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1 Prolems Ted Eisenerg, Section Editor ********************************************************* This section of the Journl offers reders n opportunity to echnge interesting mthemticl prolems nd solutions. Plese send them to Ted Eisenerg, Deprtment of Mthemtics, Ben-Gurion University, Beer-Shev, Isrel or f to: Questions concerning proposls nd/or solutions cn e sent e-mil to <eisent@01.net>. Solutions to previously stted prolems cn e seen t < Solutions to the prolems stted in this issue should e posted efore June 15, : Proposed y Kenneth Korin, New York, NY Solve the eqution + 9 = + 9 with > : Proposed y D. M. Bătinetu-Giurgiu, Buchrest, Romni nd Neculi Stnciu, George Emil Plde School, Buzău, Romni If n 1)!! = n 1), then evlute n+1 n + 1)!n + 1)!! lim n n + 1 ) n n!n 1)!!. n 599: Proposed y Ángel Plz, University of Ls Plms de Grn Cnri, Spin Let,, c e positive rel numers. Prove tht : Proposed y Arkdy Alt, Sn Jose, CA Prove the inequlity m + m + c m c 1 R r), where,, c nd m, m, m c re respectively sides nd medins of ABC, with circumrdius R nd inrdius r. 5401: Proposed y José Luis Díz-Brrero, Brcelon Tech, Brcelon, Spin Let,, c e three positive rel numers such tht + + c =. Prove tht ) + c c) c + ) 49. 1

2 540: Proposed y Ovidiu Furdui, Technicl University of Cluj-Npoc, Cluj-Npoc, Romni Clculte ) cos) cos) d, where nd re rel numers. 0 Solutions 579: Proposed y Kenneth Korin, New York, NY Solve: + 1) 4 1) =. Solution 1 y Ed Gry, Highlnd Bech, FL Cross-multiplying nd simplifying gives = 0. Oviously 0, so dividing the polynomil y 40 gives ) = 0, )) + 1 = 0. Letting t = + 1, squring t = nd then sustituting into the ove gives 1 t ) 1t ) + 1 = 0 40 t 1t + 8 = 0, so t 1 = 1 t = ) 1 ). Since t = + 1, we hve t + 1 = 0, nd solving for gives 1 = 1 ) t 1 + t 1 4 = 1 ) t + t 4 = 1 ) t 1 t = 1 ) t t 4. Sustituting in the respective vlues of t nd simplifying gives: 1 =

3 = = = Solution y Brin D. Besley, Presyterin College, Clinton, SC Given rel numer k, we seek ll rel solutions of We require + 1) 4 1) = k k) k) + 4 k) + 1 = + + 1) + + 1) = 0, where = 4 k + k 16k)/ nd = 4 k k 16k)/. Hence there re no rel solutions unless k, 0] [16, ). Solving for, we otin = ± 4)/ or = ± 4)/. We note tht if k = 0, then there is one rel solution; if k < 0 or k = 16, then there re two rel solutions; nd if k > 16, then there re four rel solutions. For the given eqution with k =, we hve four rel solutions: Letting = 1 + )/ nd = 1 )/, we otin = ± 4)/ 4.00 or 0.8; = ± 4)/ 8.44 or Comments: Arkdy Alt of Sn Jose, CA noted in his solution tht the in the sttement of the prolem could e replced with ny of the three numers 15, 16, or 18 to otin more elegnt nswer. For emple, the eqution + 1) 4 = 18 gives the solutions 1) = 5 ± ) 6 = ± = ± = ) 6 ±. Kenneth Korin, proposer of the prolem, stted: If >, then the eqution + 1) 4 1) = 4 ) gives the solutions 1 = + + c + c = 1,

