Ted Eisenberg, Section Editor *********************************************************

Transcription

1 Problems Ted Esenberg, Secton Edtor ********************************************************* Ths secton of the Journal offers readers an opportunty to exchange nterestng mathematcal problems and solutons Please send them to Ted Esenberg, Department of Mathematcs, Ben-Guron Unversty, Beer-Sheva, Israel or fax to: Questons concernng proposals and/or solutons can be sent e-mal to Solutons to prevously stated problems can be seen at < Solutons to the problems stated n ths ssue should be posted before January 5, : Proposed by Kenneth Korbn, New York, NY Let N be a postve nteger Fnd trangular numbers x and y such that x + 4xy + y 7N N 5464: Proposed by Ed Gray, Hghland Beach, FL Let ABC be an equlateral trangle wth sde length s that s colored whte on the front sde and black on the back sde Its orentaton s such that vertex A s at lower left, B s ts apex, and C s at lower rght We take the paper at B and fold t straght down along the bsector of angle B, thus exposng part of the back sde whch s black We contnue to fold untl the black part becomes / of the exstng fgure, the other half beng whte The problem s to determne the poston of the fold, the dstance defned by x as a functon of s whch s the dstance from B to the fold 5465: Proposed by Arsalan Wares, Valdosta State Unversty, Valdosta, GA Quadrlateral ABCD s a rectangle wth dagonal AC Ponts P, R, T, Q and S are on sdes AB and DC and they are connected as shown Three of the trangles nsde the rectangle are shaded pnk, and three are shaded blue Whch s larger, the sum of the areas of the pnk trangles or the sum of the areas of the blue trangles?

2 5466: Proposed by DM Bătnetu Gurgu, Mate Basarab Natonal College, Bucharest, Romana and Necula Stancu, Geroge Eml Palade School, Buzău, Romana Let f :, +, + be a contnuous functon Evaluate lm n n+ n+ n+! n n n! f x n dx 5467: Proposed by José Lus Díaz-Barrero, Barcelona Tech, Barcelona, Span In an arbtrary trangle ABC, let a, b, c denote the lengths of the sdes, R ts crcumradus, and let h a, h b, h c respectvely, denote the lengths of the correspondng alttudes Prove the nequalty a + bc b + c + b + ca c + a + c + ab a + b and gve the condtons under whch equalty holds 3abc R 3 h a h b h c, 5468: Proposed by Ovdu Furdu and Alna Sîntămăran, both at the Techncal Unversty of Clu-Napoca, Clu-Napoca, Romana Fnd all dfferentable functons f : R R wth f such that f x f xfx, for all x R Solutons 5445: Proposed by Kenneth Korbn, New York, NY Fnd the sdes of a trangle wth exrad 3, 4, 5 Soluton by Soluton by Davd E Manes, Oneonta, NY Denote the trangle by ABC wth vertces A, B and C Let a BC, the sde opposte the vertex A, b AC and c AB Let r a 3, the exradus of the crcle tangent to sde BC Smlarly, r b 4 s the exradus of the crcle tangent to AC and r c 5 s the exradus of the crcle tangent to AB If r s the nradus of trangle ABC, then + + r a r b r c r mples mples r If s the area of trangle r ABC, then r r a r b r c Therefore, 6 If s a + b + c mples 6 6 s the sempermeter of ABC, then s r Therefore, s 6 6

3 Usng the formula r a s a, one obtans a s Therefore, r a Smlarly, Ths completes the soluton a 6 3 b s r b c s r c , 35 Soluton by Bruno Salguero Fanego, Vvero, Span In the publshed solton by Howard Eves to problem 786 n the Journal Crux Mathematcorum 984,, a more general result of the above problem s proved For arbtrarly chosen postve real numbers r, r, r 3 there s one and only one trangle whose exrad are r, r, r 3, and that s the one whose sdes are: a r r + r 3 r r + r r 3 + r 3 r, b r r 3 + r r r + r r 3 + r 3 r c For the exrad values of r 3, r 4 and r 3 5 we fnd that a , b r 3 r + r r r + r r 3 + r 3 r , c Soluton 3 by Ed Gray, Hghland Beach, FL Lettng r be the n-radus of the gven trangle, r, r, r 3 the ex-rad, s ts sem-permeter, K ts area and a, b, c ts sde lengths, then followng relatonshps, that were developed by Feuerbach, hold: r r + r + r 3 K r r r r 3 3 s K r r r 3 4 a s K r, b s K r, c s K r 3 Makng the substtutons we fnd that a 7, b 3, c 35 Comment by Davd Stone and John Hawkns of Georga Southern Unversty: An nterestng connecton to ths problem from Wolfram Mathworld s that the curvature of the ncrcle equals the sum of the curvatures of the excrcles: r + + rb r c + r a r c + r a r b whch equals Thus the area can be wrtten as r a r b r c r a r b r c r a r b r c 3

