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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@03et> Solutios to previously stated problems ca be see at < Solutios to the problems stated i this issue should be posted before April 5, : Proposed by Keeth Korbi, New Yor, NY Solve the equatio: 4 x + x 4 x + 4 x x, with x > : Proposed by Titu Zvoaru, Comesti, Romaia ad Neculai Staciu, George Emil Palade Geeral School, Buzău, Romaia Calculate, without usig a calculator or log tables, the umber of digits i the base 0 expasio of : Proposed by Valcho Milchev, Peto Rachov Slaveiov Secoday School, Bulgaria Fid all positive itegers a ad b for which a4 + 3a + ab is a positive iteger 5436: Proposed by Arady Alt, Sa Jose, CA Fid all values of the parameter t for which the system of iequalities 4 x + t y A 4 y + t z 4 z + t x a has solutios; b has a uique solutio 5437: Proposed by José Luis Díaz-Barrero, Barceloa Tech, Barceloa, Spai Let f : C {} C be the fuctio defied by fz 3z z If f z f f fz, the compute f }{{} z ad lim f z : Proposed by Ovidiu Furdui, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia

2 Let 0 be a iteger ad let α > 0 be a real umber Prove that x x α + y y α + z z α x y xy α + y z yz α + x z xz α, for x, y, z, Solutios 545: Proposed by Keeth Korbi, New Yor, NY Give equilateral triagle ABC with iradius r ad with cevia CD Triagle ACD has iradius x ad triagle BCD has iradius y, where x, y ad r are positive itegers with x, y, r Part : Fid x, y, ad r if x + y r 00 Part : Fid x, y, ad r if x + y r 0 Solutio by Ed Gray, Highlad Beach, FL Editor s commet: Ed s solutio to this problem was 8 pages i legth Listed below is my greatly abbreviated outlie of his solutio method All formulas listed below were proved ad/or refereced i Ed s complete solutio He started his solutio with the secod part of the problem ad the applied the methods costructed there to the first part of the problem The reaso for this will soo become apparet Followig is Ed s solutio The followig equatios will be used i the solutio x y 6r 3rp 4r p + 4r + p rp 3rp p + r + 4r + p pr pr p 3 x + y r r + 4r 4 Solutio to Part x + y r 0 Substitutig the value x + y r 0 ito 4 ad solvig for we see that r + 4r 0 404r 00, ad substitutig this ito 3 above we see that 404r 00 pr p p r r 404r + 00

3 Lettig D equal the value uder the square root we have D r 404r + 00 r 404r + 00 D 0 Solvig for r gives r 0 ± D Lettig b D we have b Db + D This implies that there are three possible factorizatios: Case I : Case II : 3 00 Case III : Case I: So, { { b D b 530 b + D D 530 Therefore, r 5504 ad p 03 r 0 + b r 0 b < 0 p r D For these values of r ad p, we evaluate x ad y by usig formulas ad above x 6r 3rp 4r p + 4r + p rp y 3pr p + r + 4r + p pr 5 So for Case I, r 5504, x 5453, y 5, ad x + y r 0 Sice x, y, r have o commo factor, they represet a solutio { { r b D 3 b Case II: b + D 00 D 5099 So, r 0 50 < 0 p r D 05 r 5304 x 55 Followig the path i Case I, we fid that y 53, ad x + y r 0 These terms have o commo factor ad so represet a solutio 3

4 { { r b D 0 b Case III: b + D 303 D 0 So, r 0 0 0, ot viable p r D Give r 404, p 303, ad calculatig as before, we have for Case III, r 404, x 303, y 0, x + y r 0 However 0 divides all three terms, violatig x, y, r, so we do ot have a solutio I summary, ad taig ito accout the iterchageability of x ad y, there are four solutios for Part of the problem: x y r , , , Solutio to Part x + y r 00 I solvig Part of the problem we employ the same techiques that were used i Part We start off by fidig that if r + 00 the 400r 0, 000 Substitutig this ito Equatio 3, gives 4r us 400r 0, 000 pr p ad solvig for p gives us p r r 4004r + 0, 000 The discrimiat, D is give by D r 400r Writig this as a quadratic i r ad solvig for r gives us r 400r + 0, 000 D 0 r 00 ± 30, D Ad as before, lettig b 30, D we obtai b D+b + D 30, Hece there are factors which eed to be writte as the product of two factors Sice b must equal the sum of the two factors, they caot be of opposite parity Followig is a table listig all factorizatios We elimiate those factorizatios that have a odd factor by placig a asteris i frot of them 30, 000 5, , , The remaiig factorizatios represet potetial solutios We will do the first oe i detail but the others we will oly chec to see if x, y, r { { r b D b b + D 5000 D 7499 So, r < 0 p r D Give p r D For r 770, p 0 we calculate x ad y usig the stadard formulas 6r 3rp x 4r p + x r + p rp 4