4 = + c c = 1 4, with = nd with c = 4. In the given eqution 4 =. Then = 4, = >, = =, c = 4 = 1 4. Also solved y Htef I. Arshgi, Guilford Technicl Community College, Jmestown, NC; Bruno Slgueiro Fnego, Viveiro, Spin; G.C. Greuel, Newport News, VA; Pul M. Hrms, North Newton, KS; Kee-Wi Lu, Hong Kong, Chin; Moti Levy, Rehovot, Isrel; Dvid E. Mnes, SUNY College t Oneont, Oneont, NY; Polo Perfetti, Deprtment of Mthemtics, Tor Vergt University, Rome, Itly; Boris Rys, Brooklyn, NY; Henry Ricrdo, New York Mth Circle, NY. Toshihiro Shimizu, Kwski, Jpn; Neculi Stnciu, George Emil Plde Generl School, Buzău, Romni nd Titu Zvonru, Comănesti, Romni; Alert Stdler, Herrlierg, Switzerlnd; Dvid Stone nd John Hwkins, Georgi Southern University, Sttesoro, GA; Nicusor Zlot, Trin Vui Technicl College, Focsni, Romni, nd the proposer. 580: Proposed y Arkdy Alt, Sn Jose, CA Let, y, z) = y + yz + z) + y + z ) nd,, c e the side-lengths of tringle ABC. Prove tht F 16,, c ),, c), where F is the re of ABC. Solution 1 y Toshihiro Shimizu, Kwski, Jpn From the Heron s formul, F + + c) + c )c + ) + c) = 16 =,, c ) 16 Thus, it suffices to show tht,, c ),, c),, c ) 0 cn e written s ) c)q,, c), cyc ). The l.h.s) where q,, c) = c) + c) 0. Moreover, since q,, c) q, c, ) = )c + c + c + c ), 4

5 the reltion, which is lrger q,, c) or q, c, ), depends on the vlue of or. Without loss of generlity, we ssume c. Then, q,, c) q, c, ) qc,, ). Thus, ) c)q,, c) = ) c)q,, c) c)q, c, ) ) + qc,, ) c) c) cyc 0. Therefore, ) is true. Note: It is similr to the proof of Schur s inequlity. It seems tht ) is vlid for ny,, c, even if the constrint tht,, c re the side-lengths of tringle is not stisfied. Solution y Alert Stdler, Herrlierg, Switzerlnd Denote y s the semiperimeter of the tringle put s = s, s = s, s c = s c. By the tringle inequlity, s 0, s 0, s c 0. Also = s + s c, = s c + s, c = s + s. Furthermore, we note tht,, c) = s + s c, s c + s, s + s ) = 4 s s + s s c + s c s ) 0. By Heron s formul F = s s s s c = s + s + s c ) s s s c. Therefore we need to prove tht 64 s + s + s c ) s s s c s s + s s c + s c s ) s + s c ), s c + s ), s + s ) ) which is equivlent to 7 s 4 s + 1 s s + 5 s s s c 7 s 4 s s c + 6 s s s c 1) s is seen y simply multiplying out). By Schur s inequlity which is equivlent to s s s s s s c ) s s s c s ) 0 cycl s s + s s s c s s s c ) s is seen gin y multiplying out). We hve the following inequlities 5 s s + 5 s s s c 10 s s s c, y)), 7 s 4 s 7 s 4 s s c, y Muirhed s inequlity, 16 s s 16 s s s c, y Muirhed s inequlity. 5

6 1) follows y dding these three inequlities. Solution y proposer Let s := t 1 + t + t. Since t i < s, i = 1,, tringle inequlities) then our prolem is: Find m s for which there re positive integer numers t 1, t, t stisfying t i min { i, s 1}, i = 1,,, t 1 + t + t = s. First note tht s, t i, i = 1,,. Indeed, since t i s 1, then 1 s t i, i = 1,, nd, therefore, t 1 = s t t = s t ) + s t ). Cyclic we otin t, t. Hence, s 6 s. Since t = s t 1 t, t min {, s 1} then 1 s t 1 t min {, s 1} m {s t 1, s + 1 t 1 } t s 1 t 1 nd, therefore, for t we otin the inequlity 1) m {s t 1, s + 1 t 1, } t min {s 1 t 1,, s 1} with conditions of solvility : ) s t 1 s 1 s t 1 s + 1 t 1 s 1 t 1 s + 1 t 1 s t 1 s + 1 t 1 t 1 s Since s 1 s then ) together with t 1 min { 1, s 1} it gives us ounds for t 1 : ) m {s + 1, s, s + 1, } t 1 min { 1, s 1}. Since i, i =, then s + 1 s 1, s + 1 s 1 nd solvility condition for ) ecomes s s 1 + 1, s 1 s, s s 1 + 1, s s 1 s + 1. { } Thus, s = min, 1 + 1, + 1, is the lrgest vlue of integer semiperimeter. Solution 4 y Andre Fnchini, Cntú, Itly We know tht F = ss )s )s c) where s is the semiperimeter of ABC. Now mking the sustitutions nd clering the denomintors we hve to prove 16ss )s )s c) [ + c + c) + + c ) ] [ + c + c ) c 6 ) ] now we mke the following sustitutions with, y, z > 0) = y + z, = z +, c = + y 6.

