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1 Problems Ted Eisenberg, Section Editor ********************************************************* This section of the Journal offers readers an opportunity to exchange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fax to: Questions concerning proposals and/or solutions can be sent to Solutions to previously stated problems can be seen at < Solutions to the problems stated in this issue should be posted before May 5, 05 54: Proposed by Kenneth Korbin, New York, NY Four different Pythagorean Triangles each have hypotenuse equal to 4p 4 + where p is prime. Express the sides of these triangles in terms of p. 544: Proposed by Y. N. Aliyev, Qafqaz University, Khyrdalan, Azerbaijan Let ABC be isosceles with AB AC. Let D be a point on side BC. A line through point D intersects rays AB and AC at points E and F respectively. Prove that ED DF BD DC. 545: Proposed by Arkady Alt, San Jose, CA Let a, b > 0. Prove that for any x, y the following inequality holds and find when equality occurs. a cos x + b cos y a + b + ab cosx + y, 546: Proposed by D.M. Bătinetu-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade School, Buzău, Romania Show that in any triangle ABC, with the usual notations, the following hold, h b + h c h a ra + h c + h a rb h + h a + h b rc s, b h c where r a is the excircle tangent to side a of the triangle and s is the triangle s semiperimeter. 547: Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Let 0 < a < b be real numbers and let f, g : [a, b] R + be continuous functions. Prove

2 that there exists c a, b such that fc + b gc + gt dt c c a ft dt 4 R + represents the set of non-negative real numbers. 548: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania Let k be an integer. Prove that 0 ln k x ln x dx k+ k!k + ζ ζ ζk +, where ζ denotes the Riemann zeta function. Solutions 55: Proposed by Kenneth Korbin, New York, NY Given the sequence x, 7, 4, 9, 9, 89,..., with x n 6x n x n. Let y x n + x n x n. Find an explicit formula for y expressed in terms of n. Solution by D.M. Bătinetu-Giurgiu, National College Matei Basarab, Bucharest, Romania The recurrence sequence x n has the equation r 6r + 0 with solutions r +, r, so x n ur n + vr n + n u + n v, and because x, x 7 yields that + u, v,. Therefore, x n n n +, and y n x n + x n x n + n + n,

3 and we are done. Solution by Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX In our solution to Problem 508 see Dec. 04 issue of this column, we used the techniques for solving homogeneous linear difference equations to show that the closed form expression for x n is x n for all n. It follows that for all n, + n n 4n 4n 4n 4n + + x n + x n + 4n [ ] [ 4n ] + Therefore, for all n. 4n [ ] 4n [ ] + + 4n + + 4n [ 4n ] 4n + [ n ] [ n n ] n [ n ] n x n + +. y x n + x n x n [ n ] n + + Solution by G. C. Greubel, Newport News, VA First consider the difference equation x n+ 6x n x n which has the general solution x n Aa n + Bb n where a + and b. For the initial conditions x 0 and x 7 the sequence x n has the solution x n Q n+ /, where Q n are the Pell-Lucas numbers with the recurrence relation Q n+ Q n+ + Q n. The element x n+ + x n can be determined to be 4P 4n, where P n are the Pell numbers. This leads to the desired quantity being sought as y n 8P 4n Q n+.

