Australia); Michael W. Botsko and Larry J. Mismas; Paul Budney; Bruce S. Burdick; Charles Burnette; Robert Calcaterra; John Christopher; Con Amore Problem Group (Denmark); Charles Degenkolb; Patrick Devlin; Robert L. Doucette; Dmitry Fleischman; William R. Green; Lee O. Hagglund; Mowaffaq Hajja and Mostafa Hayajneh (Jordan); Robert Jones;


154


John C. Kieffer; Michael Knapp; Omran Kouba (Syria); Elias Lampakis (Greece); Longxiang Li and Luyuan Yu (China); Jody M. Lockhart; Raymond Maruca; Cristinel Mortici (Romania); Raymond Mortini (France); Edwin Gonzalo and Murcia Rodr´ıguez (Colombia); Northwestern University Math Problem Solving Group; Joel Schlosberg; Allen Stenger; Philip Straffin; Haryono Tandra; Marian Tetiva (Romania); Texas State University Problem Solvers Group; Dave Trautman; Alfred Witkowski (Poland); Stuart V. Witt; and the proposers.

Tiling a square ring with dominoes

April 2011

1868. Proposed by Donald E. Knuth, Stanford University, Stanford, CA. Let n ≥ 2 be an integer. Remove the central (n − 2)2 squares from an (n + 2) × (n + 2) array of squares. In how many ways can the remaining squares be covered with 4n dominoes? Solution by John Bonomo and David Offner, Department of Mathematics and Computer Science, Westminster College, New Wilmington, PA. The total number of coverings is 16Fn4 + 16(−1)n Fn2 + 4 = 4(2Fn2 + (−1)n )2 , where Fn is the n th Fibonacci number, defined recursively as F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. For every n, there are exactly two ways to cover the array so that no pair of dominoes covers a 2 × 2 sub-array.

The remaining cases can be divided according to the orientation of the dominoes covering the four corner squares of the array. Since each corner square of the array may be covered by a domino oriented horizontally or vertically, there are 16 possible configurations. These configurations can be partitioned into four symmetry classes: Two of type A, eight of type B, four of type C, and two of type D according to the figure below. To count the number of coverings for each configuration, it remains to count the number of ways to cover the rectangular arrays outlined by dashes in the figure.

It is well known that the number of ways to cover a 2 × m array of squares with m dominoes is Fm+1 . So, for configurations of type A, there are two 2 × n rectangles, and two 2 × (n − 2) rectangles, and the total number of coverings of this type is 2 2 2Fn+1 Fn−1 . Similarly, there are 8Fn+1 Fn2 Fn−1 coverings of type B, 4Fn+1 Fn2 Fn−1 coverings of type C, and 2Fn4 coverings of type D, for a total number of coverings of 2 2 2Fn+1 Fn−1 + 12Fn+1 Fn2 Fn−1 + 2Fn4 + 2.

This total is equivalent to the answer given above, as is shown by using Cassini’s Identity Fn+1 Fn−1 − Fn2 = (−1)n .


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Editor’s Note. Some solvers pointed out that this problem was solved in Roberto Tauraso, A New Domino Tiling Sequence, Journal of Integer Sequences 7 (2004), Article 04.2.3. The sequence of solutions also appears as sequence A061646 in the Sloane’s On-Line Encyclopedia of Integer Sequences. Also solved by Elton Bojaxhiu (Germany) and Enkel Hysnelaj (Australia), Bruce S. Burdick, Con Amore Problem Group (Denmark), Marty Getz and Dixon Jones, Amanda Goodrick, G.R.A.20 Problem Solving Group (Italy), Harris Kwong, David Nacin, Joel Schlosberg, John H. Smith, Philip Straffin, and the proposer. There was one incorrect submission.

A triangle inequality for triangle side-lengths

April 2011

1869. Proposed by Marian Dinc˘a, Bucharest, Romania. Let f : R → R be an increasing and concave-down function such that f (0) = 0. Prove that if x, y, and z are real numbers, and a, b, and c are the lengths of the sides of a triangle, then (x − y)(x − z) f (a) + (y − x)(y − z) f (b) + (z − x)(z − y) f (c) ≥ 0. Solution by Charles Burnette, Philadelphia, PA. Recall that if a function f : R → R is concave-down, and f (0) ≥ 0, then f is subadditive. Combining this with the fact that f is increasing and that a, b, and c are side lengths of a triangle, we have that f (a) < f (b + c) ≤ f (b) + f (c). Similarly f (b) < f (a) + f (c) and f (c) < f (a) + f (b). Hence f (b) + f (c) − f (a) > 0, f (c) + f (a) − f (b) > 0, and f (a) + f (b) − f (c) > 0. We now employ the following notation: X (x − y)(x − z) f (a) cyc

= (x − y)(x − z) f (a) + (y − x)(y − z) f (b) + (z − x)(z − y) f (c), where the sum is taken over all cyclicP permutations of (a, b, c) and (x, y, z) where (a, b, c) corresponds to (x, y, z). Thus cyc (x − y)(x − z) f (a) equals X1 cyc

2

(x − y)(x − z) [( f (c) + f (a) − f (b)) + ( f (a) + f (b) − f (c))] =

=

1X (x − y)(x − z) ( f (c) + f (a) − f (b)) 2 cyc 1X (x − y)(x − z) ( f (a) + f (b) − f (c)) + 2 cyc 1X (z − x)(z − y) ( f (b) + f (c) − f (a)) 2 cyc 1X + (y − z)(y − x) ( f (b) + f (c) − f (a)) 2 cyc

=

1X [(z − x)(z − y) + (y − z)(y − x)] ( f (b) + f (c) − f (a)) 2 cyc

=

1X (y − z)2 (( f (b) + f (c) − f (a)) ≥ 0, 2 cyc

as desired.