4 Also solved by Arkady Alt, San Jose, CA; Kee-Wa Lau, Hong Kong, Chna; Charles McCracken, Dayton, OH; Danel Staru, Mathematcs Department, Natonal Economc College Theodor Costescu, Drobeta Turnu - Severn, Mehednt, Romana; Albert Stadler, Herrlberg, Swtzerland; Davd Stone and John Hawkns, Georga Southern Unversty, Statesboro, GA, and the proposer 5446: Proposed by Arsalan Wares, Valdosta State Unversty, Valdosta, GA Polygons ABCD, CEF G, and DGHJ are squares Moreover, pont E s on sde DC, X DG EF, and Y BC JH If GX splts square CEF G n regons whose areas are n the rato 5:9 What part of square DGHJ s shaded? Shaded regon n DGHJ s composed of the areas of trangle Y HG and trapezod EXGC Soluton by Get Stoked Problem Solvng Group, Mountan Lakes Hgh School, Mountan Lakes, NJ Snce B H, JY B GY H, JBY GHY, and because JBY and DAJ have a shared angle, B A JBY DAJ, 4

5 and because AD DC, JD GD, A DCG, DAJ DCG and snce CG EX, DCG DEX and because DEX F, DXE GXF, DEX GF X Therefore, GY H GXF Wthout loss of generalty, set the area of Y HG 5 and trapezod EXGC 9 Addng the areas of Y HG and trapezod EXGC and fndng each sde obtans: Drawng a perpendcular lne from pont X to CG creates lne IX Because the area of XF GI s double of that of GXF and F G EC XI, XF : CG 5 : Snce F G CG, t can be concluded hat GXF s a trangle Because DGC GXF, CG : DG 5 : 3 Now that we know the rato between the two squares and that the rato of the area between two smlar polygons s the square of the rato of the sdes, t s apparent that areaexgc areajdgh Addng the two peces results n the part of square DCHJ that s shaded Soluton by Kenneth Korbn, New York, NY Answer: Let XF 5 and EX 35 Then each segment n the dagram wll have postve nteger length AD 44, AJ 6, JB 89, BY 35, Y H 65, HG 56, GF 6, F X 5, XE 35, ED 84, DJ 56, JY 9, Y G 69, CE 6, CG 6, XG 65 DX 9 5

6 Every trangle n ths dagram s smlar to the Pythagorean Trangle wth sdes 5,, 3 Area of square DGHJ Area of trangle Y HG Area of trapezod EXGC So the desred rato s Also solved by Jeremah Bartz and Ncholas Newman, Unversty of North Dakota and Troy Unversty respectvely, Grand Forks, ND and Troy, AL; Bruno Salguero Fanego, Vvero, Span; Mchael N Fred, Ben-Guron Unversty of the Negev, Beer-Sheva, Israel; Ed Gray, Hghland Beach, FL; Kee-Wa Lau, Hong Kong, Chna; Davd E Manes, Oneonta, NY; Danel Staru, Mathematcs Department, Natonal Economc College Theodor Costescu, Drobeta Turnu - Severn, Mehednt, Romana; Sacht Msra, Nelh, Inda; Bors Rays, Brooklyn, NY; Davd Stone and John Hawkns, Georga Southern Unversty, Statesboro, GA, and the proposer 54: Proposed by Iulana Trască, Scorncest, Romana Show that f x, y, and z s each a postve real number, then x 6 z 3 + y 6 x 3 + z 6 y 3 x y z x3 + y 3 + z 3 + 3x y z Soluton by Albert Stadler, Herrlberg, Swtzerland The stated nequalty s equvalent to x 6 z 3 + y 6 x 3 + z 6 y 3 x 5 y z + x y 5 z + x y z 5 + 3x 3 y 3 z 3 By the AM-GM nequalty, x 6 z 3 3 x6 z y6 x 3 cycl cycl cycl x 6 z 3 3x 3 y 3 z 3 cycl Statement follows by addng these two nequaltes Soluton by Arkady Alt, San Jose, CA Note that, x 3 6 z 3 6 y 3 6 x 3 3 cycl x 5 y z, x 6 z 3 + y 6 x 3 + z 6 y 3 x y z x3 + y 3 + z 3 + 3xyz x 6 z 3 + y 6 x 3 + z 6 y 3 6