4 Comment by Henry Ricardo, New York Math Circle, NY. The numbers x n in the proposed problem are the NSQ numbers named for Newman, Shanks, and Williams, authors of an influential 980 group theory paper. The On-Line Enopedia of Integer Sequences OEIS lists the sequence as entry A005 and gives the formula without proof x n + + n + n + n+ + n+, for non-negative integers n. In addition to many comments on the sequence itself, the connection between this sequence and other OEIS entries are also pointed out. Also solved by Arkady Alt, San Jose, CA; Brian D. Beasley, Presbyterian College, Clinton, SC; Bruno Salgueiro Fanego, Viveiro, Spain; Ethan Gegner student at Taylor University, Upland IN; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton, KS; Tsvetelina Karamfilova, Petko Rachov Slaveikov Secondary School, Kardzhali, Bulgaria; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel; Carl Libis of the University of Tennessee at Martin, TN; David E. Manes, SUNY College at Oneonta, Oneonta, NY; Corneliu Mănescu-Avram, Transportation High School Ploiesti, Romania; Angel Plaza, Universidad de Las Palmas, de Gran, Canaria, Spain; Henry Ricardo, New York Math Circle, NY; Neculai Stanciu, George Emil Palade School, Buzău, Romania jointly with Titu Zvonaru, Comănesti, Romania; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA; Albert Stadler, Herrliberg, Switzerland, 56: Proposed by Armend Sh. Shabani, University of Prishtina, Republic of Kosova Find all positive integer solutions to m! + 4k l. Solution by Ed Gray, Highland Beach, FL We note that 4k always seems to end with the integer 8 for all values of k. We prove this by induction. The statement is obviously true for k. Assume that the statement is true for all positive integers up to and including k. I.e. 4k ends in with the integer 8. Does this 4k imply that 4k+ also ends with the integer 8? 4k+ 4k+ 4 4k 6 4k. But by the induction hypothesis, 4k ends with integer 8 and so 6 4n also ends with the integer 8. Now we note that for all integers m 5, the integer m! ends with the integer 0. So, m! + 4k ends in 8 for all integers m 5. But there is no square number whose units digit is 8. So if there are any integer solutions to m! + 4k l, the value of the positive integer m must be,,, or 4. If m 4, then m! 4 ends in a 4 and so m! + 4k ends with the unit s digit in 4+8, so l must end in, but there is no integer whose square ends with a So m 4. If m, then m! 6 and so m! + 4k ends with a unit s digit of 6+8. That is, the units digit of l must be 4, which implies that l must be even. Suppose that l r. Then 6 + 4k l 4

5 6 + 4k r 6 + 4k 4r But 4 divides the right hand side and 4 divides 4k, but 4 does not divide 6 so, m. If m, then m! 4 and so m! + 4k ends with a unit s digit of, but there is no integer square has a units digit of. So,m. Finally, if m then m! + 4k becomes + 4k l 4k l 4k l l +. So, both factors l and l + must be a power of. Let l a and l + b. Subtracting gives b a whose only solution is b and a. So l and l + Since 4k l l + 4k 4 4k, so, k. The only solution to m! + 4k l is when m, k and l. Solution by Jerry Chu, student at Saint George s School, Spokane, WA We note that 4k mod is. And l mod is either 0 or. So, m! must not be a multiple of. Therefore, m or. When m, 4k l l + l. Because l and l + can only be powers of, l must equal. So m, k, l. When m, we take all terms mod 4 and see that , which is impossible. Therefore the only solution is m, k, l. Solution by Adnan Ali student in A.E.C.S-4, Mumbai, India Assume that for m, there exist solutions. Then putting the equation modulo, we see that l m! + 4k 4k mod 5

6 but is not a quadratic residue modulo. So we conclude that m. But now we may assume that there is a solution for m, then we simply realize the fact that l! + 4k + 0 mod 4, and since is not a quadratic residue modulo 4, we are left with the only option m. So, we have 4k + l l + l 4k and so we must have both l +, l as powers of, so we let l + a > l b for integers a, b so that a + b 4k + and see that a b b a b forcing b and a b which has the only solution a, b, and + 4k implies that k. So we conclude that the only possible solution is l, m, k,,. Solution 4 by Henry Ricardo, New York Math Circle, NY The triple m, k, l,, is the only solution in positive integers. To prove this assertion, we use the following easily established facts: 4k 8 mod 0 for positive integers k; If l r mod 0, then r S {0,, 4, 5, 6, 9}. First, if m 5, then m! 0 mod 0 so that m! + 4k 8 mod 0. But 8 / S. Thus m < 5. If m, then N m! + 4k + 4k, which can t be a perfect square since + 4k is odd, implying that the prime divisor does not appear with an even exponent in the prime power factorization of N. Similarly, if m, then m! + 4k + 4k, which can t be a perfect square. Finally, we eliminate m 4 since m! + 4k 4 + 4k mod 0 and / S. Also solved by Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX; Brian D. Beasley, Presbyterian College, Clinton, SC; Ethan Gegner student, Taylor University, Upland, IN; Paul M. Harms, North Newton, KS; Bruno Salgueiro Fanego, Viveiro, Spain; Jahangeer Kholdi and Farideh Firoozbakht, University of Isfahan, Khansar, Iran; Kee-Wai Lau, Hong Kong, China; David E. Manes, SUNY College at Oneonta, Oneonta, NY; Corneliu Mănescu-Avram, Transportation High School Ploiesti, Romania; Haroun Meghaichi student, University of Science and Technology Houari Boumediene, Algeria; Neculai Stanciu, George Emil Palade School, Buzău, Romania jointly with Titu Zvonaru, Comănesti, Romania; Albert Stadler, Herrliberg, Switzerland, and the proposer. 57: Proposed by D.M. Bătinetu-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade School, Buzău, Romania Show that in any triangle ABC, with the usual notations, that ab bc ca + + 9r. a + b b + c c + a 6

