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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 February, 8: Proposed by Kenneth Korbin, New York, NY Part I: An isosceles right triangle has perimeter P and its Morley triangle has perimeter x. Find these perimeters if P x +. Part II: An isosceles right triangle has area K and its Morley triangle has area y. Find these areas if K y + 83: Proposed by Kenneth Korbin, New York, NY A convex pentagon ABCDE, with integer length sides, is inscribed in a circle with diameter AE. Find the minimum possible perimeter of this pentagon. 84: Proposed by Neculai Stanciu, Buzău, Romania If x, y and z are positive real numbers, then prove that x + y)y + z)z + x) x + y + z)xy + yz + zx) : Proposed by José Luis Díaz-Barrero, Barcelona, Spain Calculate, without using a computer, the value of [ )] sin arctan + arctan + arctan + arctan + arctan + arctan. 3) ) 7) ) 3) 86: Proposed by John Nord, Spokane, WA k Find k so that b ) n a x + b dx a b ) n a x + b dx. 87: Proposed by Ovidiu Furdui, Cluj, Romania

2 Let f : [, ], ) be a continuous function. Find the value of lim n n f n f ) + n n ) + + n n f n n ) n. Solutions 64: Proposed by Kenneth Korbin, New York, NY A triangle has integer length sides a, b, c) such that a b b c. Find the dimensions of the triangle if the inradius r 3. Solution by Albert Stadler, Herrliberg, Switzerland If a, b and c are the side lengths of the triangle then the inradius r is given by the formula r b + c a)c + a b)a + b c). see, e.g., http//mathworld.wolfram.com/inradius.html). a + b + c By assumption, c b a. So 3b a)a b) 3, or equivalently 3 3b a)a b) 6. Obviously b is even. If b were odd, then both 3b a and a b are odd, and therefore their product would be odd, which is not true.) So b b and this gives the equation 3b a)a b ) 39. Note that 39 xy is the product of two integers. So, x, y), 39), 3, 3), 3, 3), 39, ),, 39), 3, 3), 3, 3), 39, )}. If 3b a x and a b y, then b x + y, and a x + 3y. We find a, b, c) 9, 4, ),, 6, ),, 6, ),, 4, 9)}, and we easily verify that each triplet satisfies the triangle inequality. Solution by Arkady Alt, San Jose, CA Let F and s be the area and semiperimeter. Since a + c b then s a + b + c 3b,

3 and using F s s a) s b) s c) sr we obtain s a) s b) s c) sr ) ) ) 3b 3b 3b a b c 3 3b 3b a ) 3b c ) 39 9b 4 ) 3b a + c) + ac 39 9b 4 ) 3b b + ac 39 4ac 3b 3. Thus we have a + c b 4a b a) 3b 4ac 3b 6 6 c b a if, and only if, 4a b a) 3b 6 c b a 8ab a 3b 6 c b a. Since 8ab a 3b 3b a) a b) and a < s b < s c < s a < 3b c < s a < 3b b a) < 3b b < a < 3b then the problem is equivalent to the system ) 3b a) a b) 6 b < a < 3b. Since 3b a a b mod ) and then ) in positive integers is equivalent to 3b a k a b m b k + m 4a k + 3m where k, m), 78), 78, ), 6, 6), 6, 6)}. Noting that the inequality b < a < 3b k + m holds for any positive k, m we finally obtain a k + 3m 4 b k + m < k + 3m, < 3 k + m) a, b) 9, 4),, 4),, 6),, 6)}. 3