7 nd epnding out into etric sums the given inequlity yields LHS: 7 4 y + 4 z + y 4 z + y 4 + z 4 + y z 4 ) + 4 y + y z + z )+ +6 yz + y z + y z + y z + yz + y z ) + 78 y z RHS: so it remins to prove tht 8 4 yz + y 4 z + yz 4 ) or 7 4 y + 4 z + y 4 z + y 4 + z 4 + y z 4 ) 8 4 yz + y 4 z + yz 4 ) which is true ecuse it follows from Muirhed s Theorem, q.e.d. 7[4,, 0] 19[4, 1, 1] 19[4,, 0] 19[4, 1, 1] Solution 5 y Polo Perfetti, Deprtment of Mthemtics, Tor Vergt University, Rome, Itly As well known F = ss )s )s c) nd s = + + c)/. Upon setting = y + z, = + z, c = + y, the inequlity ecomes sym 7 4 y + 1y) + 5yz) ) sym The third degree Schür inequlity is + + c ) + c sym, 7 4 yz + 6 y z ). which pplied to the triple y), yz), z), yields 5 y) + 5 yz) 10 y z. sym sym sym The inequlity ecomes 7 4 yz + 16 y z ), sym 7 4 y + 16y) ) sym nd the proof is complete upon oserving tht y the AGM we hve 4 y + 4 z 4 yz, y) + y) + z) y z. Also solved y Bruno Slgueiro Fnego, Viveiro, Spin; Ed Gry, Highlnd Bech, FL; Pul M. Hrms, North Newton, KS; Kee-Wi Lu, Hong Kong, Chin; Moti Levi, Rehovot, Isrel, nd Nicusor Zlot, Trin Vui Technicl College, Focsni, Romni 7

8 581: Proposed y D.M. Btinetu-Giurgiu, Mtei Bsr Ntionl College, Buchrest, nd Neculi Stnciu George Emil Plde School, Buzău, Romni Prove: In ny cute tringle ABC, with the usul nottions, holds: ) cos A cos B m+1 m+1, where m 0 is n integer numer. Solution 1 y Nikos Klpodis, Ptrs, Greece We first recll Brrow s Inequlity: If, y, z re positive rel numers nd A + B + C = π then yz + z y + y z cos A + y cos B + z 1) This inequlity first ppered in [1]. For solution see [] or [] inequlity.0, pp. -4)). Applying inequlity 1) for = cos A, y = cos B, nd z = note tht cos A, cos B, > 0, since ABC is n cute tringle) we otin cos A cos B By the following well-known trigonometric identities cos A + cos B + = sin A sin B sin C cos A + cos B + cos C) ) nd sin A sin B sin C = r 4R nd Euler s inequlity r R) we otin tht cos A + cos B + = 1 + r R Using the AM-GM inequlity nd inequlity ) we hve ) cos A + cos B + 1 cos A cos B ) = 1 4) 8 Furthermore, y the identity cos A + cos B + cos C + cos A cos B = 1 nd inequlity 4) we otin ) cos A + cos B + cos C 4 5) By ) nd 5), we hve cos A cos B Finlly, pplying Rdon s Inequlity nd using inequlity 6) we hve tht 6) ) cos A cos B m+1 = ) m+1 cos A cos B 1 m + ) cos B m+1 cos A 1 m + 8 ) cos A m+1 cos B 1 m