7 Solution by Kee-Wai Lau, Hong Kong, China By the Cauchy-Schwarz inequality, we have ab bc ca + + ab a + b b + c c + a a + b + Hence it suffices to show that ab a + b + bc b + c + bc b + c + ca c + a r. ca. c + a Let s be the semiperimeter and F the area of triangle ABC. It is well known that F rs ab sin C bc sin A ca sin B. So is equivalent to a + b + c a + b sin C + b + c sin A + c + a sin B Let S sin A + sin B + sin C and S a sin Ab + c + b sin Bc + a +. c sin Ca + b so that the left side of can be written as S + S. By the convexity of the function sin x, for 0 < x < π, we have S sin. By the sine formula, A+B+C we have S sin B + sin C + sin C + sin A + sin A + sin B sin B+C sin + B+C sec cos B C A + sec + sin C+A B + sin C+A cos C A + sin A+B C. + sin A+B cos A B Hence by the convexity of the function sec x for 0 < x < π, we have S A + B + C sec. 6 Thus holds and this completes the solution. 7

8 Solution by Perfetti Paolo, Department of Mathematics, University Tor Vergata, Rome, Italy The Cauchy Schwarz inequality yields ab bc ca + + a + b b + c c + a where r s as bs c/s, and s a + b + c/. ab + bc + ca a + b + b + c + c + a 9r. Letting x b + c a/, using the symmetry in the statement of the problem and upon clearing the denominators we obtain 7x y + A x5 + 7x 4 y 9x yz x y z 0 sym and A x + y + z + 5xy + yz + zxx + y + z > 0. Muirhead s theorem concludes the proof. Indeed The underlying AGM s are [,, 0] [,, ], [5, 0, 0] [,, ], [4,, 0] [,, ] x y + x z x yz, x 5 + y 5 + z 5 5x yz, 9x 4 y + y 4 z + z 4 x x yz and symmetry. Also solved by Arkady Alt, San Jose, CA; Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Moti Levy, Rehovot, Israel; Nikos Kalapodis four solutions, Patras, Greece; Albert Stadler, Herrliberg, Switzerland; Nicusor Zlota two solutions Traian Vuia Technical College, Focsani, Romania, and the proposer. 58: Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Without the aid of a computer, find the positive solutions of the equation x+ + x + x x + x + x+. Solution by Junho Chang, Colegio Hispano-Inglés, and Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain Multiplying both terms of the given equation by + + x + x + + x+ and simplifying we obtain + x+ + x + x x+ x + x x x. Taking logarithms, the last equation may be written as ln x x ln. Let us consider the function fx ln x x defined for positive real numbers x. Since f x ln x x, fx is 8