4 Thus, a, b, c) 9, 4, ),, 4, 9),, 6, ),, 6, )} are all solutions of the problem. Comment by David Stone and John Hawkins, Statesboro, GA. In their featured solutions to SSM 46 May issue) both Kee-Wai Lau and Brian Beasley found all integral triangles with in-radius 3. Note that the condition a b b c is equivalent to b a + c)/. That is, irrespective of how one might label or order the sides, the side b must be the middle-length side, the average of the other two sides. Also solved by Brain D. Beasley, Clinton, SC; Valmir Bucaj student, Texas Lutheran University), Seguin, TX; Elsie M. Campbell, Dionne T. Bailey and Charles Diminnie jointly), San Angelo TX; Bruno Salgueiro Fanego, Viveiro, Spain; Tania Moreno García, University of Holguín UHO), Holguín, Cuba jointly with José Pablo Suárez Rivero, University of Las Palmas de Gran Canaria ULPGC), Spain; Paul M. Harms, North Newton, KS; Enkel Hysnelaj, University of Technology, Sydney, Australia jointly with Elton Bojaxhiu, Kriftel, Germany; Sugie Lee, John Patton, and Matthew Fox jointly; students at Taylor University), Upland, IN; Kee-Wai Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Charles McCracken, Dayton, OH; Boris Rays, Brooklyn, NY; David Stone and John Hawkins jointly), Statesboro, GA; Jim Wilson, Athens, GA, and the proposer. 6: Proposed by Thomas Moore, Bridgewater, MA Dedicated to Dr. Thomas Koshy, friend, colleague and fellow Fibonacci enthusiast. Let σn) denote the sum of all the different divisors of the positive integer n. Then n is perfect, deficient, or abundant according as σn) n, σn) < n, or σn) > n. For example, and all primes are deficient; 6 is perfect, and is abundant. Find infinitely many integers that are not the product of two deficient numbers. Solution by Kee-Wai Lau, Hong Kong, China Let p, p 3, p 3,... be the sequence of primes. We show that for any n+ positive integer n, the integer p k is not the product of two deficient numbers. Suppose, on the contrary, that k n+ k Clearly a and b are relatively prime and so n+ ) 4 p i 4ab > σa)σb) σab) σ Hence, k 4 > n+ k + ) pk k p k ab, where both a and b are deficient numbers. + pk ) which is a contradiction. This completes the solution. n+ k p k ) n+ k + p k ) , Solution by Stephen Chou, Talbot Knighton, and Tom Peller students at Taylor University), Upland, IN All negative numbers have the same numerical divisors as their positive counterparts; 4

5 however, the negatives also include all the negative forms of those divisors. For instance, 6 has divisor of,, 3, 6,, 3, 6. Therefore σn) because the divisors will all negate themselves. Knowing that n of any negative will result in a lower negative, we see that all the negatives are abundant. Since the negatives are all abundant numbers and the only way to have a negative product is to multiply a negative by a positive, then at most a negative number can have only one deficient factor. Therefore, there are infinitely many integers, namely the negatives, that are not the product of two deficient numbers. Editor s comment: Once again the students have out smarted the professors; the intent of the problem was to find infinitely many positive integers that are not the product of two deficient numbers. But the problem wasn t explicitly stated that way, and so the students win; mea culpa. Solution 3 by Pat Costello, Richmond, KY Let n k 378 k k+ 94 for any non-negative integer k. We want to show that for any divisor d of n and pair d, n/d), one of the two values is either perfect or abundant. Since the σ function is multiplicative, we have σ94) σ3 3 7) σ3 3 ) σ) σ7) > 94. So 94 is abundant. Then in the pair 94, n/94), the 94 is abundant. By multiplicativity, ) σ k+ 94 σ k+ ) σ94) > σ k+) 94, by the above > k+ 94, since σm) > m for m > ) k+ 94. This means all the n values are themselves abundant so in the pair, n), the value n is the abundant value. This argument also shows that in the pair j, n/ j ), the second value is the abundant value. In the following table, we list the divisors d > of 94 and the values of the fractions σd)/d. d σd)/d The key thing we want to see from the table is that the minimum value in the second row corresponds to d 7. Suppose that d is a divisor of k+ 94 that is of the form j m where j and m is a divisor of 94 greater than. The fractions σ j )/ j are easily seen to be strictly

