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1 Problems Ted Eisenberg, Section Edit ********************************************************* This section of the Journal offers readers an opptunity to exchange interesting mathematical problems and solutions Proposals are always welcomed Please observe the following guidelines when submitting proposals solutions: 1 Proposals and solutions must be legible and should appear on separate sheets, each indicating the name and address of the sender Drawings must be suitable f reproduction Proposals should be accompanied by solutions An asterisk * indicates that neither the proposer n the edit has supplied a solution Send submittals to: Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel fax to: Questions concerning proposals and/ solutions can be sent to: <eisen@mathbguacil> to <eisenbt@013net> Solutions to the problems stated in this issue should be posted befe May 15, : Proposed by Kenneth Kbin, New Yk, NY Given isosceles trapezoid ABCD with ABD = 60 o, and with legs BC = AD = 31 Find the perimeter of the trapezoid if each of the bases has positive integer length with AB > CD 5009: Proposed by Kenneth Kbin, New Yk, NY Given equilateral triangle ABC with a cevian CD such that AD and BD have integer lengths Find the side of the triangle AB if CD = 179 and if AB, 179 = : Proposed by José Gibergans-Báguena and José Luis Díaz-Barrero, Barcelona, Spain Let α, β, and γ be real numbers such that 0 < α β γ < π/ Prove that sin α + sin β + sin γ sin α + sin β + sin γcos α + cos β + cos γ : Proposed by José Luis Díaz-Barrero, Barcelona, Spain Let {a n } n 0 be the sequence defined by a 0 = a 1 = and f n, a n = a n 1 1 a n Prove that p a p+q + a q p = p a p a q where p q are nonnegative integers

2 501 Richard L Francis, Cape Girardeau, MO Is the incenter of a triangle the same as the incenter of its Mley triangle? 5013: Proposed by Ovidiu Furdui, Toledo, OH Let k be a natural number Find the sum n 1,n,,n k 1 1 n 1+n + +n k n 1 + n + + n k Solutions 4990: Proposed by Kenneth Kbin, New Yk, NY Solve with 0 < x < 1 40x x = x x Solution by Bis Rays, Chesapeake, VA Let x = cos α, where α 0, π/ Then 40 cos α cos α = cos cos α = cos α + = cos α + 1 cos α 1 cos α 1 = 9 cos α + 1 sin α = 58 cos π 4 cos α + sin π 4 sin α = 58 cos π 4 α Therefe, 40 cos α + 4 sin α = 58 cos π 4 α 40 4 cos α sin α = cos π 4 α 0 1 cos α sin α = cos π 4 α Let cos α 0 = 0 9 Then sin α 0 = 1 0 = π cos α 0 cos α + sin α 0 sin α = cos 4 α

3 Therefe we obtain from the above, π cosα 0 α = cos 4 α 1 α 0 α 1 = π 4 α 1 α 1 = α 0 π, where α 0 = arccos 0 9 α 0 α = π 4 α 3 α = α 0 + π 4 = α π 4 α = 3 α 0 + π 6, where α 0 = arccos 0 9 Therefe, 1 x 1 = cos α 0 π = cos α 0 cos π + sin α 0 sin π The solution is: = sin α 0 cos α 0 1 = = arccos 9 x = cos 3 α 0 + π = cos 6 x 1 = x = cos + π arccos + π 9 6 Remark: This solution is an adaptation of the solution to SSM problem 4966, which is an adaptation of the solution on pages of Mathematical Miniatures by Savchev and Andreescu Also solved by Brian D Beasley, Clinton, SC; Elsie M Campbell, Dionne T Bailey and Charles Diminnie jointly, San Angelo, TX; Paul M Harms, Nth Newton, KS; José Hernández Santiago student at UTM, Oaxaca, México; Kee-Wai Lau, Hong Kong, China; Peter E Liley, Lafayette, IN; John Nd, Spokane, WA; Paolo Perfetti, Math Dept, U of Rome, Italy; David Stone and John Hawkins jointly, Statesbo, GA, and the proposer 4991: Proposed by Kenneth Kbin, New Yk, NY Find six triples of positive integers a, b, c such that 9 a + a b + b 9 = c Solution by David Stone and John Hawkins, Statesbo, GA,with comments by edit David Stone and John Hawkins submitted a six page densely packed analysis of the problem, but it is too long to include here Listed below is their solution and the gist of