9 increasing for x 0, e and it is decreasing for x e, +. Since lim fx, and x 0 + fe /e > ln / there is a unique root to the equation ln x x ln x. Also, since in 0, e, which is ln x lim fx 0, there is a unique root to the equation x + x ln in e, +, which is x 4. So, x and 4 are the only positive solutions to the problem. Solution by Haroun Meghaichi student, University of Science and Technology Houari Boumediene, Algeria For convenience we set a x+, b x + x then is equivalent to Since ab 0, we get a + b b + a ab + + b ab + + a. + + b + + a + + b + + a a b Which means that x+ x + x then x taking ln of both sides we get ln x x ln, x > Let f :, R be defined by fx ln x, then f x x ln x then f cannot have more than two roots since f increases on, e and decreases on e, + and since, 4 are obvious roots we conclude that the only positive solutions to the equation are, 4. x ln Also solved by Adnan Ali student in A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, CA; Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX; Jerry Chu, student at Saint George s School, Spokane, WA; Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton, KS; Jahangeer Kholdi and Farideh Firoozbakht, University of Isfahan, Khansar, Iran; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel; David E. Manes, SUNY College at Oneonta, Oneonta, NY; Paolo Perfetti, Department of Mathematics, University Tor Vergata, Rome, Italy; Henry Ricardo, New York Math Circle, NY; Neculai Stanciu, George Emil Palade School, Buzău, Romania jointly with Titu Zvonaru, Comănesti, Romania; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer. 59: Proposed by Arkady Alt, San Jose, CA Find the smallest value of xy + yz + zx. x x + y + y y + z + Solution by Kee-Wai Lau, Hong Kong, China 9 z z where real x, y, z > 0 and + x

10 Since x x + y + x y + y y + z + z z + x yx y x + y + y z + zy z y + z + z x + xz x z + x x y + y z + z x x + y + z, 6xy + yz + zx + x y + y z + z x 6 and equality holds when x y z, so the smallest value required is. Solution by Bruno Salgueiro Fanego, Viveiro, Spain Since real x, y, z > 0 and xy + yz + zx, there is an acute triangle ABC such that cot A x, cot B y and cot C z so x x + y + y y + z + z z + x cot A cot A cot B cot A + cot B + cot B cot B cot C cot B + cot C + cot C cot C cot A cot C + cot A cot A + cot B + cot C cot A cot B cot A cot B cot B cot C cot B cot C cot C cot A cot C cot A cot A + cot B + cot C with equality iff cot A cot B cot C and A B C π/, that is iff x y z, 0

11 where we have used that cot A, cot B > 0, cot A cot B 0 with equality iff cot A cot B and ically, and inequality.8 page 8, Geometric Inequalities, Bottema O., Djordjević, R. Z., Janić, R.R., Mitrinović, D.S. Vasić, P.M., Wolters-Noordhoff,, Groningen, 969. Solution by Henry Ricardo, New York Math Circle, NY The AGM inequality gives us Similarly, we get x x + y x xy x + y x xy xy x y. y y + z y z Adding these three inequalities, we see that and z z + x z x. Now we have fx, y, z x x + y + y y + z + z z + x x + y + z. A x + y + z x + y + z + xy + yz + zx x + y + z +, so x + y + z x + y + z +, where we have used the well-known inequality x + y + z xy + yz + zx. Thus fx, y, z, with equality if and only if x y z /. Editor s comment : The author also provided a second solution to the above problem. It starts off exactly as the one above up until statement A. Then: x fx, y, z x + y + y y + z + z z + x x + y + z x + y + z xy + yz + zx /. Thus fx, y, z, with equality if and only if x y z /. Solution 4 by Albert Stadler, Herrliberg, Switzerland Suppose that xy + y + zx. We claim that x x + y + y y + z + z z + x, with equality if and only if x y z. By homogeneity, is equivalent to the unconditional inequality x xy + yz + zx x + y + y y + z + z z + x.

12 We first note that x + y + z x + y + z + xy + yz + zx xy + yz + zx, since by the Cauchy-Schwarz Inequality, x + y + z xy + yz + zx, with equality if and only if x y z. So x xy + yz + zx x + y + y y + z + z z + x x x + y + z x + y + y y + z + z z + x. To prove it is therefore enough to prove that x x + y + z x + y + y y + z + z z + x with equality if and only if x y z. Clearing denominators we see that is equivalent to x 5 y + x y 5 + x 4 y x y 4 + x 4 y z + By the weighted AM-GM inequality, x y + x4 y x y 4, 9 x y y z z x 5 x 4 yz, x5 y + x z 4 x 4 y z, 0 9 x5 y y5 z z5 x + 4 x4 y x 4 y z. x 4 yz. 4 We conclude that x y 5 + x 4 y x y 4, 5 4 x y 5 x y y z z x 5 x 4 yz, 6 x 5 y + x 4 y x 5 y + 4 x z 4 x 4 yz, 7 4 4

