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1 Problems Ted Eisenberg, Section Editor ********************************************************* This section of the Journl offers reders n opportunit to exchnge interesting mthemticl problems nd solutions Plese send them to Ted Eisenberg, Deprtment of Mthemtics, Ben-Gurion Universit, Beer-Shev, Isrel or fx to: Questions concerning proposls nd/or solutions cn be sent e-mil to <eisenbt@0net> Solutions to previousl stted problems cn be seen t < Solutions to the problems stted in this issue should be posted before Mrch 5, 09 55: Proposed b Kenneth Korbin, New York, NY For ever prime number P, there is circle with dimeter 4P 4 + In ech of these circles, it is possible to inscribe tringle with integer length sides nd with re (P (P + (P (P Find the sides of the tringles if P nd if P 554: Proposed b Michel Brozinsk, Centrl Islip, NY A billird tble whose sides obe the lw of reflection is in the shpe of right tringle ABC with legs of length nd b where > b nd hpotenuse c A bll is shot from the right ngle nd rebounds off the hpotenuse t point P on pth prllel to leg CB tht hits let CA t point Q Find the rtio AQ QC 555: Proposed b Dniel Sitru, Ntionl Economic College Theodor Costescu, Drobet Turnu-Severin, Mehedinti, Romni Find rel vlues for x nd such tht: 4 sin (x cos x + 4 cos 556: Proposed b Ionnis D Sfiks, Ntionl nd Kpodistrin Universit of Athens, Greece The lengths of the sides of right tringle re, 6 nd 0 Determine the number of stright lines which simultneousl hlve the re nd the perimeter of the tringle 557: Proposed b José Luis Díz-Brrero, Brcelon Tech, Brcelon, Spin Let, b nd c be positive rel numbers such tht + b + c Prove tht for ll rel α > 0, holds: ( α+ b α α b α + bα+ c α b α c α + cα+ α c α α

2 ( α+ α + bα+ b α + cα+ c α ( α b α c α α b α c α 558: Proposed b Ovidiu Furdui, Technicl Universit of Cluj-Npoc, Cluj-Npoc, Romni Let > 0 Clculte dxd x 6 (x + Solutions 5505: Proposed b Kenneth Korbin, New York, NY Given Primitive Pthgoren Triple (, b, c with b > Express in terms of nd b the sides of Heronin Tringle with re b(b (A Heronin Tringle is tringle with ech side length nd re n integer Solution b Stnle Rbinowitz, Chelmsford, MA One w of doing this would be to form n obtuse tringle ABC s shown with bse of length b nd ltitude of length b, so tht the re of ABC is b(b s desired If the line segment from B to D, the foot of the ltitude from C, hs length, then hpotenuse BC in BDC would hve length c, since this tringle would be similr to n b c right tringle, scled up b Then AD would hve length b, nd b the Pthgoren Theorem, AC would hve length + b Thus, ABC is the desired Heronin Tringle, with sides b, + b, nd + b C + b c b A b B D Solution b Anthon J Bevelcqu, Universit of North Dkot, Grnd Forks, ND Given primitive Pthgoren triple (, b, c with b >, let x b, + b, z + b Note tht c nd z c Since c 4 b > 0 we hve c >

3 We clculte x + + z (b + c + ( + b 4 + c > 0, x + z (b c + c (b + c(c > 0, nd x + z (b + c ( + b (c > 0 Thus x + > z, x + z >, nd + z > x so (x,, z gives the sides of Heronin tringle Let s be the semiperimeter nd A the re of this tringle B Heron s formul we hve A s(s x(s (s z We hve s x + + z b + c, s x b + c (b c +, s b + c c b c, nd s z b + c ( + b c So A (b + c(c + (b c(c [(b (c ][(c ( ] Now (b (c b 4 b + 4 ( + b b 4 b b (b nd (c ( (c 4 (b

