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1 Problems Ted Eiseberg, Sectio Editor ********************************************************* This sectio of the Joural offers readers a opportuity to exchage iterestig mathematical problems ad solutios. Please sed them to Ted Eiseberg, Departmet of Mathematics, Be-Gurio Uiversity, Beer-Sheva, Israel or fax to: Questios cocerig proposals ad/or solutios ca be set to <eisebt@03.et>. Solutios to previously stated problems ca be see at < Solutios to the problems stated i this issue should be posted before February 5, : Proposed by Keeth Korbi, New Yor, NY Give isosceles ABC with cevia CD such that CDA ad CDB are also isosceles, fid the value of AB CD CD AB. 535: Proposed by Keeth Korbi, New Yor, NY Fid a, b, ad c such that with a < b < c. ab + bc + ca = 3 a 2 b 2 + b 2 c 2 + c 2 a 2 = 9 a 3 b 3 + b 3 c 3 + c 3 a 3 = : Proposed by Daiel Lopez Aguayo studet, Istitute of Mathematics, UNAM), Morelia, Mexico Prove that for every positive iteger, the real umber ) / ) / is irratioal. 537: Proposed by José Luis Díaz-Barrero, Barceloa, Spai Let a, b, c be positive umbers such that abc. Prove that a 5 + b 5 + c cyclic 538: Proposed by José Luis Díaz-Barrero, Barceloa, Spai Let 2 be a positive iteger. Prove that F F + )F 2 + F )F 2 + F 2 F 2 = where F is the th Fiboacci umber defied by F 0 = 0, F = ad for all 2, F = F + F 2.,

2 539: Proposed by Ovidiu Furdui, Cluj, Romaia Calculate = m= where ζ deotes the Riema Zeta fuctio. ζ + m) + m, Solutios 56: Proposed by Keeth Korbi, New Yor, NY Give square ABCD with poit P o side AB, ad with poit Q o side BC such that AP P B = BQ QC > 5. The cevias DP ad DQ divide diagoal AC ito three segmets with each havig iteger legth. Fid those three legths, if AC = 84. Solutio by David E. Maes, Oeota, NY Let E ad F be the poits of itersectio of AC with DP ad DQ respectively. The AE = 40, EF = 37 ad F C = 7. Sice ABCD is a square with diagoal of legth 84, it follows that the sides of the square have legth Let AP P B = BQ = t > 5. The AP = t P B ad QC AP + P B = AB = Therefore, Similarly, QC = t Coordiatize the problem so that P Bt + ) = 42 2 P B = t, ad AP = 42 2 t. + t 42 2 t ad BQ =. + t A = 0, 0), B = 42 2, 0), C = 42 2, 42 2), D = 0, 42 2), P = ) 42 2 t, 0, ad Q = 42 2, 42 ) 2 t. + t + t Let L be the lie through the poits D ad P. The the equatio of L is y 42 + t 2 = x. The poit of itersectio of L ad the lie y = x is the poit t E. Therefore, x 42 + t 2 = x, ad so t 2

3 x = 42 2 t 2t +. Thus, E = 42 2 t 2t +, 42 ) 2 t 2t + so that AE = ) t 2 = 84 t 2t + 2t +. Let L 2 be the lie through D ad Q. The the equatio of L 2 is y 42 2 = + t x = 42 2t + ). Thus, t + 2 ) x. Sice F is the poit of itersectio of L 2 ad y = x, we obtai F = AF = 42 2t + ) t t + ) t + 2. Usig the distace formula, oe obtais As a result, CF = t + ) t + 2, 42 ) 2t + ) so that t + 2 ) 2 = 84 t + 2. AE = 84 t 84t + ) 84, AF =, ad CF = 2t + t + 2 t + 2 If t = 0, the AE = 40, AF = 77, ad CF = 7. Therefore EF = AF AE = 37, yieldig the claimed values. Fially, oe checs that for these values all triagles i the figure are defied. Also solved by Shai Covo, Kiryat-Oo, Israel; Paul M. Harms, North Newto, KS; Boris Rays, Brooly, NY; David Stoe ad Joh Hawis joitly), Statesboro, GA, ad the proposer. 57: Proposed by Keeth Korbi, New Yor, NY Fid positive acute agles A ad B such that si A + si B = 2 si A cos B. Solutio by David Stoe ad Joh Hawis joitly), Statesboro, GA There are ifiitely may solutios, give by 3

