*********************************************************

Transcription

1 Problems Ted Eiseberg, Sectio Editor ********************************************************* This sectio of the Joural offers readers a opportuity to echage iterestig mathematical problems ad solutios. Please sed them to Ted Eiseberg, Departmet of Mathematics, Be-Gurio Uiversity, Beer-Sheva, Israel or fa to: Questios cocerig proposals ad/or solutios ca be set to <eisebt@.et>. Solutios to previously stated problems ca be see at < Solutios to the problems stated i this issue should be posted before February 5, 7 54: Proposed by Keeth Korbi, New York, NY A equilateral triagle is iscribed i a circle with diameter d. Fid the perimeter of the triagle if a chord with legth d bisects two of its sides. 54: Proposed by Arsala Wares, Valdosta State Uiversity, Valdosta, GA Polygo ABCDE is a regular petago. Petago P QRST is bouded by diagoals of petago ABCDE as show. Fid the followig: the area of petago PQRST the area of petago ABCDE.

2 54: Proposed by Oleh Fayshtey, Leipzig, Germay Let a, b, c be the side-legths, r a, r b, r c be the radii of the e-circles ad R, r the radii of the circumcircle ad icircle respectively, ad s the semiperimeter of ABC. Show that r a r + r b r c s bs c + r b r + r c r a s cs a + r c r + r a r b s as b. 544: Proposed by Nicusor Zlota, Traia Vuia Techical College, Focsai, Romaia Let a, b, c ad d be positive real umbers such that abc + bcd + cda + dab 4. Prove that a 8 a 4 + 4b 7 b + 4c 6 c + 4d 5 d : Proposed by José Luis Díaz-Barrero, Barceloa Tech, Barceloa, Spai Let F be the th Fiboacci umber defied by F, F, ad for all, F F + F. If is a odd positive iteger the show that + deta is

3 the product of two cosecutive Fiboacci umbers, where F F F F F... F F F F F F F F F A F F F F F F F F F F F F F F 546: Proposed by Ovidiu Furdui, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia Let a be a strictly icreasig sequece of atural umbers. Prove that the series a [a, a + ] coverges. Here [, y] deotes the least commo multiple of the atural umbers ad y. Solutios 54: Proposed by Keeth Korbi, New York, NY Let φ + 5. Solve the equatio + φ φ + φ with > φ. Solutio by Dioe Bailey, Elsie Campbell, Charles Dimiie, ad Karl Havlak, Agelo State Uiversity, Sa Agelo, TX Let a + φ ad b φ. We may write a b φ a b φ a a b + ab b φ a b aba b φ + φ φ φ φ φ. Simplifyig this last equatio we obtai φ φ/. Uder the coditio > φ, φ the solutio to this equatio is 7 + φ φ. 9 Solutio by Bria D. Beasley, Presbyteria College, Clito, SC. Give ay real umber a >, we solve the equatio + a a + a with > a. Similarly, give ay real umber a <, we may solve the equatio + a a + a with < a.

4 Rewritig the give equatio ad cubig both sides yields + a + a a + + a a a a, or a a + a a. The a a a, so cubig oce more produces 7a a a. Hece 8 7 a, so requirig > a yields a. I particular, whe a φ, we 9 obtai the solutio + 5 φ. 9 9 Solutio by David E. Maes, SUNY College at Oeota, Oeota, NY The value of > φ that satisfies the equatio is [ + ] φ Oe otes that > φ ad does satisfy the equatio. Let v φ. The v φ so that v + φ. Sice we wat the solutio > φ, it follows that must be positive. The origial equatio i terms of v is + φ φ + v. Cubig both sides of this equatio, we get φ v + φ v φ. Dividig by φ reduces this equatio to the moic quadratic equatio v + φ v φ with roots v φ ± 7 φ. Rejectig the egative root yields Hece, v φ + φ 7 + φ 6 [ + ] v + φ φ + 6. φ. 9 Editor s commet : D. M. Bătietu-Giurgiu of Matei Basarab Natioal College, Bucharest, Romaia with Neculai Staciu of George Emil Palade School, Buzău, Romaia geeralized the problem as follows: Let a, b, c, >, with a + b c the it ca be show that the uique real-valued solutio to the equatio + a b + c, where > c is b a + 4c