4 so A b (b Thus if (, b, c is primitive Pthgoren triple with b > then (x,, z with x b, + b, z + b is Heronin tringle with re b(b NB For prticulr (, b, c there cn be other Heronin tringles with re b(b For exmple, for the primitive Pthgoren triple (5,, we re looking for Heronin tringle with re 440 The formuls bove give the tringle (69, 0, 69, but (4, 0, 07 is nother tringle with re 440 Solution b Tre Smith, Angelo Stte Universit, Sn Angelo, TX Let x b, + b, nd z + b be the lengths of the three sides of the tringle We first observe tht ll of these re positive integers; x nd z obviousl so, nd since + b c, so tht The perimeter of the tringle is + b c c x + + z (b + ( + b + ( + b (c 4 + c + c c + c 4 Then the semiperimeter is s c + c Appling Heron s formul to find the re A, we hve A s(s x(s (s z s(s (b (s ( + b (s ( + b s(s (c 4 (s c(s c (c + c ((c + c (c 4 ((c + c c((c + c c (c + c (c + (c c (c [(c + (c ][(c + ][(c + (c ][(c ] (c (c 4 b (b 4

5 Thus A b(b Editor s Comment : Dvid Stone nd John Hopkins of Georgi Southern Universit dded the following comment to their solution to this problem: So how did we find x,, z? We first tired the simplest possible exmple; (, b, c (, 5, After some lgebr nd some computer help, we found the tringle (x,, z (69, 69, 0 hs the pproprite re From this we conjectured the form for rbitrr x,, z x 69 c 69 5 b z 0 5 c Then it onl required simple lgebr to verif this construction Some Excel computtions lso led us to the broder result (when b < The perfect exmple of computing power ssisting person! Also solved b Brin D Besle, Presbterin College, Clinton, SC; Ed Gr, Highlnd Bech, FL; Ionnis D Sfiks, Ntionl nd Kpodistrin Universit of Athens, Greece; Dvid Stone nd John Hwkins, Georgi Southern Universit, Sttesboro, GA, nd the proposer 5506: Proposed b Dniel Sitru, Theodor Costescu Ntionl Economic College, Drobet Turnu-Severin, Mehedinti, Romni [ ( 00 ( ] Find Ω det Solution b Michel Btille, Ronen, Frnce ( ( ( ( Let A, B, O 5 5 5, I It is redil checked tht AB BA O nd A + B 6I Since AB BA, the binomil theorem gives 00 ( 00 (A + B 00 A k B 00 k ( k Now, if k {,,, 50}, then k0 A k B 00 k A k B k B 00 k (AB k B 00 k O B 00 k O (note tht A k B k (AB k since AB BA nd similrl, if k {5, 5,, 99}, then A k B 00 k A k 00 (AB 00 k O As result, ( gives (A + B 00 A 00 + B 00, tht is, 6 00 I A 00 + B 00 We cn conclude: Ω det(6 00 I 6 00 Solution b Jeremih Brtz, Universit of North Dkot, Grnd Forks, ND Observe ( ([ 5 ] [ ] 00 5 [ 5 5 ] ( [ 5 ] [ 5 ] 99 ( [ ]

6 nd ( ([ 5 It follows tht [ ( Ω det ] [ ] 00 [ ( ] ] ( [ 5 ] [ 5 [( det ] 99 ( [ ] ] 6 00 Solution b Dvid A Huckb, Angelo Stte Universit, Sn Angelo, TX ( ( Let A nd B Mtrices A nd B re ech smmetric, hence orthogonll digonlizble Solving the eqution det (λi A 0 ields λ 0 nd λ 6 s the eigenvlues of A Solving the eqution (λi A x ( 0 successivel for λ 0 nd λ 6 ields 5 ( x 6 nd x 6 5 s corresponding unit eigenvectors So ( (0 ( A Similrl, ( (0 ( B Since for both A nd B the mtrix of eigenvectors is orthogonl, we hve ( 5 A 00 (0 ( B 00 ( (0 ( So Ω det [ ( A 00 + B 00] det ( ( ( ( 6 99, nd ( ( ( ( Solution 4 b Ionnis D Sfiks, Ntionl nd Kpodistrin Universit of Athens, Greece A w to clculte A n for mtrix is to use the Hmilton-Cle Theorem: A Tr(A A + det A I 0 ( ( For exmple, if we hve mtrix A (or A with det A 0 nd Tr(A +, then the Hmilton-Cle theorem becomes: A Tr(A ( + A 6