156


Also solved by Dionne Bailey, Elsie Campbell, and Charles Diminnie; Michel Bataille (France); Bruce S. Burdick; Robert Calcaterra; Con Amore Problem Group (Denmark); Dmitry Fleischman; Paul Loomis; Scott Pauley, Natalya Weir, and Andrew Welter; Paolo Perfetti (Italy); Joel Schlosberg; Haohao Wang and Jerzy Wojdyło; Stuart V. Witt; and the proposer. There were two incorrect submissions.

A double zeta sum

April 2011

1870. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Calculate ∞ X ∞ X m(ζ (n + m) − 1)

(n + m)2

n=1 m=1

,

where ζ denotes the Riemann Zeta function. Solution by Joel Schlosberg, Bayside, NY. For x ∈ (0, 1), the power series for x/(1 − x) and ln(1 − x) converge absolutely, so ∞

∞

∞

∞

X X xk Xk−1 Xk−1 x + ln(1 − x) = xk − = xk = xk. 1−x k k k k=1 k=1 k=1 k=2 The terms of the double series are all positive. So we may rearrange the order of summation. Then ∞ X ∞ X m(ζ (n + m) − 1) n=1 m=1

=

(n + m)2 ∞ X ∞ X ∞ X j=2 m=1 n=1

=

∞ X ∞ X n=1 m=1

∞ X m 1 (n + m)2 j=2 j n+m

∞ X ∞ X ∞ X m m = 2 jk (n + m)2 j n+m k j=2 m=1 k=m+1

∞ X ∞ X k−1 ∞ X ∞ 1 ∞ ∞ X X (k − 1)k 1 X X k − 1 −1 k m 2 = = = (j ) k2 j k k2 j k 2 j=2 k=2 k j=2 k=2 m=1 j=2 k=2

∞ ∞ 1X j −1 1X 1 −1 = + ln(1 − j ) = + ln( j − 1) − ln j 2 j=2 1 − j −1 2 j=2 j − 1 N X 1 1 = lim + ln( j − 1) − ln j 2 N →∞ j=2 j − 1    N X 1 1 1 γ = lim  − ln N  −  = , N →∞ 2 j N 2 j=1 where γ = lim N →∞ (

PN

n=1

1/n − ln N ) is the Euler-Mascheroni constant.

Also solved by Michel Bataille (France), Bruce S. Burdick, Charles Burnette, Dmitry Fleischman, Michael Goldenberg and Mark Kaplan, G.R.A.20 Problem Solving Group (Italy), Eugene A. Herman, Enkel Hysnelaj (Australia) and Elton Bojaxhiu (Germany), Omran Kouba (Syria), Peter W. Lindstrom, Rituraj Nandan, Paolo Perfetti (Italy), Nicholas C. Singer, Satyanand Singh, Michael Vowe (Switzerland), and the proposer.


VOL. 85, NO. 2, APRIL 2012

157

Answers Solutions to the Quickies from page 151. A1019. We prove that f is continuous on [a, b]. Define the function g : 0 → [a, b] such that g((x, f (x))) = x. Clearly g is both one-to-one and onto and thus the inverse function g −1 : [a, b] → 0 exists. Using the sequential definition of continuity, it follows that g is continuous on 0. Thus g is a homeomorphism because it is a one-to-one continuous mapping of a compact space onto a Hausdorff space. Since g is a homeomorphism, g −1 is continuous on [a, b]. However, since g −1 (y) = (y, f (y)), it follows that f is continuous on [a, b]. A1020. The only solutions are n = 1 and n = 4. Indeed 1 = 100 and 1 + 2 + 3 + 4 = 101 . To prove no others exist, suppose Tn := 1 + 2 + · · · + n = 21 n(n + 1) = 10k for some nonnegative integer k. If n ≥ 2, then Tn > 1 and thus we may assume that k ≥ 1. Then n(n + 1) = 2 · 10k = 2k+1 · 5k . Since n and n + 1 relatively prime, we must have n = 2k+1 and n + 1 = 5k , as 5k > 2k+1 precludes the case of n + 1 = 2k+1 and n = 5k . Thus, 5k − 2k+1 = 1. However, 5k = (22 + 1)k ≥ 22k + 1 ≥ 2k+1 + 1 where the equalities hold if and only if k = 1.

To appear in College Mathematics Journal, May 2012 Articles Rediscovering Pascal’s Mystic Hexagon by Michael Augros The Finite Lamplighter Groups: A Guided Tour by Jacob A. Siehler Proof Without Words: The Square of a Balancing Number is a Triangular Number by Michael A. Jones When Abelian = Hausdorff by Timothy Kohl The Catenary as Roulette by Javier Sánchez-Reyes Partitioning Pythagorean Triangles Using Pythagorean Angles by Carl E. Swenson and André L. Yandl Proof Without Words: The Pythagorean Theorem with Equilateral Triangles by Claudi Alsina and Roger B. Nelsen Push-To Telescope Mathematics by Donald Teets The Distribution of the Sum of Signed Ranks by Brian Albright Harmonic Series Meets Fibonacci Sequence by Hongwei Chen and Chris Kennedy The Basel Problem as a Telescoping Series by David Benko