5 3rp y p + r + y 5 4r + p pr So x 7650, y 5, r 770 These have o commo factor ad so represet a solutio { b D 4 We ow move to the ext case x 3900, y 5, r 395 Sice b + D 7500 x, y, z, this is ot a solutio { b D 6 Ad the ext case x 650, y 53, r 703 Sice b + D 5000 x + y r 00 ad x, y, r this is a solutio { b D 8 Ad the ext x 05, y 54, r 079 Sice x + y r 00 b + D 3750 ad x, y, r this is { a solutio 0pt Worig our way through the table of potetial b D 4 solutios we fid that x 775, y 6, r 837 ad sice b + D 50 x + y r 00 ad x, y, r this is a solutio Systemically worig our way the table we see that may values did ot result i a aswer to the problem Summarizig Part of the problem, ad taig ito accout the iterchageability of x ad y, we see that there are exactly eight solutios x 7650, y 5, r 770 x 650, y 53, r x 05, y 54, r x 775, y 6, r x 5, y 7650, r x 53, y 650, r x 54, y 05, r x 6, y 775, r 837 Also solved by Kee-Wai Lau, Hog Kog, Chia; David Stoe ad Joh Hawis, Georgia Souther Uiversity, Statesboro, GA, ad the proposer 546: Proposed by Arsala Wares, Valdosta State Uiversity, Valdosta, GA Two cogruet itersectig holes, each with a square cross-sectio were drilled through a cube Each of the holes goes through the opposite faces of the cube Moreover, the edges of each hole are parallel to the appropriate edges of the origial cube, ad the ceter of each hole is at the ceter of the origial cube Lettig the legth of the origial cube be a, fid the legth of the square cross-sectio of each hole that will yield the largest surface area of the solid with two itersectig holes What is the largest surface area of the solid with two itersectig holes? 5

6 Solutio by Paul M Harms, North Newto, KS Let the side legths of the drilled squares be x at the surface of the origial cube The surface area of the oe side of the origial cube, with a square hole cut out of it, is a x There are four of these sides o the origial cube O a side of the origial cube the shortest distace betwee a edge of the origial cube 6

7 ad a parallel side of the drilled square hole is a x Now cosider the surface area iside the cube made by the part of the drilled square that starts at a side of the origial cube ad eds whe the drilled square meets the other drilled square origiatig from a adjacet side of the cube This surface area looig at oe side of the cube icludes four rectagles with oe side legth of x ad depth legth of a x 4xa x, so this surface area is a x There are four of these aroud the origial cube The surface area of each of the two sides of the origial cube which have o holes is a I the middle of the origial cube at the itersectio of the two drilled square holes, there are two squares of side legth x with are parallel to the sides of the origial cube with o holes The area of each square is x The total surface area of the problem is 4a x + 4 xa x + a + x 6a + 8ax 0x The maximum surface area occurs whe 8a 0x 0 or x a The maximum surface 5 area is 38a 5 whe a side of the drilled square holes as a legth of a 5 Editor s commet: David Stoe ad Joh Hawis, both from Georgia Souther Uiversity, Statesboro, GA accompaied their solutio by placig the statemet of the problem ito a story settig They wrote: A iterpretatio: i the aciet Martia civilizatio, the rulers favorite meditatioal spot was a levitatig cube havig a cubical ier sactum formed by two horizotal square tuels, meetig at the ceter of the cube, from which he could see out i all four directios The desigers were charged to costruct the ship with a maximum amout of wall space for iscriptios ad carved lieesses of His Highess There are four short hallways leadig from the ier room to the outside walls They let x be the side legth of the square tuels that are drilled through the origial cube ad oted that each tuel has a x x cross sectio ad has legth a The ier most cubical room is x x x They the metioed that by drillig the tuels ad opeig up a iterior chamber, the surface area has icreased from 6a to 38 5 a, a icrease of 8 5 a or 7% So the Kig has his private getaway ad more space for pictures ad wall hagigs Also solved by Jeremiah Bartz, Uiversity of North Daota, Grad Fors, ND ad Nicholas Newma, Fracis Mario Uiversity, Florece SC; Michael N Fried, Be-Gurio Uiversity, Beer-Sheva, Israel; David A Hucaby, Agelo State Uiversity, Sa Agelo, TX; David Stoe ad Joh Hawis, Georgia Souther Uiversity, Statesboro, GA, ad the proposer 547: Proposed by Arady Alt, Sa Jose, CA Prove that for ay positive real umber x, ad for ay atural umber, + x + + x + + x + + x Solutio by Hery Ricardo, New Yor Math Circle, NY 7