7 A ( + A ( + A, A n ( + A n ( + n A So we hve: ( nd finll we hve: Ω det ( 00 ( ( 5 ( , ( ( ( ( ( ( 00 (( ( ( 00 ( det ( , 5 ( ( , 5 0 ( Solution 5 b Polo Perfetti, Deprtment of Mthemtics, Tor Vergt Universit, Rome, Itl Let c 6 We know tht ( ( ( ( 5 5/c /c 0 0 5/c /c AΛA 5 5 /c 5/c 0 6 /c 5/c Thus ( ( ( /c 5/c 0 0 5/c /c 0 6 ( /c 5/c BΛB 5/c /c Ω AΛ 00 A + BΛ 00 B ( AΛ 00 A ( BΛ 00 B ( Ω det Also solved b Arkd Alt, Sn Jose, CA; Ashlnd Universit Undergrdute Problem Solving Group, Ashlnd Universit, Ashlnd, Ohio; Brin D Besle, Presbterin College, Clinton, SC; Anthon J 7

8 Bevelcqu, Universit of North Dkot, Grnd Forks, ND; Dionne Bile, Elsie Cmpbell nd Chrles Diminnie, Angelo Ste Universit, Sn Angelo, TX; Pt Costello, Estern Kentuck Universit, Richmond, KY; Dvid Diminnie, Texs Instruments Inc, Dlls, TX; Michel Fleski, Universit Center, MI; Bruno Slgueiro Fnego Viveiro, Spin; Ed Gr, Highlnd Bech, FL; Kee-Wi Lu, Hong Kong, Chin; Moti Lev, Rehovot, Isrel; Crl Libis, Columbi Southern Universit, Ornge Bech, AL; Ismil Mmmdzd (student, ADA Universit, Bku, Azerbijn; Pedro Pntoj, Ntl/RN, Brzil; Rvi Prksh, Oxford Universit Press; New Delhi, Indi; Neculi Stnciu George Emil Plde School, Buzău, Romni nd Titu Zvonru, Comănesti, Romni; Henr Ricrdo (four different proofs, Westchester Are Mth Circle, NY; Tre Smith, Angelo Stte Universit, Sn Angelo, TX; Albert Stdler, Herrliberg, Switzerlnd; Dvid Stone nd John Hwkins, Georgi Southern Universit, Sttesboro, GA; Mrin Ursărescu, Romn Vodă College, Romn, Romni; Dniel Văcru, Pitesti, Romni, nd the proposer 5507: Proposed b Dvid Benko, Universit of South Albm, Mobile, AL A cr is driving forwrd on the rel xis strting from the origin Its position t time 0 t is s(t Its speed is decresing function: v(t, 0 t Given tht the drive hs finite pth (tht is lim t s <, tht v(t/v(t hs rel limit c s t, find ll possible vlues of c Solution b Moti Lev, Rehovot, Isrel We will show tht the set of ll possible vlues of c, is the intervl [ 0, ], ie, 0 c Let us summrize the conditions on the speed function v (t: v (t 0, v (t is decresing function for ll t 0, 0 v (t dt < v(t 4 lim t v(t c, c is rel number Since v(t 0, then clerl c 0 Since v (t is decresing function, then c It follows tht 0 c Now we show tht c cn ttin n vlue in the intervl [ 0, ] Let r be rel number nd r > Then v (t +t r stisfies ll four requirements from the speed function, in prticulr nd 0 dt <, for r >, + tr v (t lim t v (t lim + t r t + r t r r c It follows tht c ( 0, To see tht c cn ttin lso the vlue zero, choose v (t e t To see tht c cn ttin lso the vlue, choose v (t { ln t ln t, for 0 t,, for < t 8

