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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 December 5, 6 549: Proposed by Kenneth Korbin, New York, NY Given isosceles trapezoid ABCD with AB < CD, and with diagonal AC = AB + CD. Find the perimeter of the trapezoid if ABC has inradius and if ACD has inradius : Proposed by Arkady Alt, San Jose, CA For the given integers a, a, a 3 find the largest value of the integer semiperimeter of a triangle with integer side lengths t, t, t 3 satisfying the inequalities t i a i, i =,, 3. 54: Proposed by D.M. Bătinetu-Giurgiu, Matei Basarab National College, Bucharest, Romania and Neculai Stanciu, George Emil Palade General School, Buzău, Romania Let a n n, b n n be real valued positive sequences with lim n a n = lim n b n = a R + If lim n n a n a = b R and lim lim n n b n a = c R compute n a n+ n+ n +! b n n n!. Note: R + means the positive real numbers without zero. 54: Proposed by Michal Kremzer, Gliwice, Silesia, Poland Given positive integer M. Find a continuous, non-constant function f : R R such that f fx = f [x], for all real x, and for which the maximum value of fx is M. Note: [x] is the greatest integer function. 543: Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Compute lim n n i j n n + i + jn + ij.

2 544: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania Let A, B M C be such that 5AB 6BA = 7I. Prove that Here, C is the set of complex numbers. AB BA = O. Solutions 539: Proposed by Kenneth Korbin, New York, NY A triangle with integer length sides 49, b, b + has integer area. Find two possible values of b. Solution by Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX If we let s and A denote the triangle s semi-perimeter and area, respectively, then and Heron s Formula yields s = 49 + b + b + = b + 5 A = s s 49 s b s b + = b + 5 b = 6 b + 5 b 4. Since A is a positive integer, we must have b + 5 b 4 = 6k for some positive integer k. If we expand the left side of and complete the square, we ultimately obtain b + 4k = 4 = 49. One way to find acceptable values for b, k, and A is to solve the Pell Equation x 4y = and then set b + = 49x, k = 49y, and A = 6k. Since the solution of with the smallest value of x is x = 5, y =, we get all solutions x n, y n of by setting x n + y n 4 = n. For each solution, we let b = 49x n, k = 49y n, and A = 6k. The first five solutions of and the corresponding values of b, k, and A are listed in the following table. x y b k A 5 49, 94 49, 49 9, , 88 4, 85 9, 6 4, , 64 48,, 88, 47, 55 9, 7, 64, , 349 8, 5, 94

3 Editor s note: Some readers extended the above table a bit further. David Stone and John Hawkins of Georgia Southern University added the following to the above table. x y b k A 47, 55 9, 7, 64, , 349 8, 5, 94,56, 8,3,8 8,38, 4,95,64 79,476,6,794,76,6,9,43,44,766,58,4 7,665,8,4,8,8,6 7,385,86,94 73,858,6,94 Also solved by Ulrich Abel, Technische Hochschule Mittelhessen, Germany; Jeremiah Bartz, University of North Dakota, Grand Forks, ND; Brian Beasley, Presbyterian College, Clinton, SC; Bruno Salgueiro Fanego, Viveiro, Spain; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; Carl Libis, Columbia Southern University, Orange Beach, AL; Charles McCracken, Dayton, OH; John Nord, Spokane, WA; Toshihiro Shimizu, Kawasaki Japan; Neculai Stanciu George Emil Palade School, Buzău, Romania and Titu Zvonaru, Comănesti, Romania; David Stone and John Hawkins, Georgia Southern University, Statesboro, GA, and the proposer. Editor s note: The problem solving column is sometimes used in a problem solving course offered at Taylor University in Upland, IN, where the students in the course often work in small groups. Each of the following students at Taylor University should also be credited with having solved 539. Group : Madison Massot, Julia Noonan and Benjamin Thayer Group : Amish Mishra, Raquel Helton, and Allie Ternet Group 3: Matt Garringer, Sarah Glett, and Erin Song Group 4: Caleb Belmont, Caleb Holleman, and Nick Iorio A comment and a question about 539 from the proposer, Kenneth Korbin. It can be seen that there are infinitely many Primitive Heronian Triangles that have a side with length 49. Question: Is there another positive integer less than 5 that can also be the length of a side of infinitely many Primitive Heronian Triangles? 539: Proposed by Titu Zvonaru, Comănesti, Romania and Neculai Stanciu, George Emil Palade School, Buzău, Romania Prove that if x, y, z >, then 4 x + y + z 7 xy + yz + zx + x 7x + y + z + y x + 7y + z + Solution by Kee-Wai Lau, Hong Kong, China z x + y + 7z 3 7. By homogeneity, we assume without loss of generality that x + y + z =. 3