8 Let α + x + + x / + ad defie The, for x > 0 ad, we see that F x + x + x + + x + x + x + + x α α α α F x F + Now we show that F x attais its absolute maximum value at x For x, we have F x x x + x + + x + + x + + x x Gx {}}{ x x + x xx Hx {}}{ x + x + + x + x Notig that Gx is egative for 0 < x < ad positive for x >, we examie the factor Hx to see that Hx x + x x x [ x + x + + x + x [ x + x + + x + is egative for all x > 0 by the AM-GM iequality ] x ] x x x Thus F x > 0 for 0 < x < ad F x < 0 for x >, implyig that F x has a absolute maximum value at x that is, F x F o 0,, which proves the proposed iequality COMMENT: This was proposed by Walther Jaous as problem , p 06 i Crux Mathematicorum My solutio is based o the published solutio of Chris Wildhage Solutio : by Moti Levy, Rehovot, Israel If x the the iequality holds, sice + x + + x + x + + x + We assume that x > Let us defie the cotiuous fuctios g t, ad f t, t R, t >, as follows, g t : xt+ x t +, f t : g t t 8

9 +x+ +x Clearly, + fuctio f is + x + x f f, for +x+x 3 +x follows from +x+x 3 +x f The origial iequality i terms of the x 0 For, Therefore, it suffices to prove that f t is mootoe icreasig fuctio for t We will show this by provig that the derivative of l f t is postive for t The derivative is give by t d dt l f l g + t dg dt g The first step is showig l g + t dg dt g > 0 for t dg dt l g + t g + x l 4 t + x l x x To show that l +x 4 + x l x > 0 for x > 0, we see that x lim x 0 l +x 4 + x l x l 4 > 0 x Now we show that the derivative of l +x 4 d l +x 4 dx + x l x x We use the well ow iequality: l x x x + x l x x x x l x x 0 is positive: x x l x x for x > 0 to show that The secod step is showig that the derivative of l g + t dg dt g is positive for t > 0, d l g + t dg dt g dt dg dt dg g + dt g + d dt dg dt d g dt After some tedious calculatio we arrive at, dg d dt x t+ x t+ l x t+ dt g x t+ t + dg dt g To show that x t+ x t+ l x t+, or that l x t+ x t+ x t+, we use agai the iequality l y y y for y > 0, l y y y y + y But y+ y ; hece, l y y, y y > 0 9

10 Now set y x t+ to fiish the proof Solutio 3 by Kee-Wai Lau, Hog Kog, Chia Deote the iequality of the problem by It is easy to see that if holds for x t the it also holds for x Hece it suffices to prove for 0 < x t Let fx l 0 x l 0 x + l + By taig logarithms, we see that is equivalet to fx 0 We have f 0 ad for 0 < x <, fx l x + l x + l x + l Hece to prove, we eed oly prove that f x < 0 for 0 < x < Sice f x gx x x x +, where, where 0 < x + gx x x x x +, it suffices to show gx > 0, for 0 < x < Now g x x + x + + x x, g x x + 3 x + + x x 3, ad g x 4 x x + + x 3 3 x 4 Thus g g g g x 0 so that is a root of multiplicity 4 of the equatio gx 0 By Descartes rule of sigs, the equatio gx 0 has o other positive roots Sice g0 > 0, so gx > 0 for 0 < x < This completes the proof Solutio 4 by Paolo Perfetti, Departmet of Mathematics, Tor Vergata Uiversity, Rome, Italy Let ft /x The iequality goes uchaged because + t + + t t + + t + + t + + t + + t t + t + + t 0