4 their analysis as to how they solved it Interested readers may obtain their full analysis by writing to David at to me at Others who solved the problem programmed a computer David and John began by listing what they believe to be all ten solutions to the problem a b c The analysis in their wds: Rewriting the equation, we seek positive integer solutions to 1 81b + 9a + ab = 9abc Theem A solution must have the fm a = 3 i A, b = 3 j A, where A, 3 = 1, i, j 0 At least one of i, j must be 1 Proof From equation 1, we see that 9 divides all terms but ab, so 9 divides ab, so 3 divides a b so at least one of i, j must be 1 Also from equation 1, it is clear that if p is a prime different from 3, then p divides a if and only if p divides b Suppose p is such a prime and a = 3 i p m C, b = 3 j p n D, where m, n 1, and C and D are not divisible by 3 p Then equation 1 becomes 81 3 j p n D i p C m + 3 i p m C 3 j p D n = 9 3 i p m C 3 j p n D c, # 3 j+4 p n D + 3 i+ p m C + 3 i+j p m+n CD = 3 i+j+ p m+n CDc If n < m, we can divide equation # by p n to obtain 3 j+4 D + 3 i+ p m n C + 3 i+j p m+n CD = 3 i+j+ p m CDc But then p divides each term after the first, so p divides 3 j+4 D, which is impossible If n > m, we can divide through equation # by p m to obtain 3 j+4 p n m D + 3 i+ C + 3 i+j p n m CD = 3 i+j+ p n m CDc 81p m n D + 9C + p m CD = 9p m n CDc Noting that n > 4m > m and n > m > m, we see that p divides each term except 3 i+ C, so p divides 3 i+ C, which is impossible Therefe n = m That is, a and b have the same prime diviss, and in b, the power on each such prime is

5 twice the cresponding power in a; therefe, in b, the product of all diviss other than 3 is the square of the analogous product in a So the proof is complete They then substituted this result into equation 1 obtaining 81 3 j A i A + 3 i A 3 j A = 9 3 i A 3 j A c, j i+ + 3 i+j A 3 = 3 i+j+ Ac and started looking f values of i, j, A and c satisfying this equation Analyzing the cases 1 where 3 divides b but not a; where 3 divides a but not b; and 3 where 3 divides a and b led to the solutions listed above They ended their submission with comments about the patterns they observed in solving analogous equations of the fm N a + b + c N = c f various integral values of N Also solved by Charles Ashbacher, Marion, IA; Britton Stamper student at Saint Gege s School, Spokane, WA, and the proposer 499: Proposed by Elsie M Campbell, Dionne T Bailey and Charles Diminnie, San Angelo, TX A closed circular cone has integral values f its height and base radius Find all possible values f its dimensions if its volume V and its total area including its circular base A satisfy V = A Solution by R P Sealy, Sackville, New Brunswick, Canada 1 3 πr h = πr + πr r + h rh = 6r + 6 r + h Squaring and simplifying gives r h = 36 h 1 Therefe, h is a square, and h 1 h {1, 4, 9, 16, } Note that fh = h is a decreasing function of h f h 1 h 1 h > 1 and that h16 = 4 Note also that f13, f14 and f15 are not squares of integers Therefe h, r = 16, 4 is the only solution Also solved by Paul MHarms, Nth Newton, KS; Peter E Liley, Lafayette, IN; John Nd, Spokane, WA; Bis Rays, Chesapeake, VA; Britton Stamper student at Saint Gege s School, Spokane, WA; David Stone and John Hawkins jointly, Statesbo, GA, and the proposer 4993: Proposed by José Luis Díaz-Barrero, Barcelona, Spain Find all real solutions of the equation 16x 7 17x = 0 Solution by N J Kuenzi, Oshkosh, WI