7 x 5 y z + x y 5 z + x y z 5 + 3x 3 y 3 z 3 By AM-GM Inequalty, x 6 z 3 + y 6 x 3 + z 6 y x 6 z 3 y 6 x 3 z 6 y x 9 y 9 z 9 3x 3 y 3 z 3 And agan by AM-GM Inequalty x 6 z 3 + y 6 x x 6 z 3 y 6 x x 5 y 6 z 6 3x 5 y z, and therefore, 3 x 6 z 3 x 6 z 3 + y 6 x 3 3x 5 y z cyc cyc cyc cyc x 6 z 3 cyc x 5 y z Thus, x 6 z 3 x 6 z 3 + x 6 z 3 x 5 y z + 3x 3 y 3 z 3 cyc cyc cyc cyc Soluton 3 by Mot Levy, Rehovot, Israel By Murhead nequalty 6, 3, maorzes 5,,, sym x 6 x 3 z sym x 5 y z, or explctly, x 6 z 3 + y 6 x 3 + z 6 y 3 + x 6 y 3 + v 6 z 3 + z 6 x 3 x 5 y z + x y 5 z + x y z 5 Agan, by Murhead nequalty 5,, maorzes 3, 3, 3, or explctly, sym x 5 y z sym x 3 y 3 z 3 x 5 y z + x y 5 z + x y z 5 3x 3 y 3 z 3 Gven three postve numbers a, b, c We can always assgn ther values to x, y and z respectvely, such that x 6 z 3 + y 6 x 3 + z 6 y 3 x 6 y 3 + y 6 z 3 + z 6 x 3 Hence, wthout loss of generalty, we can assume that then by, and 3 x 6 z 3 + y 6 x 3 + z 6 y 3 x 6 y 3 + y 6 z 3 + z 6 x 3, 3 x 6 z 3 + y 6 x 3 + z 6 y 3 x 6 z 3 + y 6 x 3 + z 6 y 3 + x 6 y 3 + v 6 z 3 + z 6 x 3 whch s equvalent to x 5 y z + x y 5 z + x y z 5 x 5 y z + x y 5 z + x y z 5 + 3x 3 y 3 z 3 x 6 z 3 + y 6 x 3 + z 6 y 3 x y z x3 + y 3 + z 3 + 3xyz 7