5 If a b φ, the φ ad the equatio + φ φ+ φ with > φ, has the solutio φ φ + 4φ 7 6 φ. 9 Also solved by Ada Ali studet, A.E.C.S-4, Mumbai, Idia; Arkady Alt, Sa Jose, CA; Ashlad Uiversity Udergraduate Problem Solvig Group, Ashlad, OH; D. M. Bătietu-Giurgiu of Matei Basarab Natioal College, Bucharest, Romaia with Neculai Staciu of George Emil Palade School, Buzău, Romaia; Bria Bradie, Christopher Newport Uiversity, Newport News, VA; Bruo Salgueiro Faego, Viveiro, Spai; Ed Gray, Highlad Beach, FL; Kee-Wai Lau, Hog Kog, Chia; Moti Levy, Rehovot, Israel; Boris Rays, Brookly, NY; Albert Stadler, Herrliberg, Switzerlad; David Stoe ad Joh Hawkis, Georgia Souther Uiversity, Statesboro, GA; Nicusor Zlota, Traia Vuia Techical College, Focsai, Romaia; the proposer, ad Studets from Taylor Uiversity see below; Studets at Taylor Uiversity, Uplad, IN. Group. Be Byrd, Maddi Guillaume, ad Makayla Schultz. Group. Caleb Kuth, Michelle Frach ad Savaah Porter. Group. Laure Morelad, Aa Souzis, ad Boi Hermadez 544: Proposed Arkady Alt, Sa Jose, CA For ay give positive iteger, fid the smallest value of the product of..., where,,,... > ad Solutio by Ed Gray, Highlad Beach, FL Suppose each term had the value of. Sice there are terms, the sum is equal to, satisfyig the problem restrictio. I the evet for each k, k. + k, so k, ad the value of the product is:.. If this is ot the smallest product, at least oe value of k must be less tha. Suppose k e where e >. The the series cotais the therm. We must icrease the value of aother + k e term so that the sum maitais the value of. We must have:. e + + m m e e + m e 6. + m e e 7. + m e e e e e e 5

6 e 8. m e 9. The ew product is: e e e e + e e k m. If the ew product is to be smaller, we must have:. e e <, or dividig by e. e e + e < e,. e e+ e < e +, which simplifies to:. e + e < e. Dividig by e, 4. + <, which is a cotractio. Therefore, we did ot decrease the product, but e icreased it. So is the miimum product. Solutio by Ramya Dutta studet, Cheai Mathematical Istitute Idia Cosider the polyomial P j + j, the P P Deotig the j-th symmetric polyomial by, σ j σ, P σ j j ad P jσ j j j j Therefore, the coditio P P is equivalet to, Usig, AM-GM iequality: σ j I.e., writig σ / j α, we have, j k <k < <k j σ j σ j j σ j/ for j. j α j σ j j α j j j α j j j j + j, i.e., P P. k k kj α j j + α α α + α α α + α + + α that is, + α + α α sice, α > So, the miimum value of is. for j ad Solutio by David Stoe ad Joh Hawkis, Georgia Souther Uiversity, Statesboro, GA 6

7 We shall use the Method of Lagrage Multipliers to show that the smallest value of the product is, achieved whe each i. First suppose that all but oe of the i are equal: let i b for i ad choose so that the costrait is satisfied: + i i i + i + b +, + b > to make >. The the product f,,..., i i b b. b, where We ote that as b becomes ubouded positive, the product of the is becomes ubouded positive, ad as b approaches from above, the product of the is also becomes ubouded positive. Thus if the product has a absolute etremum subject to the give costrait, it must be a miimum sice the product is ubouded above. For b, we see that, so every i ad the product is equal to, We cosider this as a Lagrage Multiplier problem where we miimize the product f,,..., i subject to the costrait i i + i That is, subject to the costrait g,,..., i i i i By the Method of Lagrage Multipliers, we ll fid the miimum of f where f,,..., λ g,,..., for k. i i We see that: f,,..., i ad g,,..., i i k. Thus we wat to solve the system, i ik i i ik Solvig each equatio for λ gives λ + k λ, for k. + k i i k Hece, for ay j, k we must have λ + i i for k. i i k i + j + i for i i i k 7