6 increasing with a limit of. Then ) σ j m / j m σ j) σm)/ j m σ j )/ j σm)/m from the table and that j Hence the divisor j m is perfect or abundant. Suppose that d is a divisor of 94 and less than 94, say d 94/m for an m. Then the pair is 94/m, k+ m) and the second value is perfect or abundant. All pairs d, n/d) have at least one value which is perfect or abundant. Since k is an arbitrary nonnegative integer, we have the desired infinite set. Solution 4 by Brian D. Beasley, Clinton, SC We make use of the following three facts: ) 94 is abundant in fact, it is the smallest odd abundant number); ) any nontrivial multiple of a perfect number is abundant; 3) any multiple of an abundant number is abundant. Given any integer k, we show that n k k 94 is not the product of two deficient numbers. For contradiction, if n k k xy for deficient numbers x and y, then the perfect number 6 divides neither x nor y, so without loss of generality, we assume that k divides x and 3 3 divides y. Next, we consider cases: a) If divides x, then x is abundant, since it is a multiple of the abundant number. b) If 7 divides x, then x is either perfect or abundant, since it is a multiple of the perfect number 8. c) If neither nor 7 divides x, then y is abundant. Since each case leads to a contradiction, we are done. In fact, it follows that for k 3, if n k xy, then at least one of x or y is abundant. Addendum. Facts ) and ) above may be found in Burton s Elementary Number Theory 6th edition) on page 3, while fact 3) follows by applying an argument similar to that used to prove fact ). Solution by proposer A computer program shows that there are such numbers below, the smallest being 378. The canonical factorization of these numbers is revealing. One notices that the list includes all numbers of the from 378p were p, 3, 7, 9, 3}. This suggest that N 378p is such a number, for all primes p. To prove this, lent N 378p ab with a b N. Now 378p p has 96 divisors, many of which are multiples of. But is an abundant number and so is any multiple of. More generally, any multiple of an abundant number is also 6

7 abundant.) So we need only consider factorizations N ab where neither a nor b is a multiple of. We list all these factorizations in the tables below showing the companion factors a and b, along with their type P: perfect; D: deficient; A: abundant). a type b type a type b type D 378p A p D 378 A D 89p A p D 89 A 4 D 94p A 4p D 94 A 6 P 63p A 6p A 63 A D 378p A p D 378 A 4 D 7p A 4p D 7 A 8 A p D 8p A D A 89p D p A 89 D 7 D 4p A 7p D 4 A 8 P 3p D 8p A 3 D 3 A 6p D 3p A 6 D 4 A 9p A 4p A 9 A 4 A 7p A 4p A 7 A Also solved by David E. Manes, Oneonta, NY; Albert Stadler, Herrliberg, Switzerland, and David Stone and John Hawkins jointly), Statesboro, GA. 66: Proposed by José Luis Díaz-Barrero, Barcelona, Spain Let a, b, c be lengths of the sides of a triangle ABC. Prove that 3 a+b + c b 3 b ) 3 b+c + a c 3 c ) 3 c+a + b a 3 a ) 8. Solution by Boris Rays, Brooklyn, NY By the Arithmetic-Geometric-Mean Inequality for each expression in the parentheses above we have: 3 a+b + c b 3 b 3 a+b c c b 3 b b 3a 3 b+c + a c 3 c 3 b+c a a c 3 c c 3b 3 c+a + b a 3 a 3 c+a b b a 3 a a 3c. Therefore, 3 a+b + c b 3 b ) 3 b+c + a c 3 c ) 3 c+a + b a 3 a ) c 8 b a c b a 3a 3 b 3 c 8 3 a+b+c 7