4 Here s why. t A = si 2, B = cos t, where 4 2t 5 < t <. The give coditio is equivalet to 2 si A2 cos B ) = si B so we see that 2 cos B > 0, that is, 0 < B < π 3. Solvig for si A, we must have si A = Upo squarig, this is equivalet to si B 2 cos B, which requires 0 si B 2 cos B. si 2 B 4 cos 2 B 4 cos B + cos 2 B 4 cos 2 B 4 cos B + cos B 4 5. So if we choose agle B to mae cos B 4, the we ca choose agle A to mae 5 si B si A = 2 cos B. Sice cosie is) decreasig i the first quadrat, the size coditio o cos B forces 4 B cos I fact, for ay t, with 4 5 t, we ca let B = cos t, i which case si B = t t 2, ad let A = si 2. 2t Note that the edpoit solutio give by t = is A = 0, B = 0, which we disregard. Also, the edpoit solutio give by t = 4 5 is A = π 2, B = cos 4 5. It is worth otig that we thus have a right triagle solutio, but it does t quite meet the problem s criteria, so we ll disregard this oe. Thus, there are ifiitely may solutios, give i terms of the parameter t for 4 5 < t <. We also ote that oe could also say that all solutios are give by si A = where agle B is chose so that cos B > 4 5. si B 2 cos B, Also solved by Dioe Bailey, Elsie Campbell, ad Charles Dimiie joitly), Sa Agelo, TX; Michael Brozisy, Cetral Islip, NY; Shai Covo, Kiryat-Oo, Israel; Paul M. Harms, North Newto, KS; David E. Maes, Oeota, NY; Charles McCrace, Dayto, OH; Raúl A. Simó, Satiago, Chile; Taylor Uiversity Problem Solvig Group; Uplad, IN, ad the proposer. 4

5 58: Proposed by David E. Maes, Oeota, NY Fid the value of Solutio by Shai Covo, Kiryat-Oo, Israel The value is More geerally, for ay iteger 3 we have = + 2) + 2) + 3) + ) + 4) + + 5) + = 2009 correspods to the origial problem.) The claim follows from a iterative applicatio of the idetity = + 2) + 2) + ), as follows: = + 2) + 2) + ) = = + 2) + 2) + 3) + ) + 2) + 2) + 2) + 3) + ) + 4) + + 3) =. Solutio 2 by Taylor Uiversity Problem Solvig Group, Uplad, IN We use Ramauja s ested radical approach. Begiig with we see that x + + a) 2 = x a 2 + 2ax + 2x + 2a, x + + a = x a 2 + 2ax + 2x + 2a = ax a 2 + 2a + x x a) = ax + + a) 2 + x x a). However, the x a) term o the right is basically of the same form as the left with replaced by 2). We ca mae the correspodig substitutio, ad cotiue this process idefiitely, util we are left with x + + a = ax + + a) 2 + x ax + ) + + a) 2 + x + ) a x + 2) + + a) 2 + x + 2) Substitutig i x = 2007, = a = produces 2009 = =

6 Hece, the value is Also solved by Scott H. Brow, Aubur Uiversity, Motgomery, AL; G. C. Greubel, Newport News, VA; Paul M. Harms, North Newto, KS: Keeth Korbi, NY, NY; Charles McCrace, Dayto, OH; Paolo Perfetti, Departmet of Mathematics, Uiversity of Rome, Italy; Boris Rays, Brooly, NY; David Stoe ad Joh Hawis joitly), Stateboro GA, ad the proposer. 59: Proposed by Isabel Díaz-Iriberri ad José Luis Díaz-Barrero, Barceloa, Spai Let be a o-egative iteger. Prove that csc < F + F where F is the th Fermat umber defied by F = for all 0. Solutio by Charles R. Dimiie, Sa Agelo, TX To begi, we ote that for x 0, π ), cos x is decreasig ad the Mea Value Theorem 3 for Derivatives implies that there is a poit c x 0, x) such that si x = si x si 0 = cos c x x 0) > cos π 3 x = x 2. As a result, whe x 0, π ), 3 x csc x < 2. Sice F F 0 = 3 for all 0, it follows that 0 < F 3 < π 3 csc < 2, or F F csc < 2F ) Let P ) be the statemet F csc < 2 + F + 2) 2) F ad hece, By ), csc < 2F 0 = 2 3 = 2 F 2) F0 6

7 ad P 0) is true. If P ) is true for some 0, the by ), + csc F = csc < csc F + F + ) csc F ) 2 + F + 2) < 2F F + 2) = = = 2 +2 F +2 2) ad P + ) follows. By Mathematical Iductio, P ) is true for all 0. Sice 2) is equivalet to the give iequality, the proof is complete. Also solved by Shai Covo, Kiryat-Oo, Israel; Bruo Salgueiro Faego, Viveiro, Spai; David Stoe ad Joh Hawis joitly), Statesboro, GA, ad the proposers. 520: Proposed by José Luis Díaz-Barrero, Barceloa, Spai Calculate lim 2 ) log Solutio by Ovidiu Furdui, Cluj, Romaia The limit equals 0. More geerally, we prove that if f : [0, ] R is a cotiuous fuctio the lim 2 ) f = 0. ) Before we give the solutio of the problem we collect the followig equality from [] Formula 0.543), p.4): If p 0 is a oegative iteger, the the followig equality holds ) p = 0. ) Now we are ready to solve the problem. First we ote that for a polyomial m P x) = a j x j we have, based o ), that j=0 2 m ) P = ) j=0 a j j 2 ) ) j = 0. 2) Let ɛ > 0 ad let P ɛ be the polyomial that uiformly approximates f, i.e. fx) P ɛ x) < ɛ for all x [0, ]. We have, based o 2), that 7