8 Algebra gives j + j k, j, k. + k We claim this forces i k. Suppose that k i for some k j. Now cosider the fuctio h Note that h i h k for j, k for >. + By calculus, h is strictly icreasig for < < to a maimum of /4 at, ad is the strictly decreasig for >. That is, h ecept for the peak at is two- to- oe fuctio for >. Moreover, h has the reflective property h h. Hece, for j k, h j h k ad j k j k. The jour costrait becomes + + other positive terms + k + j + k + + k + other positive terms which is impossible. Therefore, k j. + k + k + k + other positive terms + other positive terms Hece, to achieve the etreme value, which must be a miimum, all of the i are equal ad must equal, forcig the miimum value of the product to be. Solutio 4 by Nicusor Zlota, Traia Vuia Techical College, Focsai, Romaia Deote by y i i y i, y i >, i,,, + i y i By the AM-GM, we get... i y i y i y + y y y... y + y y y y y... y. y y... y So,.... Equality occurs for.... Editor s commet : I additio to a geeral solutio to this problem, the problem s author, Arkady Alt of Sa Jose, CA, also provided 4 differet solutios for the cases. Solutio A. Let. We have Sice + + 8

9 the + +. Solutio B. Sice Similarly we obtai Hece, + + +, Solutio C. Let a : +, b : +, c : + the a, b, c,, a + b + c ad a a b + c a Therefore, bc a Solutio D. bc a, b c + a b b ca b ab c 8. ca, c a + b b c c ab. c First ote that at least oe of the products,, must be greater the. Ideed,assume that,,. The sice ad we obtai a cotradictio Let it be > ad let t :, r :. The becomes r t r ad, sice + t, t >, we obtai r r t t r t t t + r t t t , because t + t t. t + t Commet by Editor: Neculai Staciu of George Emil Palade School, Buzău, Romaia ad Titu Zvoaru of Comăesti, Romaia, stated that there is a paper i the Romaia Mathematical Gazette, Volume CXX, umber, 5 pp by Euge Păltăea that presets five solutios ad etesios for the followig propositio: Let,,..., >,. If , the They preseted a ew solutio to this propositio ad the applied it to problem 544. Also solved by Ada Ali, Studet i A.E.C.S-4, Mumbai, Idia; Bruo Salgueiro Faego, Viveiro, Spai; Kee-Wai Lau, Hog Kog, Chia; Moti Levy, Rehovot, Israel; Hery Ricardo, New York Math Circle, NY; Albert Stadler, Herrliberg, 9

10 Switzerlad; ad the authors. 545: Proposed by D. M. Bătietu-Giurgiu, Bucharest, Romaia ad Neculai Staciu, George Emil Palade School, Buzău, Romaia If a, b R such that a + b, e + ad c l +, the compute k k lim + a + +!c b a b e. Solutio by Ramya Dutta studet, Cheai Mathematical Istitute Idia Usig log + log! For >, + k k k, for < < ad the Stirlig Approimatio: k log + log π + O!e b/ b log! ep + b ep b log + b log b log e b b ep e b b + b log e b b + b log Agai, c H log γ + + O Therefore, c b/+ b log π b + + b log π + O + b log π + O + O + + b b log + b + O + b log π + b log + O b log c ep + b log γ ep + + b + log b log γ ep + O + γ + O

11 Similarly, +!c b/+ b log + e b + b b log π ep O b log e b + b b log π b log γ ep + + O ep e b + b + b log + b logπγ log + O c b/+ + O Thus, lim + a + +!c b!e a b lim e b + + b log + b logπγ + O e b + b log lim e b + O lim e b + b log γ b + O log b log π + + b log + O log b log γ + e b b log + O log e b a + b log γ Solutio by Albert Stadler, Herrliberg, Switzerlad The th harmoic umber admits the asymptotic epasio k k l + γ + O, as. See for istace umber. Stirlig s formula states that! π e + O, as. See for istace %7s approimatio. So + a + +!c b b + a+b π e b + O + e b + b b + γ + O b + b + logπ + b + log + + b + log γ + O e b + + b logπ + b log log + + b log γ + O, log