6 Both 1 and 1/ are easily seen to be positive rational roots of the given equation So x 1 and x 1 are both facts of the polynomial 16x 7 17x Facting yields 16x 17x = x 1x 163x x c 3 + 7x + 3x + 1 The equation 63x x c 3 + 7x + 3x + 1 does not have any rational roots Rational Roots Theem n any positive real roots Descartes Rule of Signs Using numerical techniques one can find that is the approximate value of a real root The four other roots are complex with approximate values: i i i i So the real solutions of the equation 16x 7 17x = 0 are 1, 1/ and Also solved by Paul M Harms, Nth Newton, KS; Peter E Liley, Lafayette, IN; Charles McCracken, Dayton, OH; Bis Rays, Chesapeake, VA; David Stone and John Hawkins jointly, Statesbo GA, and the proposer 4994: Proposed by Isabel Díaz-Iriberri and José Luis Díaz-Barrero, Barcelona, Spain Let a, b, c be three nonzero complex numbers lying on the circle C = {z C : z = r} Prove{ that the roots of the equation az + bz + c = 0 lie in the ring shaped region D = z C : 1 5 z 1 + } 5 Solution by Kee-Wai Lau, Hong Kong, China By rewriting the equation as az = bz c, we obtain a z = az = bz + c b z + c z z z + z 0 so that z By rewriting the equation as c = az bz, we obtain c = az bz a z + b z z + z z + z 0 so that z 5 1 This finishes the solution Also solved by Michael Brozinsky, Central Islip, NY; Elsie M Campbell, Dionne T Bailey, and Charles Diminnie jointly, San Angelo, TX; Russell Euler and Jawad Sadek jointly, Maryville, MO; Bis Rays, Chesapeake, VA; José Hernández Santiago student at UTM Oaxaca, México; R P Sealy, Sackville, New Brunswick, Canada; David Stone and John Hawkins jointly, Statesbo, GA, and the proposers

7 4995: Proposed by K S Bhanu and M N Deshpande, Nagpur, India Let A be a triangular array a i,j where i = 1,,, and j = 0, 1,,, i Let a 1,0 = 1, a 1,1 =, and a i,0 = T i + 1 f i =, 3, 4,, where T i + 1 = i + 1i + /, the usual triangular numbers Furtherme, let a i,j+1 a i,j = j + 1 f all j Thus, the array will look like this: Show that f every pair i, j, 4a i,j + 9 is the sum of two perfect squares Solution 1 by Dionne Bailey, Elsie Campbell, and Charles Diminnie, San Angelo, TX If we allow T 0 = 0, then f i 1 and j = 0, 1,, i, it s clear from the definition of a i,j that a i,j = a i,0 + T j = T i T j = i + 3i + j + j Therefe, f every pair i, j, 4a i,j + 9 = i + 3i + j + j + 9 = i + 3i + j + j + 5 = i + j + + i j + 1 Solution by Carl Libis, Kingston, RI F every pair i, j, 4ai, j + 9 = i j i + j + since [ 4ai, j + 9 = 4 ai, 0 + ] [ jj + 1 i + 1i = 4 + = i + 1i jj = i + 6i j + j + 1 = i j i + j + ] jj Also solved by Paul M Harms, Nth Newton, KS; N J Kuenzi, Oshkosh, WI; R P Sealy, Sackville, New Brunswick, Canada; David Stone and John Hawkins jointly, Statesbo GA; José Hernándz Santiago student at UTM, Oaxaca, México, and the proposers

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Problems Ted Eisenberg, Section Editor ********************************************************* This section of the Journal offers readers an opportunity to exchange interesting mathematical problems