4 Let t = x + y + z so that xy + yz + zx = x + y + z x + y + z 4x + y + z 7xy + yz + zx = 8t 7 t. Since the function + 6s Jensen s inequality, we have x 7x + y + z + y x + 7y + z + z x + y + 7z = = t is convex for s >, so by and x + 6x + y + 6y + z + 6z + 6x + y + z = + 6t. Hence the left side of the inequality of the problem is greater than or equal to as desired. 8t 7 t + + 6t = 43t 7 t + 6t , Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata, Rome, Italy By Cauchy Schwarz reversed, we get cyc x 7x + y + z = cyc The inequality is implied by x x7x + y + z x + y + z 7x + y + z + xy + yz + zx. 4x + y + z 7xy + yz + zx + x + y + z + xy + yz + zx 7x + y + z + xy + yz + zx 3 7 Since the l.h.s. of is equal to 756x + y + z xy yz zx the original inequality is proven. Solution 3 by Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania 3 x 7x + y + z = 9 x 7x + y + z = y + z x 97x + y + z = y x 97x + y + z + z x 97x + y + z = y x 97x + y + z + x y 9x + 7y + z = x y 37x + y + zx + 7y + z, and 4x + y + z 7xy + yz + zx 4 7 = x y 7xy + yz + zx. 4

5 So, the inequality becomes x y 37x + y + zx + 7y + z x y 7xy + yz + zx. To show this it suffices to show that x y 37x + y + zx + 7y + z x y 7xy + yz + zx I.e., which is obviously true. Equality holds if x = y = z. 4x + 4y + z + 9xy + 7yz + 7zx, Solution 4 by Moti Levy, Rehovot, Israel This inequality deserves brute force attack by Muirhead s inequality. 4 x + y + z xy + yz + zx + 7x 7x + y + z + 7x+y+zx+7y+z 9xy+yz+zx. 7y x + 7y + z + After some tedious manipulations, our inequality is equivalent to: 8 cyc x cyc x 4 y+64 cyc xy cyc x 3 yz 8 cyc 7z 3. x + y + 7z x 3 z +8 cyc x z cyc x y z, or to 4 x x 4 y x 3 yz 8 x 3 z + 46 x y z. Now we prepare the inequality for application of Muirhead s inequality by splitting some terms in left and right hand sides: 4 x x 4 y+98 x 4 y+364 x 3 yz 4 x 3 z +66 x 3 z +98 x y z+364 x y z. We use the following majorization relations: 5,, 3,,, 4,, 3,,, 4,,,,, 3,,,, to show that x 5 x 3 z, x 4 y x 3 z, x 4 y x y z, x 3 yz x y z. 5

6 A nice tutorial on the application of Muirhead s inequality can be found at: Also solved by Arkady Alt, San Jose, CA; Michael Brozinsky, Central Islip, NY; Ed Gray, Highland Beach, FL; Toshihiro Shimizu, Kawasaki Japan; Albert Stadler, Herrliberg, Switzerland, and the proposer. 5393: Proposed by José Luis Díaz-Barrero, Barcelona, Tech, Barcelona, Spain Through the midpoint of the diagonal BD in the convex quadrilateral ABCD we draw a straight line parallel to the diagonal AC. This line intersects the side AD at the point E. Show that [ABC] + [AEC] 4 [CED]. Here [XY Z] represents the are of XY Z. Solution by Arkady Alt, San Jose, CA 6