9 Then v (t stisfies ll the four requirements from the speed function, in prticulr nd 0 v (t dt ln + ln, v (t lim t v (t lim t ln t t t ln (t To finish the proof, we hve to show tht c / (, ] Suppose lim t v(t v(t c, then for ever ε > 0, there is rel number t 0 such tht t > t 0 implies v(t v(t > c ε Now we define stircse function s (t, s follows: s (t : (c ε k v (t 0, for k t 0 t 0 t < k t 0, k,, Since the function v (t is positive decresing function for ll t 0, then v (t s (t, hence t 0 v (t dt Integrting the stircse function, we get k t 0 s (t dt s (t dt v (t 0 (c ε k k v (t 0 (c ε ( (c ε k t 0 If c ε then t 0 s (t dt diverges nd so t 0 v (t dt diverges We conclude tht if c > then t 0 v (t dt diverges, contrdicting propert of the speed function Solution b Kee-Wi Lu, Hong Kong, Chin k0 We show tht 0 c ( Let lim s(t L < Then lim s(t L nd 0 s(t < s(t < L for t > 0 Hence, t t b L Hôpitl s rule, we hve lim t L s(t L s(t lim t ds(t dt ds(t dt lim t v(t v(t c Thus ( holds B tking s(t e t, s(t (t + ln(c ln, s(t ln( + e c 0, 0 < c <, c we see tht ech c in ( is in fct dmissible Solution b Albert Stdler, Herrliberg, Switzerlnd We clim tht the set C of possible vlues of c is the closed intervl 9 ccording s [ 0, ]

10 Indeed, if v(t v 0 e t, then v(t is decresing function, v(t lim 0 So 0 C t v(t If nd v(t nd lim t c > v(t v(t lim t then c/ C v 0 + t ln ( + t + t ln ( + t + t ln ( + t So then v(t is decresing function, 0 v(tdt <, nd 0 v(tdt < ( 0, ] C It remins to prove tht if v(t Suppose if possible tht lim t v(t c, where c > c / Let ɛ : > 0 Then there is number T T (ɛ > 0 such tht ɛ < v(t c < ɛ, whenever t > T We conclude v(tn tht T v(rtdt T v(tdt > (c ɛ T v(tdt (c ɛ which is contrdiction, nd the proof is complete Also solved b the proposer 5508: Proposed b Pedro Pntoj, Ntl RN, Brzil T v(tdt (+ɛ Let, b, c be positive rel numbers such tht + b + c Find the minimum vlue of f(, b, c b + b + b bc + c + c c + Solution b Solution b Dionne Bile, Elsie Cmpbell, nd Chrles Diminnie, Angelo Stte Universit, Sn Angelo, TX To begin, we note tht since, b, c > 0 nd + b + c, the Arithmetic - Geometric Men Inequlit implies tht + b + c ( + b + c ( + b + c As result of (, we hve + b + c + b + bc + c + b + b c + c ( + b + ( b + bc + ( c + c + b + b c + c 4 b + b 4 c + c 4 + b + b c + c T v(tdt > ( b + b c + c ( ( + b + c + b + c + (b + bc + c ( b + b c + c + (b + bc + c ( b + b + ( b c + bc + ( c + c ( T v(tdt, 0

11 Then, using propert (, the convexit of g (x on (0,, nd Jensen s Theorem, x we obtin f (, b, c b + b + b bc + c + c c + g (b + b + bg (bc + c + cg (c + g [ (b + b + b (bc + c + c (c + ] g [( b + b + ( b c + bc + ( c + c ] ( b + b + (b c + bc + (c + c f (,, It follows tht( under the conditions, b, c > 0 nd + b + c, the minimum vlue of f (, b, c is f,, Solution b Dvid E Mnes, Oneont, NY We will show tht the minimum vlue of f is B the Arithmetic Men-Geometric Men inequlit, we get f(, b, c b( + b c(b + c (c + ( + (b + (c + We gin use the AM-GM inequlit to obtin ( + + (b + + (c + ( + (b + (c + ( + b + c + 6 Hence, so tht f(, b, c ( + (b + (c + (/ ( + (b + (c + with equlit if nd onl if b c Solution b Bruno Slgueiro Fnego, Viveiro, Spin From Bergström s nd the Arithmetic men -Geometric men inequlities, ( ( ( b ( c b c b f(, b, c + + c b + + c + b + c + b c + + b + + c + b + c +