12 !e b a+b π b b e b + O e b + b logπ + b log + b + O b + b log e b + b logπ + b log log + b + O., Thus lim + a + +!c b!e a b lim e b + + b logπ + b log + + b logγ b logπ b log log b + O e b + b log γ b e b a + b log γ. Also solved by Arkady Alt, Sa Jose, CA; Bria Bradie, Christopher Newport Uiversity, Newport News, VA; Kee-Wai Lau, Hog Kog, Chia; Moti Levy, Rehovot, Israel, ad the proposers. 546: Proposed by Corel Ioa Vălea, Timis, Romaia Calculate: where H k k H deotes the harmoic umber. ζ, Solutios ad by Ramya Dutta studet, Cheai Mathematical Istitute Idia Solutio : Chagig the order of summatio i ad usig H ζ k k k k k k k H H k k H k + H k, we have: H 4 H H k + H k k H 4 H 4

13 Lemma: Proof: k H k k + k H + H + ζ H k H k k + k k k j j kj+ j k j k k k jk + k j kj jk + k + j jk + k j + j jk + j + k + j + jk + k + j + H +k k + k + j H +k k + k + j j j + j ζ H j + j j + j Justificatios: Iterchaged order of summatio, made the chage i variable k k + j, 4 used the symmetry of the summatio w.r.t. k ad j, j k jk + j + k + j j k j k 5 used the idetity, H m m m 6 used partial fractio, j k jk + j + k + j + jk + k + j, j j j m + j kk + j + k + j j k j kk + j + k + j jm + j ad j + j j j + j j ζ H.

14 Agai, k H +k k + k k k k k Hk k H +k + + k H k k + H k k + H k k j k j k k H + H H+k H k k k 7 k + j 8 j k k + j 9 Thus, combiig lies 6 ad, k H k k + k H + H + ζ H Now, dividig both sides of the idetity with ad summig over, H + H + ζζ H 4 k k H k k H k k + k ζ H k k where, we used partial fractio decompositio from lie 6 earlier. That is, H + H ζ H ζζ + H 4 + H I Now we provide a evaluatio of the Euler sum: Cosider the partial fractio decompositio, k k k k k H. k + k k + k + H + H k + k 4

15 Dividig both sides by ad summig over, H + H k m k m k m k ζζ k k chage of variable m + k k m k + m km + k k k m k + m k m m m k H m m 4 km k + m i.e., H ζζ 5 H 4 + ζ5 II Thus, combiig the results from I ad II, H ζ k H + H k Usig Euler s summatio formula: ζ ζ H 4 H ζζ H 6 ζζ H 4 + H 4 + ζ5 H q + q ζq + q ζj + ζq j, for q we have the particular cases, i.e., Solutio : H H ζ j ζ ad k H 4 ζ5 ζζ, k ζζ 7 ζ5 H 5

16 We start with evaluatig the itegral for a >, Thus, So, a log d lim b b log d H + H H ζ k a b d ΓaΓb lim b Γa + b γ + ψa + + ζ ψ a + a k b Usig the reflectio formula for Dilogarithm, Li + Li ζ log log we may rewrite the itegral as, Li log d ζ The first itegral I: log d }{{} I The secod itegral II: Usig H + H H 4 log d Li log log log d }{{} II log log log log d d log d log d ζ H, log log d d H log d 6 6 H H 4 H 4 Li log d }{{} III H 4 6ζ5

17 The third itegral III: Li log Combiig the results, H ζ k k d Li log d H log d ζζ 4 ζζ H + H H 4 + 4ζ5 ζζ 7 ζ5 H 4 + ζ5 ζ5 H Editor s commet : Albert Stadler of Herrliberg, Switzerlad metioed i his solutio H k that the epressio k 4 π ζ + ζ5 is due to Euler ad that Euler wet o to 6 k geeralize this formula as follows: m H m m + ζm + ζm ζ +, m,,... The referece he gave for this is: L.Euler, Meditatioes circa sigulare serierum geus, Novi Comm. Acad. Sci. Petropolitaae 775, Reprited i Opera Omia, ser. I, vol. 5, B.G. Teuber, Berli, 97, pp Solutio by Moti Levy, Rehovot, Israel We calculate the sum by epressig it as a sum of defiite itegrals ivolvig polylogarithmic fuctio ad the make use of results by Prof. Pedro Freitas []. The tail of ζ is ζ + k. The followig defiite itegral is kow []: l d k +. Substitutig i ad chagig the order of summatio ad itegratio give, ζ +k l d k l 7 k d k l d.