7 We assume that midpoint M of diagonal BD does not coincide with K the point of intersection of AC and BD because otherwise [AEC] =. Also, w.l.o.g. assume that KD > BK. Let r = BM = MD and d := KM. Since ME AC then AE ED = KM MD = d r and, therefore, [AEC] = [ACD] AE AD = [ACD] d ED and [CED] = [ACD] r + d AD = [ACD] r r + d. Let BP, DQ AC.Then KP B KQD = BP DQ = BK and since DK BK = r d, KD = r + d we have [ABC] AC BP = [ADC] AC DQ = r d r + d. Thus, [ABC] + [AEC] 4 [CED] [ABC] + r + d 4 r + d d [ACD] r [ACD] [ACD] [ABC] + r + d d and we have 4 r + d r r d + d 4 r r + d r d + r + d 4 r + d d r = r d dr r d. r d + d 4 r Solution by Toshihiro Shimizu, Kawasaki Japan Let M be the midpoint of BD, F be the intersection of AC and BD and S be the area of the quadrilateral ABCD. Let x = F B/DB. Then, [ABC] = xs, [ADC] = xs and DM/DF = / x. Thus, [AEC] = [ADC] AE/AD = [ADC] MF/DF = xs / x/ x = / x S. Similarly, [CED] = [ADC] ED/AD = [ADC] MD/DF = xs // x = /S. Therefore we need to show that From Cauchy-Schwarz s inequality, x + x 4 / = 8. x + x.it s equivalent to the desired inequality. x + x Soltuion 3 by Kee-Wai Lau, Hong Kong, China Without loss of generality, let BD =. Let O =,, A = x A, y A, B =,, C = x C, y C, D,, where x A > and x C <. Suppose that AC and BD interset at F =, f. Since OE AC and E lies on AD, so f >. Since quadrilateral ABCD is convex, so f <. Suppose that slope of AC = slope of OE = m. We readily obtain y A = mx A + f, y C = mx C + f and xa E = + f, mx A. By the standard formula, we obtain [ABC] = fx A x C. + f [AEC] = fx A x C and [CED] = x A x C. Hence the inequality of the problem is 7

8 equivalent to f + f This completes the solution. 4. But this follows from the fact that f + f 4 = f ff. Solution 4 by Bruno Salgueiro Fanego, Viveiro, Spain Let M be the midpoint of the diagonal BD and let us draw a straight line parallel to the diagonal AC through B. NB is parallel to AC, triangles ABC and ANC have the same basis AC and the same altitude, so [ABC] = [ANC]. By the Arithmetic Mean -Harmonic Mean Inequality applied to the positive numbers [ANC] and [AEC] [ABC] + [AEC] = [ANC] + [AEC] 4 [ANC] + [AEC] = 4 [CNE], with equality iff [ANC] = [AEC], that is, iff AN = AE, or equivalently, NE = AE; that DE = AE, or what is the same AD = 3AE, i.e., DP = 3MP, where P denotes the intersection of the diagonals AC and BD. Since M is the midpoint of BD, this is equivalent to DP = 3BP. ME and BNare parallel lines because they are both parallel to the diagonal AC and M is the midpoint of BD, so E is the midpoint of DN and, hence triangles CNE and CED have the same area because they have the same base-lengths NE = DE, and the same altitude distance C, AD. Thus, [ABC] + [AEC] 4, and equality occurs iff the diagonal AC divides the [DEC] diagonal BD in the ratio 3 :. Also solved by Boris Rays, Brooklyn, NY; Yaqub Aliyev, Qafqaz University, Baku, Azerbaijan, and the proposer. 5394: Proposed by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain Let a, b and c be positive real numbers such that ab + bc + ca = 3 and n >. Prove that n a + abc + n b + abc + n c + abc 3 n. Solutions and by Henry Ricardo, New York Math Circle, NY. We have, using the AM-GM inequality several times, n a + [ abc 3 a + b + c + ] /3n abc abc abc [ 3 /bc /ac ] /3n /ab = 3 8 /3n = abc 3 n abc /3n 3 n 8