9 m+1 cos A cos B ) m ) m m = m+1. References: [1] L. J. Mordell nd D. F. Brrow, Solution 740, The Americn Mthemticl Monthly Vol. 44, No. 4 Apr., 197) pp [] R. R. Jnic, On A Geometric Inequlity Of D. F. Brrow, Univ. Beogrd. Pul. Elektrotehn. Fk. Ser. Mt. Fiz. No ), pp [] O. Bottem, R. Z. Djordjevic, R. R. Jnic, D. S. Mitrinovic, nd P. M. Vsic, Geometric Inequlities, Wolters-Noordhoff Pulishing, Groningen, The Netherlnds, Remrk: Inequlities ), 4), nd 5) lso pper respectively s inequlities.16,. nd.1 in reference []. Solution yángel Plz, University of Ls Plms de Grn Cnri, Spin By the RMS-AM inequlity it is enough to prove tht cos A cos B. Tking into ccount tht A + B + C = π, then = cos π A + π B) = sin A sin B cos A cos B, so the inequlity to e proved my e written with cotngents s cot A cot B 1 cot A cot B, or 1 1 cot A cot B 9. It is well known tht if α = cot A, β = cot B, nd γ = cot C, then αβ + βγ + γα = 1. Therefore, tking = αβ, y = βγ, nd z = γα we hve to prove tht which follows y Jensen s inequlity, since function f) = 1 1 Solution y Henry Ricrdo, New York Mth Circle, NY Elementry clcultions show tht for A, B, C 0, π/) is conve for 0, 1). Furthermore, we hve cos A cos B = tn C tn A + tn B. 1) tn C tn A + tn B. ) y Nesitt s inequlity. 9

10 Finlly, 1), ), nd the power men inequlity give us ) cos A cos B m+1 = ) tn C m+1 tn A + tn B 1 tn C tn A + tn B ) 1 m+1 = m+1. Solution 4 y Nicusor Zlot, Trin Vui Technicl College, Focsni, Romni Using the inequlity m+1 + m+1 + c m+1 1 m + + c)m+1 *) we hve ) cos A cos B m+1 = ) tn C m+1 1 ) tn C m+1 tn A + tn B y ) m. ) tn A + tn B Setting tn A =, tn B = y, tn C = z, nd using Nesitt s inequlity, we hve tn C tn A + B = z, ) + y y Nesitt) The sttement of the prolem follows from ) nd ). Also solved y Bruno Slgueiro Fnego, Viveiro, Spin; Ed Gry, Highlnd Bech, FL; Kee-Wi Lu, Hong Kong, Chin; Moti Levi, Rehovot, Isrel; Toshihiro Shimizu, Kwski, Jpn, nd the proposer. 58: Proposed y Ángel Plz, University of Ls Plms de Grn Cnri, Spin Prove tht if,, c re positive rel numers, then m+1 Solution 1 y Henry Ricrdo, New York Mth Circle, NY By the rithmetic-geometric men inequlity, ech of the sums greter thn or equl to. Thus Solution y Ed Gry, Highlnd Bech, FL Clerly, if = = c, the ove product ecomes, + 8 ) + 8 ) = 7 = )) )) = + 4) + 4) = 7 = 79 = is

11 Therefore, if we show tht the product is minimum when ll vriles re equl, then the conjecture would e true. It is sufficient to clculte the product in two different wys. First, suppose tht = nd c = Second, suppose = nd c = If oth of these products eceed 79, tht would show tht if ll vriles re not equl, we do not hve minimum. Cse 1: =, c = The product ecomes ) ) ) ) Cse : =, c = The product ecomes ) ) = QED = Solution y Dionne Biley, Elsie Cmpell, nd Chrles Diminnie, Angelo Stte University, Sn Angelo, TX We egin y pplying the etension of the Arithmetic - Geometric Men Inequlity which sttes tht if α, β,, y > 0 nd α + β = 1, then α + βy α y β, with equlity if nd only if = y. It follows tht + 8 = + 8 ) = ) 9 9 ) 1 = 9 9 ) 7 9, with equlity if nd only if =, c = c, nd c =, i.e., if nd only if = = c. c Net, pply the stndrd version of the Arithmetic - Geometric Men Inequlity to get ) 7 9 ) 8 9 ) 7 9 = 7, 1) with equlity if nd only if = c =, i.e., if nd only if = = c. c 11