11 This meas that we may assume x Let x The iequality becomes + } + + {{ + } } + + {{ + } + times times Let x > The iequality is also x + + x x x, that is x x t x This is the Power Meas iequality for itegrals x t Also solved by Ed Gray, Highlad Beach, FL; Albert Stadler, Herrliberg, Switzerlad, ad the proposer 548: Proposed by DM Bătietu-Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu, George Emil Palade Geeral School, Buzaău, Romaia Let ABC be a acute triagle with circumradius R ad iradius r If m 0, the prove that cos A cos m+ B 3 m+ R m cos m+ C m+ R + r m Solutio by Nios Kalapodis, Patras, Greece Applyig Rado s Iequality ad taig ito accout that cos A + cos B + cos C + r R ad cos A cos B 3 see Solutio of Problem cos C 538, SSMA, April 06 we have cos A cos m+ B cos m+ C 3 m+ R m m+ R + r m cos A cos B cos C cos m A m+ Solutio by Arady Alt, Sa Jose, CA cos A cos B cos C m cos A Firstly, we will prove that i ay acute triagle the iequality cos A cos B 3 cos C, holds cyc m+

12 Let α : π A, β : π B, γ : π C The α, β, γ > 0 sice A, B, C < π/, α + β + γ π ad cyc si α si β si γ 3 Let a, b, c be sideleghts of a triagle with agles α, β, γ, respectively, ad s be semiperimeter of this triagle The si α cos α b + c a s b s c ad, bc bc similarly, si β s c s a, si γ s a s b ca Hece, ab si α si β s b s c s c s a cyc si γ bc ca s c s cyc s a s b cyc c cyc c 3 ab a + b + c a + b + 3 c Notig that cos A + cos B + cos C + r ad usig a combiatio of the Weighted R Power Mea-Arithmetic Iequality with weights cos A, cos B, cos C > 0 ad iequality we obtai: cos B m+ cos A cos m+ B cos m+ cos B m+ cos A cos A cyc cos C cos A C cyc cos C cyc cos A cyc cos B cos A cyc cos C cos A cos A cyc cyc m+ cos A cos B cyc cos C m cos A cyc m+ 3 + r R m+ cyc m+ cos A cos B cos A cyc cyc cos C m+ cos A cyc m 3 m+ R m m+ R + r m Solutio 3 by Nicusor Zlota, Traia Vuia Techical College, Focsai, Romaia The iequality is equivalet to ad Rado s iequality, ad applyig it we obtai cos A cos B m+ cos A cos B m+ cos A cos m+ B cos m+ C cos C cos m A Rado cos C cos A m 3 m+ R m m+ R + r m, where cos A + r R ad cos A cos B cos C ta C ta A + ta B Deote ta A x, ta B y, ta C z Usig Nesbitt s iequality, we have

13 ta C ta A + ta B z x + y Nesbitt Solutio 4 by Hery Ricardo, New Yor Math Circle, NY the x p+ a p 3 We will use the followig ow results: Rado s iequality: If x, a > 0, p > 0, x p+ / a p ; 3/ ; 3 cos A R + r/r cos A cos B cos C Now we have cos A cos m+ B cos m+ C cos A cos B m+ cos C cos m A cos A cos B cos C m cos A m+, 3 3/ m+ R + r/r m 3 m+ R m m+ R + r m Commets: a Iequality appeared as problem 4053, proposed by Šefet Arslaagić, i Crux Mathematicorum ad reappeared i several solutios to problem 538 i this Joural; b Iequality 3 appeared i Solutio to problem 538 i this Joural It is also Lemma 5 i Iequalities: A Mathematical Olympiad Approach by R Mafrio et al; c The related iequality cos A cos B m+ cos C 3/ m+ appeared as problem 538 by the curret proposers Editor s commet: Moti Levy of Rehovot Israel stated i his solutio that: A ice article o Rado s iequality is A geeralizatio of Rado s Iequality by D M Bătieţu-Giurgiu ad Ovidiu T Pop, i CREATIVE MATH & INF 9 00, No, 6 - Also solved by Ed Gray, Highlad Beach, FL; Moti Levy, Rehovot, Israel; Albert Stadler, Herrliberg, Switzerlad, ad the proposer 549: Proposed by José Luis Díaz-Barrero, Barceloa Tech, Barceloa, Spai Let a, a,, a be positive real umbers Prove that a a t t + 4 where for all, t is the th tetrahedral umber defied by t Couter example by Moti Levy, Rehovot, Israel