8 a + b + c. 8 3 The factor 3 a+b+c)/ is an exponential expression with base 3 3 > ) and exponent a + b + c)/ >. Hence, 3 a+b+c)/ >. Therefore, 3 a+b + c b 3 b ) 3 b+c + a c 3 c ) 3 c+a + b a 3 a ) 8. Also solved by Arkady Alt, San Jose, CA; Bruno Salgueiro Fanego Viveiro Spain; Enkel Hysnelaj, University of Technology, Sydney, Australia jointly with Elton Bojaxhiu, Kriftel, Germany; Valmir Bucaj student, Texas Lutheran University), Seguin, TX; Elsie M. Campbell, Dionne T. Bailey and Charles Diminnie jointly), San Angelo TX; Kee-Wai Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Paolo Perfetti, Department of Mathematics, University Tor Vergata, Rome, Italy; Albert Stadler, Herrliberg, Switzerland: Neculai Stanciu, Buzău, Romania; David Stone and John Hawkins jointly), Statesboro, GA, and the proposer. 67: Paolo Perfetti, Department of Mathematics, University Tor Vergata, Rome, Italy Find the maximum of the real valued function fx, y) x 4 x 3 6x y + 6xy + y 4 defined on the set D x, y) R : x + 3y }. Solution by Michael Brozinsky, Central Islip, NY We note that the given constraint x + 3y implies that x and y 3. Now, f, ) 3, and to show that 3 is the maximum it suffices to show that fx, y) 3 x + 3y ). That is x 4 x 3 6x y + 6xy + y 4 3 x + 3y ) or equivalently, ) ) x x x 3 + y y + 6x ) y 6x + 9 when x, y) is in D. ) Now x x 3 if x 3 and y + 6x 3 + 6x 6x + 9 for all x, as the minimum of 6x 6x + 9 is 7.), and so ) is obvious as x x x 3 ) and y y + 6x ) y x + 9 ) when x, y) is in D. Solution by Enkel Hysnelaj, University of Technology, Sydney, Australia and Elton Bojaxhiu, Kriftel, Germany First we will look for extreme points inside the region, which is the set x, y) R : x + 3y < }, and such points will be the critical points of the function fx, y). The partial derivatives of the function fx, y) will be fx 4x 3 6x xy + 6y f y x y + xy + 4y 3 8

9 Solving the system of the equations 4x 3 6x xy + 6y x y + xy + 4y 3, we have that the only critical point inside the region will be x, y), ), which will be considered a point where the function might get the maximum value. Now we will find the extremes on the contour of the region, which is x, y) R : x + 3y }. For any point on the contour we have y x and 3 substituting this into the formula of fx, y) we obtain the function gx) such that gx) x 4 x 3 6x x + 6x ) x x x + x 9 4x3 + 8x4 9 so, we have to find the extremes of the function gx) on the segment [, ]. If x ± we have f, ) 3 and f, ), and so far, we shown that a local maximum point is when x, y), ). Now we much check to see if there is a maximum point inside the segment [, ]. Taking the derivative of the function gx) we obtain g x) 4x 9 x + x3. 9 The equation g x) has no solution inside the segment [, ], which implies that there is no extreme point inside this segment. And so we may conclude that 3 is the absolute maximum of the real valued function fx, y) on the given domain and that it is achieved at the point x, y), ). Solution 3 byángel Plaza, University of Las Palmas de Gran Canaria, Spain Function fx, y) is harmonic. Then, by the maximum principle its maximum and minimum) is attained at the boundary of compact subset D. Since the boundary of D is an ellipse, by using its parametrization the problem is reduced to a one variable optimization problem. The parametric equations of the given ellipse are x cos t; y 3 sin t and the problem yields to maximizing the function gt) cos 4 t cos 3 t cos t sin t + cos t sin t + sin4 t 9 cos 3 t + cos 4 t + sin4 t 9. Since g t) 9 cos t sin t 7 cos t + 8 cos t sin t ) it is deduced that the maximum is attained at t π with the value fπ) 3. Also solved by Pat Costello, Richmond, KY; Bruno Salgueiro Fanego, Viveiro, Spain; Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; David E. Manes, Oneonta, NY; Boris Rays, Brooklyn, NY; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins jointly), Statesboro, GA, and the proposer. 9

10 68: Proposed by G. C. Greubel, Newport News, VA Find the value of a n in the series 7t + t 36t + 4t a + a t + a t + + a n t n +. Solution by Bruno Salgueiro Fanego, Viveiro, Spain By direct division, 4t 36t + t + 7t we see that a, a. Moreover, the 4 characteristic equation of the denominator is 36r + 4r, whose roots are r 9 4, r 9 + 4, so a n Ar n + Brn for some real numbers A and B. Taking n, we obtain and taking n we obtain A + B A + B Ar + Br a, A B 9 4 Ar + Br a 4. A + B So, by solving the system of equations 8A + B) + 8 A B) we obtain A 4, B + 4. Hence, a n Ar n + Br n 4 ) n ) n Solution by Kee-Wai Lau, Hong Kong, China It can be checked readily that 7t + t 36t + 4t t t For t > ) n 9 + 4, we have and t n ) n Hence for positive integer n t n t a n + 4 ) n ) n 9 4..