12 A similr set of steps yields with equlity if nd only if = = c. + 8 Therefore, y 1) nd ), + 8 with equlity if nd only if = = c. + 8 Solution 4 y Andre Fnchini, Cntú, Itly 7, ) 7 = 9, Clering the denomintors nd mking the multiplictions we hve c c + c 4 + c 4 ) c + 4 c + c 4 ) c + c )+ or which is true ecuse +16 c + c + c + c + c + c ) 54 c 16[4,, 0] + 65[4, 1, 1] + 65[,, 0] + [,, 1] 8[,, ] 16[4,, 0] 16[,, ] 65[4, 1, 1] 65[,, ] 65[,, 0] 65[,, ] [,, 1] [,, ] ech of which follows from Muirhed s Theorem, q.e.d. Also solved y Arkdy Alt, Sn Jose, CA; Htef I. Arshgi, Guilford Technicl Community College, Jmestown, NC; Michel Brozinsky solutions), Centrl Islip, NY; Bruno Slgueiro Fnego, Viveiro, Spin; Pul M. Hrms, North Newton, KS; Kee-Wi Lu, Hong Kong, Chin; Moti Levi, Rehovot, Isrel; Nikos Klpodis, Ptrs, Greece; Polo Perfetti, Deprtment of Mthemtics, Tor Vergt University, Rome, Itly; Boris Rys, Brooklyn, NY; Neculi Stnciu, George Emil Plde School, Buzău, Romni nd Titu Zvonru, Comănesti, Romni; Toshihiro Shimizu, Kwski, Jpn; Alert Stdler, Hellierg, Switzerlnd; Nicusor Zlot, Trin Vui Technicl College, Focsni, Romni, nd the proposer. 58: Proposed y José Luis Díz-Brrero, Brcelon Tech, Brcelon, Spin Let n e positive integer. Find gcd n, n ), where n nd n re the positive integers for which 1 5) n = n n 5. Solution 1 y Ethn Gegner Undergrdute student, Tylor University), Uplnd, IN 1

13 The eqution n+1 n+1 5 = n n 5)1 5) = n + 5 n n + n ) 5 yields the recurrence reltions Thus, n+1 = n + 5 n n+1 = n + n gcd n+1, n+1 ) = gcd n + 5 n, n + n ) = gcd6 n n 1, n n 1 ) = gcd16 n + 40 n, 8 n + 16 n ) = 8 gcd n + 5 n, n + n ) = 8 gcd n, n + n ) = 8 gcd n, n ) Since 1 = 1 = 1, = 6, =, = 16, = 8, we hve gcd 1, 1 ) = 0, gcd, ) = 1, gcd, ) =. It follows inductively tht { n : n gcd n, n ) = n 1 : otherwise Solution Ángel Plz, University of Ls Plms de Grn Cnri, Spin For n = 1, 1 = 1 = 1. For n > 1, ) n n 5 = n 1 n ) 5 n = n n 1 5 n 1 + n 1 ) so n = n n 1, nd n = n 1 + n 1, or in mtri form ) ) ) ) n 1 5 n 1 n 1 5 = = n 1 n ) n 1 1 Therefore, n = n 1 L n nd n = n F n, where L n nd F n+1 respectively re the nth Lucs nd the nth Fioncci numers. Since L n = F n 1 + F n+1, then gcdl n, F n+1 ) = 1 nd hence gcd n, n ) = n 1, if L n is odd, while gcd n, n ) = n if L n is even, tht is when n is multiple of. Solution y Crl Liis, Columi Southern University, Ornge Bech, AL Let 1 5 ) n = n n 5. Then ) n+1 n+1 5 = n n 5 1 ) 5 = n + 5 n ) n + n ) 5. Thus, i) n+1 = n + 5 n, ). 1