12 b b c c Equlit is ttined iff it occurs in those two inequlities, tht is, iff b c b + c b + b c + nd b b c c These lst identities re true if nd onl if b c, tht is, if nd onl if b c In this cse equlit is lso obtined in Bergström s inequlit So, the minimum vlue of f(, b, c is, nd this occurs if nd onl if b c Solution 4 b Arkd Alt, Sn Jose,CA ( Since + b (( ( ( b 0 then triples ( b + ( b b (b + ( + +, b b +, c, c +, b, re greed in order nd, therefore, b the c Rerrngement Inequlit ( cc + ( cc + b ( + b cc + cc + cc Also, b Cuch Inequlit ( + cc cc cc + 9 cc + Thus, f(, b, c nd since f(,, we m conclude tht min f(, b, c Solution 5 b Ionnis D Sfiks, Ntionl nd Kpodistrin Universit of Athens, Greece Since c b, then we hve: f(, b, c b + b + b b( b + ( b + b ( b + Tht mens tht we m ssume the function: g(, b b + b b (b + ( + b + + b + ( + b 5 To find the sttionr points of g(, b, work out g nd g nd set both to zero b This gives two equtions for two unknowns nd b We m solve these equtions for nd b (often there is more thn one solution Let (x, be sttionr point If g > 0 nd g bb > 0 t (x, then (x, is minimum point So, g ( + b (6 + b 5 ( + b 5 b( + + b (b + ( + b + b + b + ( + b 5

13 g b (b ( + + b( + 6b (b + ( + b (b + ( + b + ( + b 5 (,, we hve: [ nd for (, b nd for (, b min g(, b min b + b (,, we hve: [ min g(, b min b + b b (b + ( + b + + b ] ( + b 5 b (b + ( + b + + b ] ( + b 5 Solution 6 b Albert Stdler, Herrliberg, Switzerlnd We will prove tht the minimum vlue equls nd the minimum is ssumed if nd onl if b c / To tht end we must prove tht f(, b, c ( + b + c b + b( + b + c + ( + b + c b + b( + b + c + ( + b + c b + b( + b + c We cler denomintors nd get the equivlent inequlit ( + b ( + b 5, 0 4 b +4 b +8 4 c +4 4 c 4 bc+5 4 b c+ 4 bc +8 b c ( ccl ccl ccl ccl ccl ccl ccl ccl B the (weightedam-gm inequlit, 4 bc, 4 b + 4 c ccl ccl ccl 5 b 5 ( b + c 5 b c, ccl ccl ccl 4 c ( 4 c + 6 b4 + 6 c4 b bc, ccl ccl ccl 9 ccl 4 b 7 b c, 9 ccl b 7 b c, 6 ccl 4 c 8 b c, 4 ccl 5 c b c, nd ( follows if we dd the lst seven inequlities In ll seven inequlities equlit holds if nd onl if b c

14 Comment b Stnle Rbinowitz of Chelmsford, MA Problems such s this re esil solvble b computer lgebr sstems these ds For exmple; the Mthemtic commnd Minimize [{/( b + b + b/(b c + c + c/(c +, > 0&& b > 0&&c > 0&&{ + b + c }, {, b responds b sing tht the minimum vlue is nd occurs when b c Also solved b Konul Aliev (student, ADA Universit, Bku, Azerbijn; Michel Btille, Rouen, Frnce; Ed Gr, Highlnd Bech, FL; Trn Hong (student, Co Lnh School, Dong Thp, Vietnm; Snong Hureri, Rttnkosinsomphothow School, Nkon, Pthom, Thilnd; Sern Ibrhimov, Bku Stte Universit, Msilli, Azerbijn; Kee-Wi Lu, Hong Kong, Chin; Moti Lev, Rehovot, Isrel; Polo Perfetti, Deprtment of Mthemtics, Tor Vergt Universit, Rome, Itl; Stnle Rbinowitz of Chelmsford, MA; Neculi Stnciu George Emil Plde School, Buză, Romni nd Titu Zvonru, Comănesti, Romni; Dniel Văcru, Pitesti, Romni; Nicusor Zlot Trin Vui Technicl College, Focsni, Romni, nd the proposer 5509: Proposed b José Luis Díz-Brrero, Brcelon Tech, Brcelon, Spin Let x,, z be positive rel numbers tht dd up to one nd such tht 0 < x, z, z x < π Prove tht ( ( ( z x x cos + cos + z cos < z x 5 5 Solution b Michel Btille, Rouen, Frnce The Cuch-Schwrz inequlit provides ( x cos + ( z cos + z x z cos ( ( x ( ( z (x++z / cos + cos + cos z x Since x + + z, it follows tht the left-hnd side L of the proposed inequlit stisfies ( ( ( ( z x / L cos + cos + cos z x Thus, it suffices to show tht ( ( ( z x cos + cos + cos < z x 9 5 ( Now, Jensen s inequlit pplied to the cosine function, which is concve on (0, π, ields ( /z + z/x + x/ ( ( z cos + cos + cos z x ( x cos ( ( x / 4