18 H ζ H l d H l d Let F : H, the df d H. The geeratig fuctio of the sequece H is well kow [] H l. It follows that df d l. To fid F we itegrate, l t F t t dt l t l t dt dt t t l + Li 4 Now we substitute 4 i ad obtai the required sum as a sum of two defiite itegrals, H ζ 4 These defiite itegrals appear i [] as etries i Table 6: H l l d + l Li d. l l d 4ζ ζ + 8ζ 5. l Li d 6ζ ζ ζ 5. ζ 6ζ ζ ζ 5 + 4ζ ζ + 8ζ 5 4 ζ ζ 7 π ζ 5 ζ 7 ζ Refereces: [] Freitas Pedro, Itegrals of Polylogarithmic fuctios, recurrece relatios, ad associated Euler sums, arxiv:math/464v [math.ca] Ju 4. [] Gradshtey ad Ryzhik, Table of Itegrals, Series ad Products 7Ed, Elsevier, 7, Etry.7-. [] Roald L. Graham, Doald E. Kuth, Ore Patashik Cocrete Mathematics, A Foudatio for Computer Sciece, d Editio 994, page 5, 7.4. Solutio 4 by Kee-Wai Lau, Hog Kog, Chia We show that the sum of the problem, deoted by S, equals We eed the facts that H + m 4ζζ 7ζ5. l d, see p. 6, of [], m+ l d, see formula.7 of [], ad 8

19 γ m m m + m. For let Li. By followig closely the method of solutio of problem.6 i m [, p. ], we obtai, S y l y l y y ddy y l y y l y Li y y dy 4 I + J, l y l y l y y where I dy ad J dy. It is kow [, p.46, Table 6] y y that I 8ζ5 4ζζ ad J 6ζζ ζ5. Hece the claimed result for the sum of the problem. Refereces:. Freitas P.: Itegrals of polylogarithmic fuctios, recurrece relatios ad associated Euler sums, Mathematics of Computatio, vol. 74, umber 5, Furdui O.: Limits, Series, ad Fractioal Part Itegrals, Spriger, New Hork,. Gradshtey, I.S. ad Ryzhik, I.M.: Tables of Itgerals, Series, ad Products, Seveth Editio, Elsevier 7. Also solved by Bruo Salgueiro Faego, Viveiro, Spai; Ed Gray, Highlad Beach, FL; Albert Stadler, Herrliberg, Switzerlad, ad the proposer. 547: Proposed by José Luis Díaz-Barrero, Barceloa Tech, Barceloa, Spai Fid all triples a, b, c of positive reals such that a + b + c, a + bc + b + ca + c + ab 4 6. Solutio by Neculai Staciu of George Emil Palade School, Buzău, Romaia ad Titu Zvoaru of Comăesti, Romaia Sice a + b + c the a + bc a + bc aa + b + c + bc a + ba + c. We deote a + b, b + c y ad c + a z the + y + z. Usig well-kow iequalities we have 4 6 y + y z + z 9

20 y yz + yz z + z y yz + y + z + y + z 9 + y + z Hece, y z a b c. Solutio by Dioe Bailey, Elsie Campbell, ad Charles Dimiie, Agelo State Uiversity, Sa Agelo, TX Assume that a, b, c > ad a + b + c. The, by the Arithmetic - Geometric Mea Iequality, b + c a a + a + bc a + a +, with equality if ad oly if b c. Sice a, b, c >, it follows that with equality if ad oly if b c. Similar steps show that with equality if ad oly if c a, ad with equality if ad oly if a b. By,, ad, we have [ a + bc + b + ca + c + ab 6 with equality if ad oly if a b c. a + bc 6 a + 4, b + ca 6 b + 4, c + ab 6 c + 4, a b c + 4 ], 4 Further, if f 4, the f > o,, ad hece, f is strictly cove o 6,. If we use Jese s Theorem, we obtai a b f a + + f b + + f c + c + [ ] a + + b + + c + f 4 f 4 56, 5