14 ii) n+1 = n + n, nd using i), nd ii) we cn show tht iii) n+1 = n + 4 n 1, iv) n+1 = n + n 1. By oservtion we note from the first few terms v) n = n 1 l n, vi) n = n 1 f n, where l n nd f n re Lucs nd Fioncci numers. We cn verify v) nd vi) y sustituting them into iii) nd iv). {, if n It is well known tht gcdf n, l n ) = 1, otherwise.. See< or < Therefore, gcd n, n ) = gcd n 1 f n, n 1 l n ) = n 1 { {, if n 1, otherwise. = n, if n n 1, otherwise. Comment y Editor: Kenneth Korin of NewYork, NY oserved connection etween this prolem nd the solution to prolem 57, tht required us to find positive integers nd y such tht = + y 5, the unique nswer of which ws, y) = He continued on s follows. Oserve tht: 1 5 ) 1 = 4096) ) = ), nd lso 161) 7 5) = 1. So, ) ) ) = ) = = ) ) 6. And dditionlly: 6 ) 6 = Also solved y Arkdy Alt, Sn Jose, CA; Dionne Biley, Elsie Cmpell, nd Chrles Diminnie, Angelo Stte University, Sn Angelo, TX; Brin D. Besley, Presyterin College, Clinton, SC; Bruno Slgueiro Fnego, Viveiro, Spin; Ed Gry, Highlnd Bech, FL; G.C. Greuel, Newport News, VA; Kenneth Korin, New York, NY; Kee-Wi Lu, Hong Kong, Chin; 14

15 Moti Levi, Rehovot, Isrel; Dvid E. Mnes, SUNY College t Oneont, Oneont, NY; Toshihiro Shimizu, Kwski, Jpn; Dvid Stone nd John Hwkins, Georgi Southern University, Sttesoro, GA, nd the proposer. 584: Proposed y Ovidiu Furdui, Technicl University of Cluj-Npoc, Cluj-Npoc, Romni Find ll differentile functions f : R R which verify the functionl eqution f ) + f ) =, for ll R. Solution 1 y Michel Brozinsky, Centrl Islip, NY We hve t once from the given eqution f ) + f ) = tht f0) = 0, nd since = ) tht f ) + f) = f ) + f) so tht f ) + f )) = f) f ) which cn e cst s G ) = G), which we lel s eqution 1) nd in which G) = f) f ). From 1) we hve G) = c for some constnt c nd thus G ) = 0, nd so f ) = f ) nd we lel this s eqution ), where we hve used the chin rule. Now, y differentiting the given eqution twice we hve nd so from ) we hve f ) + f ) + f ) + f ) = f 0) = nd f ) + f ) =. ) Letting v = f ) in ) we hve the liner differentil eqution using the integrting fctor we otin ) dv + v =, nd d dv + vd = d so tht v = + A nd f ) = v = + A 4) where the constnta = 0 since f 0) =. Integrting 4) twice we otin f) = + B + C where B nd C re constnts nd since f0) = 0, we hve C = 0. Hence, the generl solution is f) = Solution y Toshihiro Shimizu, Kwski, Jpn + B, where B is n ritrry constnt. Let P ) e the given eqution. From P ) + P ), we get d d f) + f )) + f) + f )) =. 1) 15