15 But we hve z z x x /z+z/x+x/ (b AM-GM nd 0 < /z+z/x+x/ < π π, hence ( /z + z/x + x/ cos cos( (since the cosine function is decresing on (0, π Then ( gives cos ( ( ( z + cos z x + cos x cos( There just remins to remrk tht cos( < 06 5 to obtin the desired inequlit ( Solution b Trn Hong (student, Co Lnh School, Dong Thp, Vietnm LHS BCS ( ( z ( x x + + z cos + cos + cos z x ( ( z ( x cos + cos + cos z x ( Let f(t cos t, t 0, π f (t cos t < 0 ( Using Jensen s we hve: ( ( ( z ( x z f + f + f f z x + z x + x ( z cos + z x + x cos( {( cos(, 7 < 5 5, 46 Solution b Dvid E Mnes, Oneont, NY Let J x cos ( z + cos ( ( z x + z cos x We will show tht J cos < 5 5 B the Cuch-Schwrz inequlit, one obtins ( ( ( z x J x cos + cos + z cos z x ( ( ( z x x + + z cos + cos + cos z x ( ( ( z x cos + cos + cos z x At the risk of being redundnt, note tht ( J cos (x + + z ( cos z z cc cc (x + + z cos ( ( z + (x + + z cos + (x + + z cos z x ( x 5

16 Since the cosine function is concve on the intervl (0, π/, it follows b Jensen s inequlit tht for ech of the following terms in the cclic sum under the squre root sign, we get Therefore, J ( ( z x cos + cos + z cos z x ( ( z cos + z cos z x ( ( z z cos + x cos + cos z x cos ( x z + z x + zx ( ( x x cos z + z x + xz ( ( x + x cos cos z + z x + x ( x cos( + z + x cos ( + cos z + z x + x + cos For the first term, x z + z x + zx, in prentheses bove, observe tht using the Arithmetic Men-Geometric Men inequlit, one obtins ( x z + z x z x xz, ( z x + zx xz x z, ( zx + x x z x z z Summing the bove terms ields x z + z x + zx x + + z ( Using the Cuch-Schwrz inequlit in the Engel-Titu form for the second term in prentheses in J bove, one immeditel obtins z + z x + x ( + z + x z + x + ( Since the cosine function is decresing on the ( intervl [0, π/] nd s result of inequlities ( nd (, it follows tht cos x z + z x + zx cos nd ( cos z + z x + x cos Therefore, ( ( ( z x J x cos + cos + z cos z x cos Finll, note tht for ech of the bove steps the inequlities become equlities if nd onl if x z Solution 4 b Dniel Văcru, Pitesti, Romni One hs x cos z + cos zx + z cos x x + + z cos }{{} z + cos zx + cos x cos z + cos z x + cos x 6