21 with equality if ad oly if a + b + c +, i.e., if ad oly if a b c. By combiig 4 ad 5, we see that the coditios a, b, c > ad a + b + c imply that a + bc + b + ca + 4 c + ab , with equality if ad oly if a b c. Therefore, the uique solutio for our system must be a b c. Solutio by Bruo Salgueiro Faego, Viveiro, Spai a 4bc b + c 4bc b c with equality iff b c + a a a + a + bc > with equality iff b c 4 4 equality iff b c, ad cyclically, so a + bc + b + ca + c + ab 6 a + bc 6 + a 4 with + a b c 4 with equality iff a b c. By the arithmetic mea geometric mea iequality, a + bc + b + ca + c + ab 6 + a 4 + b 4 + c a + b + c 48 + a + + b + + c with equality iff a b c, so from this ad the secod of the give equatios we coclude that a b c. Editor s commet: D.M. Bătietu-Giurgiu, of Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu of George Emil Palade School Buzău, Romaia geeralized the problem as follows: If a, b, a + b, c, d, m >,, y, z >, such that + y + z s > ad as + b m+ c + dyz m as + btm+ as + bzm+ + + cy + dz m cz + dy m m a + b m+ s, the fid all triples, y, z. c + ds m They foud the solutio that sice + y + z s, the s y + yz + z with equality iff y z s.

22 If s, m, a, b, c, d we obtai + y + z ad cyc + yz 4 6, i.e., problem 547. Also solved by Ada Ali, Studet i A.E.C.S-4, Mumbai, Idia; Arkady Alt, Sa Jose, CA; Ed Gray, Highlad Beach, FL; Ramya Dutta studet, Cheai Mathematical Istitute, Idia; Kee-Wai Lau, Hog Kog, Chia; Moti Levy, Rehovot, Israel; Albert Stadler, Herrliberg, Switzerlad; David Stoe ad Joh Hawkis, Georgia Souther Uiversity, Statesboro,GA, ad the proposer. 548: Proposed by Ovidiu Furdui, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia Calculate: l l d. Solutio by Albert Stadler, Herriliberg, Switzerlad l l d l l d l l d l d ζ. Solutio by Moti Levy, Rehovot, Israel Sice + ad l l d I : l l d l l d the l l d. Usig the Taylor series of l for < <, ad chagig the order of summatio ad itegratio, l k I d k l d. k k k k Gradshtey ad Ryzhik, etry.7-, l l d I k k k l d k ζ. k k

23 Solutio by Kee-Wai Lau, Hog Kog, Chia We show that the itegral of the problem, deoted by I equals. It is well kow that for o-egative itegers. l l d costat. Hece for < a <, we have a l d a l a l + d + a a a +, so that + l d l d a + +. Sice I +, so as asserted. l l d + l l d l l d +, Solutio 4 by Bria Bradie, Christopher Newport Uiversity, Newport News, VA A geeratig fuctio for the Harmoic umbers is H l. The radius of covergece for this series is, so the order of summatio ad itegratio ca be reversed to yield l l l d H d H l d H

24 ζ. Solutio 5 by Ada Ali, Studet i A.E.C.S-4, Mumbai, Idia Let I deote the above itegral ad let f l l ad g +. The f l l ad g l l. Evaluatig I by parts we have I [fg] f gd l [l l l l ] l l l d l l l l d l d + l d l l d l l d Let I l d, the l d I with the substitutio y. Similarly let I l l d, the l l d I with the substitutio y. So, I I I. But we also otice that itegratio of I by parts yields takig / as secod fuctio l l I d [ l l ] l l l d l d l l d I I. Thus I I ad so I I I I. Now to calculate I, we otice that I l d l d 5 Now from itegratio by parts we have by takig as the secod fuctio [ ] l d l + l d l d + [ ] l d + + d +. Substitutig the result obtaied above i, we get I I I ζ. 4 ζ. Thus, +

25 Also solved by Hatef I. Arshagi, Guilford Techical Commuity College, Jamestow, NC; Pat Costello, Easter Ketucky Uiversity, Richmod, KY; Ramya Dutta studet Cheai Mathematical Istitute, Idia; Bruo Salgueiro Faego, Viveiro, Spai; Ed Gray, Highlad Beach, FL; Albert Stadler, Herrliberg, Switzerlad, ad the proposer. Late Ackowledgmet Hery Ricardo of the New York Math Circle should have bee credited for havig solved problem