14 The idex appears twice i the left had side This seems odd The proposer has bee ased ad here is his respose: Here, idex is used i both sum ad product But idices i sums ad product are dummy variables ad they do ot eed to be distict Surely, it is coveiet but ot ecessary Followig the proposer s argumet that the idex is a dummy variable, we chage the first idex desigatio from the letter to the letter j Now the proposed iequality becomes: But j j hece the proposed iequality implies a t a a t t + 4 a t, a t t + 4 a Let us chec this iequality for the special case, for example: Now tae a 4 ad a Sice the iequality is ot true a t a t + a t a + a 4 a t 34 a 5 + a , Editor s ote : The impossibility of this problem as it origially appeared was also oted by Albert Stadler of Herrliberg, Switzerlad I, as editor, should have oticed this mistae, but did t; mea culpa I correspodece with the proposer of the problem, José Luis Díaz-Barrero, it was acowledged that the problem should have read as follows: Let a, a,, a be positive real umbers Prove that a j a t j t + 4 where for all, t is the th tetrahedral umber defied by + + t 6 4

15 However, by chagig the idex i this maer, as Moti Levy metioed, chages the meaig of the problem Below is a proof of the problem as it was iteded to be i the first place Solutio by the proposer We cosider the fuctio fx la x + ax + + ax that is covex i R, as ca be easily prove Applyig Jese s iequality to fx, we obtai p l a x + + ax l a p x + + a p x where p are positive umbers of sum oe ad x, x,, x R Taig ito accout that fx lx is ijective, the the precedig expressio becomes p l l a p x + + a p x j a x j or equivaletly, j a x j p a p x + + a p x Settig p, ad x t,, ad taig ito accout that t 4 t +, as ca be easily prove for istace by iductio, the we have j a t j / t + 4 a from which the statemet follows Equality holds whe, ad we are doe Commet: O accout of the precedig for the particular case, we have t + 4 a or j a t j a t + a t a t + a t a t 3/4 + a t 3/4 Lettig a 4, a, t, t 4, t 3 0 i the last expressio, we obtai / : Proposed by Ovidiu Furdui, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia 3 Let A Calculate 4 lim I + A 5

16 Solutio by Bria Bradie, Christopher Newport Uiversity, Newport News, VA Let [ 3 A 4 The characteristic polyomial of A is λ λ +, so λ is a eigevalue of A with algebraic multiplicity The vector [ ] v forms a basis for the eigespace of A correspodig to λ Oe solutio of the equatio A I v is the vector [ ] The matrix A ca therefore be writte i the form [ ] A T T, 0 where [ T A straightforward iductio argumet establishes that [ ] [ ], 0 0 ] ] so that Thus, A T [ 0 A ] [ T , ] ad I + A T T Aother straightforward iductio argumet establishes that , 6

17 so that I + A T T Fially, lim [ I + A e e 4e e ] Solutio by Hery Ricardo, New Yor Math Circle, NY To simplify the solutio, we ivoe a ow result that is a cosequece of the Cayley-Hamilto theorem: If A M C ad the eigevalues λ, λ of A are equal, the for all we have A λ B + λ C, where B I ad C A λ I See, for example, Theorem 5b i Essetial Liear Algebra with Applicatios by T Adreescu, Birhäuser, 04 The eigevalues of the give matrix A are both equal to, so we apply to get A A I Now we use the last expressio to see that M I + A / A + I / ; ad, sice M s eigevalues are both equal to + /, we apply agai to determie that I + A M [ + [ [ + I + + M + ] I + M + + A + I A I ea I + e e 4e e + ] I + + ] I Solutio 3 by Albert Stadler, Herrliberg, Switzerlad 0 3 Put I, J, S The AS SJ, S 9 7, A SJS 6 3, A SJS SJ S, J 7, 0

18 I + A I + SJS I + S J S S I + J S S I + J S + S S + S es S 0 0 e S 0 0 S S e, as 4 S Solutio 4 by Bria D Beasley, Presbyteria College, Clito, SC Solutio Let B I + /A It is straightforward to show by iductio that B A + /I Usig the characteristic polyomial of B, we have B + /B + / I It the follows by iductio o that for each positive iteger, B + B + I 8

19 Thus [ lim B lim + B + ] I ea ei ea I e 4 Also solved by Arady Alt, Sa Jose, CA; Hatef I Arshagi, Guilford Techical Commuity College, Jamestow, NC; Athoy J Bevelacqua, Uiversity of North Daota, Grad Fors, ND; Bruo Salgueiro Faego, Viveiro, Spai; Kee-Wai Lau, Hog Kog, Chia; Moti Levy, Rehovot, Israel; David R Stoe ad Joh Hawis, Georgia Souther Uiversity, Statesboro, GA, ad the proposer 9