9 since 3 = ab + bc + ca 3 3 abc implies that /abc. Equality holds if and only if a = b = c =. Solution : We have, using the AM-GM inequality several times, n a + abc n bc /n n /n 3 3 bc ca = 3 n abc /3n 3 n since 3 = ab + bc + ca 3 3 abc implies that /abc. Equality holds if and only if a = b = c =. /n /n ab Solution 3 by Nikos Kalapodis, Patras, Greece By the AM-GM inequality we have 3 = ab + bc + ca 3 3 abc. It follows that abc. Using again AM-GM inequality properly, we have n a + abc + n b + abc + n c + abc n + n + n 3 3 n 3 bc ca ab abc = 3 3n 3 abc 3 3n 3 = 3 n. Solution 4 by Neculai Stanciu, George Emil Palade School, Buzău, Romania and Titu Zvonaru Comănesti, Romania By the AM-GM inerquality we have n bc = n... bc n + bc, and the other two inequalities analogously. n Hence, by the AM-GM inequality and Bergström s inequality we obtain n a + abc + n b + n c + n n abc abc bc + n ca + n ab Bergström n 9 n ab + n bc + n ca n 9 n + ab + n + bc + n + ca n 9

10 = n 8n 6n = 3 n, and we are done. Also solved by Arkady Alt, San Jose, CA; Dionne Bailey, Elsie Campbell, and Charles Diminnie, Angelo State University, San Angelo, TX; Michael Brozinsky, Central Islip, NY; Bruno Salgueiro Fanego, Viveiro Spain; Ed Gray, Highland Beach, FL; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel; Boris Rays, Brooklyn, NY; Toshihiro Shimizu, Kawasaki Japan; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania, and the proposer. 5395: Proposed by Mohsen Soltanifar Ph.D. student, Biostatistics Division, Dalla Lana School of Public Health, University of Toronto, Canada. Given the sequence {σn} n= of positive numbers X Nµ, σ. Define recursively a sequence of random variables {X n } n= via X n+ X n N X n, σn+ n =,, 3,... Calculate the limit distribution X of {X n } n=. Reference: Rosenthal, J.S. 7. A First Look at Rigorous Probability nd edition, World Scientific, p. 39. Proposer s note concerning the problem: This is a Bayesian Hierarchical Model of Human Heights from Adam & Eve to the end of time. Consider a family with its children. We know that height has a normal distribution. We also know that height of children is due to genetic factors which are dependent on the height of their parents, but usually this distribution has the same mean as the mean height of their parents but may vary some children are taller, some shorter, some are average- versus their parents. So, the height of children may be modeled as the normal distribution conditioned to the height of their parents with same mean but potentially different variance. The first term in the sequence is the distribution of height of Adam & Eve. The second term is the conditional distribution of their children s height. This goes till the end of time consecutively when, according to some beliefs, the Messiah returns. Accordingly, the Messiah will return and a generation of humans will observe this return. But we do not know when this will occur. So, we may assume the Messiah will return as time approaches infinity, and that the distribution of the height of generations of humans that observe the return is X. We are interested in knowing certain features of this distribution. This problem is a mathematical modeling of the above belief. Solutions and by Moti Levy, Rehovot, Israel The random variables {X n } n= have normal distribution X n N µ n, s n.