16 From P ) P ), we get d f) f )) f) f )) = 0. ) d First, we solve 1). Let g) = f) + f ).Then, 1) cn e rewritten s dg d = g) The root of this differentil eqution is g) = C/ for constnt C R. Net, we solve ). Let h) = f) f ). Then, ) cn e rewritten s dh d = h) The root of this differentil eqution is h) = D for constnt D R. Thus, f) = g) + / + h))/ = C/ + D + / for some constnt C, D R. Since, f) should e defined for ll R, C must e 0. Therefore, f) = D + /, where D R is constnt nd this stisfies P ). Solution y Bruno Slgueiro Fnego, Viveiro, Spin f ) + f ) =, R = f ) + f )) ), R, tht is f ) + f) =, R = f ) + f ) = = f ) + f), R = f ) f )) + f) + f ) =, R, or equivlently, g ) + g) =, R, where g : R R is the function defined y g) = f) + f ), R, tht is h ) =, with h : R R, R. h) = g), R = h) = + C, for some C R, R implies f) + f ) = g) = h) = + C, {0}. f differentile implies f differentile t = 0 = f continuous t = 0. This fct nd the equlity f) + f ) = + C imply tht C = 0. Hence, f ) = f) nd thus f ) + f) = f ) + f ) =. R {0} = f ) f) = R 0} = f ) f) = 1, R {0}. = k ) 1 f), where k : R R is the function defined y k) =, {0} = k) = + D with D R, R {0} = f) = + D, R {0}. Since ) f ) f ) + f) + f ) =, R = f0) = 0 f 0) f 0)) + f0) + f 0) = 0 = 0, so f0) = 0, we conclude tht f) = + D,, where D is ny rel constnt. Solution 4 y Moti Levy, Rehovot, Isrel The derivtive of f : R R stisfies the functionl eqution 16

17 f ) = f ), 1) hence it is lso differentile function mye ecept for = 0). Differentition of the functionl eqution gives, Sustitution of 1) into ) gives, f ) + f ) f ) =. ) or f ) + f ) + f ) =, f ) + f ) f ) =. ) All the differentile functions which stisfy the functionl eqution f ) + f ) =, must stisfy ). The solutions of the differentil eqution ) re f ) = 1 + α + 1 ) + β 1 ) Now we sustitute 4) in the left side of the originl functionl eqution: d 1 + α + 1 d ) )) + β 1 = 1 α + β ) = α β). 4) + 1 α + 1 ) ) 1 + β It follows tht α must e equl to β for 4) to e solution. All the differentile functions f : R R, which stisfy the functionl eqution f ) + f ) =, for ll R re f ) = 1 + c, c R. Solution 5 y Kee-Wi Lu, Hong Kong, Chin Denote the given functionl eqution y 1). We show tht f) = where k is n ritrry constnt. Replcing t in 1), we otin Sutrcting ) fromf 1), we otin + k, ) f ) + f) =. ) f ) + f ) ) f) f )) = 0. 4)

18 Integrting 4), we otin f) f ) =, where is n ritrry constnt. By sustituting f ) = f) ck into 1). we otin f ) + f) = +. 5) Integrting 5), we otin f) = + +, where is constnt. By putting = 0 we see tht = 0. Thus ) hold for 0. By putting = 0 into 1), we otin f0) = 0 nd so ) hold for = 0 s well. Also solved y Arkdy Alt, Sn Jose, CA; Dionne Biley, Elsie Cmpell, nd Chrles Diminnie, Angelo Stte University, Sn Angelo, TX; Ed Gry, Highlnd Bech, FL; Henry Ricrdo, New York Mth Circle, NY; Dvid Stone nd John Hwkins, Georgi Southern University, Sttesoro, GA, nd the proposer. Editor s Notes The conjecture in 575* hs een revised y its uthor Kenneth Korin of NY, NY to the following: 575* revised): Prove or disprove the following conjecture. Let k e the product of N different prime numers ech congruent to 1mod 4). The totl numer of different rectngles nd trpezoids with integer length sides nd digonls tht cn e inscried in circle with dimeter k is ectly 5N N. Toshihiro Shimizu of Kwski, Jpn provided counter emple to the originl sttement of the prolem tht did not require the digonls to lso e integers. He let k = 5 = 85 nd then developed the trpezoids 4, 4, 4, 8) nd 50, 4, 50, 8). The digonls of these two trpezoids re not of integrl length. Ken commented on Toshihiro s emples y sying tht: It never occurred to me tht trpezoid with integer length sides inscried in circle with dimeter k could hve non-integer length digonls. So with the revision, 575* remins n open prolem. 18