8 2 ) P ɛ = 0. Thus, 2 ) f = 2 2 ɛ = ɛ. Thus, the limit is 0 ad the problem is solved. ) ) f 2 f ) ) P ɛ )) ) P ɛ ) [] I.S. Gradshtey ad I.M. Ryzhi, Table of Itegrals, Series, ad Products, Sixth Editio, Ala Jeffrey, Editor, Daiel Zwilliger, Associate Editor, Solutio 2 by Shai Covo, Kiryat-Oo, Israel We will show that lim 2 ) log 2 = 0. ) 2 + The log fuctio i ) has o sigificat role i the aalysis below, we could replace it by ay other cotiuous fuctio.) The lemma below follows straightforwadly from the Cetral Limit Theorem CLT). We recall that, accordig to the CLT, if X, X 2,... is a sequece of idepedet ad idetically distributed i.i.d) radom variables with expectatio µ ad variace σ 2, the P a < X ) + + X µ σ b Φb) Φa) 2) as, for ay a, b R with a < b where Φ is the distributio fuctio of the Normal 0, ) distributio i.e., Φx) = 2π) /2 x e µ2 /2 du). Lemma: For ay ɛ > 0, there exists a r > 0 such that 2 < ɛ 3) 0 /2 r /2+r < for all sufficietly large. Proof: Fix ɛ > 0. Choose r > 0 sufficietly large so that Φ2r) Φ 2r) > ɛ. Let X, X 2... be a sequece of i.i.d. variables with P X i = 0) = P X i = ) = /2. Put Y = i= X i. Thus Y has a biomial, /2) distributio. The X i s have expectatio µ = /2 ad variace σ 2 = /4. Hece by 2) with a = 2r ad b = 2r), P /2 r < Y /2 + r ) > ɛ 8

9 for all sufficietly large. I tur, by taig complemets, we coclude 3), sice the distributio of Y is give by P Y = ) = 2, = 0,...,. It follows from the lemma ad the fact that 2 ) log is bouded uiformly i 2 + say, by 2) that ) will be proved if we show that lim 2 ) log /2 r <</2+r for ay fixed r > 0. This is show as follows. We first write = ) log log = 0 4) 2 + ) ) log )) ). 5) ) + log 2 + ) ) Clearly, the expressio multipyig o the right of the equality i 5) ca be made arbitrarily small uiformly i [/2 r, /2 + r ], where r > 0 is fixed, by choosig sufficietly large. The, i view of the triagle iequality, 4) follows from 2 ɛ = ɛ / where ɛ > 0 is arbitrarily small) ad 2 uif. 0 to be used if the sum i 4) cosists of a odd umber of terms). The desired result ) is thus proved. Also proved by Boris Rays, Brooly, NY ad the proposer. 52: Proposed by Tom Leog, Scotru, PA Let, ad r be positive itegers. It is easy to show that 2 r + r =,, 2, r N r + r r= usig geeratig fuctios. Give a combiatorial argumet that proves this idetity. Solutio by Shai Covo, Kiryat-Oo, Israel Suppose we have idetical boxes ad r ) idetical balls. The stated equality is trivial if r =, hece we ca assume r >. We begi with the left-had side of the stated equality. Assumig,..., r, it gives the umber of ways to divide the boxes ito r groups the ith group havig i elemets ad put exactly balls i each group. 9

10 As for the right-had side, suppose that i additio to the boxes ad the r balls we have r separators. This gives rise to a + r )-tuple of boxes ad separators. We deote this tuple by M. We idetify a sequece i, i 2,..., i r+r ) such that i < i 2 < < i r+r + r with the followig arragemet: the i j th j =,..., r + r ) elemet of M is a separator if j is a multiple of + ad a box cotaiig a ball otherwise. ) The remaiig r elemets are empty boxes.) We thus + r coclude that gives the umber of ways to place r separators betwee the r + r boxes ad r balls ito the boxes, such that each of the resultig r groups cotais exactly balls. This establishes the equality of the left-ad right-had sides. Solutio 2 by the proposer Both sides cout the umber of possible ways to arrage r + r gree balls ad r red balls i a row. This is clearly true for the right side. I the left side, ote that ay term i the sum with i < for some i is equal to zero; so we may assume i for all i. For each compositio + + r = of, cosider the row of red ad r gree balls arraged as } RR {{ R} G } RR {{ R} G RR }{{ R} G G } RR {{ R} balls 2 balls 3 balls r balls G RR R }{{} r balls From each bloc of ) red balls, ) choose of them ad pait them gree. The umber of ways to 2 r do this is. This results i a row cosistig of r + r gree balls ad r red balls. Coversely, i ay row cosistig of r + r gree balls ad r red balls, we ca determie a uique compositio r = of by reversig the process. 0