13 4 x 5 y + 4 x 4 y 0 9 x 5 y + 76 x 5 z z 5 x + x 4 y 4 x 4 y z. 8 4 Condition 4 follows by adding 5,6,7, and 8. Equality holds if and only if x y z. This is the equality condition for weighted AM-GM inequalities. Also solved by Adnan Ali student, in A.E.C.S-4, Mumbai, India; Michael Brozinsky, Central Islip, NY; Ed Gray, Highland Beach, FL; Moti Levy, Rehovot, Israel; Paolo Perfetti, Department of Mathematics, Tor Vergata Roma University, Rome, Italy; Neculai Stanciu, George Emil Palade School, Buzău, Romania jointly with Titu Zvonaru, Comănesti, Romania; Nicusor Zlota plus a generalization Traian Vuia Technical College, Focsani, Romania, and the proposer. 50: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania x Let Bx and let n be an integer. x Calculate the matrix product BB Bn. Solution by Neculai Stanciu, George Emil Palade School, Buzău, Romania jointly with Titu Zvonaru, Comănesti, Romania We denote AnBB... Bn. We have We assume that Since An A n +! n +!, A An n +! n +! n +! n +! n +! n +!. n + n +. n +! n +! n +! n +! we have shown, by mathematical induction that holds for all integers n. Solution by Moti Levy, Rehovot, Israel Let Bx xi + A, where A is an involute matrix i.e., A I. Let P n B B B n.,

14 Since A is an involute matrix then P n α n I + β n A, P I + A. A recurrence formula for α n, β n is P n+ P n Bn + α n I + β n A n + I + A. Let x n α n β n and y n α n + β n, then The solution for x n, y n is Solving for α n, β n, α n+ n + α n + β n β n+ n + β n + α n α, β. x n+ nx n, y n+ n + y n, x, y. x n n!, y n n +!. α n 4 n +! + n!, β n 4 n +! n!. For any involutory matrix A, P n 4 n +! + n! I + 4 n +! n! A. For the special case A 0 0, the solution is B B B n 4 n +! + n! 4 n +! n! 4 n +! n! 4 n +! + n!. Also solved by Arkady Alt, San Jose, CA; Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX; Brian D. Beasley, Presbyterian College, Clinton, SC; Jerry Chu student at Saint George s School, Spokane, WA; Bruno Salgueiro Fanego, Viveiro, Spain; Ethan Gegner student at Taylor University, Upland IN; Ed Gray, Highland Beach, FL; G. C. Greubel, Newport News, VA; Jahangeer Kholdi and Farideh Firoozbakht, University of Isfahan, Khansar, Iran; Kee-Wai Lau, Hong Kong, China; Carl Libis of the University of Tennessee at Martin, TN; David E. Manes, SUNY College at Oneonta, Oneonta, NY; Haroun 4

15 Meghaichi student, University of Science and Technology Houari Boumediene, Algeria; Corneliu Mănescu-Avram, Transportation High School Ploiesti, Romania; Henry Ricardo, New York Math Circle, NY; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA; Aoxi Yao student at Saint George s School, Spokane, WA; Ricky Wang, student at Saint George s School, Spokane, WA, and the proposer. Late Solutions A late solution was received to problem #59 and to 5 by Adnan Ali student in A.E.C.S-4, Mumbai, India. Solutions to problems 5, 5 and 54 were received from Arkady Alt of San Jose, CA. They were received on time but misfiled by me, and his name was inadvertently not listed as having solved these problems in the February 05 issue of the column. Arkady, I am sorry; mea culpa once again. Solutions to problems 5, 54, 55, and 58 were also received from Carl Libis of the University of Tennessee at Martin, TN. They too were received on time but misfiled by me again, mea culpa. Solutions 50 and to 5 were also submitted on time by Albert Stadler of Herrliberg, Switzerland, and inadvertently and misfiled by me. And for the files submitted by Moubinool Omarjee of Lyce Henri IV, Paris, France my computer of its own accord, placed them into a junk file. But he deserves credit for having solved problems 557, 569, 575, 576, and 58. To Albert and to Moubinool, and to all others for whom this might have also happened, mea culpa. 5