17 sin ( π ( π + sin z x z ( π + sin x {}}{ < (π z ( π + x z ( π + x π ( z + z x + x The inequlit under the brce is true becuse sin x < x, x ( 0, π On the other hnd, one knows tht z + z x + x b the MA-MG inequlit Therefore one hs x cos z cos + zx cos + x < π ( π ( < (5 5 Also solved b Bruno Slgueiro Fnego, Viveiro, Spin; Ed Gr, Highlnd Bech, FL; Kee-Wi Lu, Hong Kong, Chin; Moti Lev, Rehovot, Isrel; Polo Perfetti, Deprtment of Mthemtics, Tor Vergt Universit, Rome, Itl; Ionnis D Sfiks, Ntionl nd Kpodistrin Universit of Athens, Greece; Albert Stdler, Herrliberg, Switzerlnd nd the proposer 550: Proposed b Ovidiu Furdui nd Alin Sîntămărin both t the Technicl Universit of Cluj-Npoc, Cluj-Npoc, Romni Clculte [4 n (ζ(n ], n where ζ denotes the Riemnn zet function Solution b Albert Stdler, Herrliberg, Switzerlnd (4 n (ζ(n n ( ( 4 n n m m n n m ( n m m n ( n m ( m ( m m m 4 m 4 m ( m m Solution b Polo Perfetti, Deprtment of Mthemtics, Tor Vergt Universit, Rome, Itl 7

18 [4 n (ζ(n ] n k n lim n k + lim n k 4 n k n k [4 n k n ] n 4 k ( 4 n [ k k n [ k + ] k + k k ] + lim 4 k 4 k n n k n k [ k k n k n ] + lim n n k [ k ] + k + Solution b Moti Lev, Rehovot, Isrel Let Then S : (4 n (ζ (n, S N : n S N n N (4 n (ζ (n n (( N ( n N k n k n ( k n N n k n k 4 k 4 ( k k N n k n N S lim N S N k k 4 k 4 ( k k + k k k k5 Now, s bonus, let us evlute prmetrized version of the bove sum: Then S (t : S (t N n (t n (ζ (n tn n, S N (t : (( N t n n ( N t n k n k n k n k n ( tn N n k n n k N n t n k n N t n n N t n k n k S (t lim S (t N t N k t 8 k (t n (ζ (n tn n N t n n n t k t ( ( t N k

19 Let us ssume tht t is not positive integer nd stisfies the inequlit t >, then ψ (t + ψ ( t + k, t t k where ψ (t is the Digmm function ψ ( z + ψ (z + π cot (πz k t ( cot (πt t t k S (t t t ( π cot (πt t t t t t 5t4 5t + 4 πt ( t 4 5t + 4 cot (πt (t 4 5t + 4 We summrize our result s follows, 5t 4 5t + 4 πt ( t 4 5t + 4 cot (πt (t 4 5t, t < + 4 S (t 5t 4 5t + 4 πt ( t 4 5t + 4 cot (πt lim t (t 4 5t 5, t, + 4 5t 4 5t + 4 πt ( t 4 5t + 4 cot (πt lim t (t 4 5t + 4 5, t Remrk: The function S (t, s defined bove, is continuous in the intervl t < Reference: Borwein, Jonthn; Brdle, Dvid M; Crndll, Richrd (000 Computtionl Strtegies for the Riemnn Zet Function J Comp App Mth ( : Solution 4 b Kee-Wi Lu, Hong Kong, Chin Denote the sum of the problem b S so tht S k n n k ( n k Since the summnds re positive, so interchnging the order of summtion, we hve ( n S 4 k k 4 For n integer M, we hve 4 k k M k 4 ( k 5 M+ k + k It follows tht S 5 9 km k

20 Comment b Editor : Ed Gr of Highlnd Bech, FL wrote: I didn t know n recursive formul tht would help, so I did the sum b brute force, computing the sum of the first 0 terms, getting result of 088This roused m curiosit, so I went to Wolfrm-lph nd sought the sum for gret number of terms, like 00 nd 00 It becme cler tht the nswer is 08 forever Converting this to frction, we get beutiful nswer of 5/4 He continued on sing tht he did not ctull solve the problem This is being mentioned here s ver useful heuristic for getting feel for the problem, nd s cvet tht there re n infinite number of different ws to express closed form representtion for specific deciml Also solved b Michel Btille, Rouen, Frnce; Bruno Slgueiro Fnego, Viveiro, Spin; Ionnis D Sfiks, Ntionl nd Kpodistrin Universit of Athens, Greece, nd the proposer Me Culp Mr Wgner-Krnkel of St Mr s Universit in Sn Antonio, TX should hve been credited with hving solved problem