11 The probability density function of X n is f Xn t = πsn e t µn s n πσn+ e t Xn.The probability σ density function of X n+ X n is f Xn+ X n t = n+. The probability density function of X n+ is given by the following integral f Xn+ t = f Xn+ X n ξ f Xn ξ dξ = = π σn+ + s n e t µn e t ξ σ n+ e ξ µn s n dξ πσn+ πsn σ n+ +s n = N µn, σn+ + s n. By induction argument, the mean of X n+ = µ and the variance of X n+ = n+ The limit distribution of {X n } n= is N µ, k= k σ. k= σ k. Solution Remark: Alternative method for finding the probability density function of X n+ is by employing characteristic functions defined as follows: ϕ X t := E [ e itx]. The characteristic function of normal random variable X N µ, σ is ϕ X t = e iµt σ t. ϕ Xn+ t = E [ E [ e itx n+ X n ]] = E [ e ixnt σ n+ t] It follows that X n+ N µ n, σ n+ + s n. = e σ n+ t E [ e ixnt] = e σ n+ t ϕ Xn t = e σ n+ t e iµnt s nt = e iµnt σ n+ +s nt Solution 3 by Toshihiro Shimizu, Kawasaki Japan Let fx µ, σ = exp x µ be the density function of the normal πσ σ distribution. Let g n x be the density function of the distribution X n. We show that g n x = f x µ, n i= σ i. n = is obvious. We asume that the statement is true for some n. Then, g n+ x = = fx t, σn+ g n tdt fx t, σ n+ f n+ = f x µ, = f x µ, Thus, the statement is true for n +. i= n+ σi i= σ i. t µ, n i= σ i dt

12 If the value i= σ i is bounded, it converges to some value σ. Then, the limit distribution X is Nµ, σ. If the value i= σ i is not bounded. There is no limit distribution. Also solved by the proposer. 5396: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania Find all continuous functions f : R R such that f x = x + x e t fx tdt, x R. Solution by Michael Brozinsky, Central Islip,NY x If we rewrite the given equation as f x = x + e x e x t ftdt and use the change of variable u = x t, the given equation can be written as x f x = x + e x e u fudu. Note that the right hand side is a differentiable function of x follows from the fundamental theorem of integral calculus and thus so is fx or equivalently, x f x x e x e u fudu. If we differentiate both sides of with respect to x and use the chain rule and the fundamental theorem of integral calculus we obtain f x x e x + e x f x = e x fx and so dividing by e x gives f x x + f x = fx which can be written as f x = f x x fx 3 The right hand side of 3 is differentiable and using 3 or its equivalent form in which all x s are replaced by x we have by differentiation f x = f x f x = f x x fx fx + x f x =. 4 If we subtract two times equation 3 from equation 4 we obtain f x f x = 3 f x + x + fx using3 = 3 f x + x + f x x f x = f x +, so that fx satisfies the differential equation.

13 f x = where the initial conditions f = and f = and f = follow from the given equation and from 3 and 4 respectively. Hence, fx = a + bx + cx and the initial conditions readily give a =, b = and c =, so that fx = x x. Solution by Albert Stadler, Herrliberg, Switzerland We perform a change of variables and get f x = x + x x e t fx tdt = x + e x d t ftdt. The right hand side is a differentiable function, since f is continuous. The f is x differentiable. Suppose f is n-times differentiable. Then x + e x e t ftdt is differentiable n + times, and so f is n + times differentiable. So f is differentiable infinitely many times. Thus, d dx fx + xe x = x e t ftdt, fx + xe x = e x f x = e x x + e x x x fxe x + f xe x + e x xe x = xe x e x e t ftdt, d dx fxe x + f xe x + e x = x fxe x + f xe x + e x = e x fx, fxe x + f x + e x = fxe x, f x = e t ftdt, e t ftdt, We deduce from and that f = and f =. So fx = x x = xx +. Also solved by Bruno Salgueiro Fanego, Viveiro, Spain; Kee-Wai Lau, Hong Kong, China; Moti Levy, Rehovot, Israel; Toshihiro Shimizu, Kawasaki Japan, and the proposer. Late Solutions Received The name of Paul M. Harms of North Newton, KS should be added to the list of those who solved 539. He also noted in his solution that the case of b =, which was allowed in the statement of the problem, led to a counter example of the statement, and 3

14 like the others who solved this problem, he showed that the problem was only true for b >. 4