11 Solution 3 by Dionne Bailey, Elsie Campbell, and Charles Diminnie, San Angelo, TX Let f n } be the Fibonacci sequence defined by f, f, and f n+ f n+ + f n for n. Also, let φ + and φ. Then, we will use Binet s Formula for n and the known results f n φn φ n φ n f n φ + f n,φ n f n φ + f n, and φφ ) for n. To begin, make the change of variable s t and simplify to get 7t + t 36t + 4t 7s + s 36s + 4. Note that ) implies that φ 6 f 6 φ + f 8φ and similarly, φ Then, the roots of s 36s + 4 are s 8 ± 8 φ 6, φ 6 and we have 7s + s 36s + 4 7s + s φ 6 ) s φ 6). If we perform a partial fraction expansion and use Binet s Formula, ), and the formula for a geometric series, we obtain 7s + s 36s + 4 7φ6 + 8 s φ 6 7φ6 + 8 s φ φ6 φ 6 ) + 7φ 6 + s φ 6 φ 6 s φ 6 ) φ 6) n φ 6) n s n 7 + φ 6) φ n 6 ) n s n 6 6 n n n [ 7 + φ 6 n φ 6n 6 ] φ 6n s n 7 + φ 7 + φ 6 ) φ 6n 7 + φ 6) φ 6n n ) 6n s n [ φ 6n φ 6n ) φ n+4 7 6n+6 φ 6n+6 )] + s n

12 n n 7f 6n + f 6n+6 n+4 s n f 6n+ + 3f 6n n+ s n n f 6n+ + 3f 6n n+ t n. Also, since φ < φ, the series converges when s < min φ 6 }, φ 6 φ 6, i.e., when Therefore, for n. t > φ 6 φ6. a n f 6n+ + 3f 6n n+ Also solved by Arkady Alt, San Jose, CA; Brian D. Beasley, Clinton, SC; Paul M. Harms, North Newton, KS; Enkel Hysnelaj, University of Technology, Sydney, Australia jointly with Elton Bojaxhiu, Kriftel, Germany; David E. Manes, Oneonta, NY; Ángel Plaza University of Las Palmas de Gran Canaria), Spain; Boris Rays, Brooklyn, NY; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins jointly), Statesboro, GA; Boris Rays, Brooklyn, NY, and the proposer. 69: Proposed by Ovidiu Furdui, Cluj, Romania Let n be an integer and let i be such that i n. Calculate: x i dx dx n. x + x + + x n Solutions and by Albert Stadler, Herrliberg, Switzerland ) Let I i I I I n. So, x i x + x + + x n dx dx n. Then by symmetry, I + I + + I n x + x + x n dx dx n, x + x + + x n and I i n for i n.

13 ) Another albeit less elegant proof runs as follows: x i dx dx n x + x + + x n x i e tx +x + x n) dx dx n dt e t ) n + t)e t) dt t n+ n n lim t d dt e t ) n dt t n e t ) n t n n. The above is so because: e tx j dx j e t, t x i e tx i dx i + t)e t t, d e t ) n dt t n n e t ) n t n+ + n e t ) n e t t n n e t ) n + t)e t ) t n+. Also solved by Michael N. Fried, Kibbutz Revivim, Israel; Enkel Hysnelaj, University of Technology, Sydney, Australia jointly with Elton Bojaxhiu, Kriftel, Germany; Kee-Wai Lau, Hong Kong, China; Paolo Perfetti, Department of Mathematics, University Tor Vergata, Rome, Italy; David Stone and John Hawkins jointly), Statesboro, GA, and the proposer. 3