8 Soluton 4 by Kee-Wa Lau, Hong Kong, Chna We prove the stronger nequalty x 6 z 3 + y 6 x 3 + z 6 y 3 x y z x 3 + y 3 + z 3 Snce x 3 + y 3 + z 3 3xyz by the AM-GM nequalty, the nequalty of the problem follows mmedately from By homogenety, we assume wthout loss of generalty that xyz By substtutng z nto, we deduce after some algebra that s equvalent to xy x 9 + x 9 y 9 + x 9 y 3 x 6 y 6 x 3 Denote the left sde of by f It can be checked readly by expandng both sdes that + x 3 + x 3 y 3 f x 9 + x 3 + y 3 y x 3 x 3 + x 3 + y 3 + x 3 y 3 x 3 y 3, whch s nonnegatve Thus holds and ths completes the soluton Soluton 5 by Hatef I Arshag, Gulford Techncal Communty College, Jamestown, NC It s well-known that for all a, b, c we have a 3 + b 3 + c 3 a b + b c + c a, and a 3 + b 3 + c 3 3abc Now for all postve real numbers x, y, and z we can wrte x 6 z 3 + y 6 x 3 + z 6 y 3 x y z [ x z y x + y x z y + z y x z ] x y z x 3 + y 3 + z 3 x 3 + y 3 + z 3 + x 3 + y 3 + z 3 x 3 + y 3 + z 3 + 3xyz Now, multplyng both sdes of the nequalty x 6 z 3 + y 6 x 3 + z 6 y 3 x y z x 3 + y 3 + z 3 + 3xyz, by, wll gve us the desred result Also solved by Bruno Salguero Fanego, Vvero, Span; Ed Gray, Hghland Beach, FL; Mot Levy, Rehovot, Israel; Davd E Manes, Oneonta, NY; Sacht Msra, Delh, Inda; Paolo Perfett, Department of Mathematcs, Tor Vergata Unversty of Rome, Italy; Danel Staru, Mathematcs Department, Natonal Economc College Theodor Costescu, Drobeta Turnu - Severn, Mehednt, Romana, and the proposer 5448: Proposed by Yubal Barros and Ángel Plaza, Unversty of Las Palmas de Gran Canara, Span Evaluate: lm n n, n +n 8

9 Soluton by Bran Brade, Chrstopher Newport Unversty, Newport News, VA The generatng functon for the central bnomal coeffcents s 4x / ; that s, It follows that, n,+n s the coeffcent of x n n the functon whch s 4 n Thus, n lm n x 4x n n n 4x 4x 4x 4x n,, n,+n lm n n n 4 n lm n 4 4 Soluton by Danel Staru, Mathematcs Department, Natonal Economc College Theodor Costescu, Drobeta Turnu - Severn, Mehednt, Romana + x + x n + + x + x n + + x 4 + x n x n + x n + + x n The coeffcent of x n n LHS and RHS are equal: n n 4 n n n n + + n n + n n n, n +n n + n n lm n n, n lm n n n + n n CAUCHY D ALEMBERT {}}{ n + 3 lm n n + 9 n+! n+! n! n! n + n + lm n n + 4

10 Soluton 3 by Kee-Wa Lau, Hong Kong, Chna It s well known that that x 4,! and Hence, 4x 4x Thus for nonnegatve ntegers n, n x, n +n so that the lmt of the problem equals 4 x 4x, wth the usual conventon that n 4 n,, n +n x n Also solved by Arkady Alt, San Jose, CA; Bruno Salguero Fanego, Vvero, Span; Ed Gray, Hghland Beach, FL; Perfett Paolo, Department of Mathematcs, Tor Vergata Unversty, Rome Italy; Albert Stadler, Herrlberg, Swtzerland; Anna V Tomova, Varna, Bulgara, and the proposer 5449: Proposed by José Lus Díaz-Barrero, Barcelona Tech, Barcelona, Span Wthout the use of a computer, fnd the real roots of the equaton x 6 6x x 39x + 3x 3x Soluton by Albert Stadler, Herrlberg, Swtzerland We need to consder only the values of x /3, snce the square root s not real for x < /3 We see that x and x are roots of the gven equaton Suppose that x / {, } We fnd that x 6 7x x 39x + x x x 4 + 3x 3 + 7x x + 5 3x 3x x 3 3x 3 x 6 3x 3x + x 3 x x x 4 + 3x 3 + 7x x + 4 3x 3x + x 3, mplyng x 4 + 3x 3 + 7x x + 5 x4 + 3x 3 + 7x x + 4 3x 3x + x 3

11 We note that x 4 + 3x 3 + 7x x + 4 x 4 + x 3x + 3x, snce x /3 So has no other real solutons than x and x Soluton by Ed Gray, Hghland Beach, FL Defne the functon: fx x 6 6x x 39x + 3x 3/ Consder the term 3x 3/ Snce the values of x and x both provde nteger values, t s worth tryng these values as a frst guess In fact, f , so n fact, x s a root 3 f , so, n fact x s also a root It may be frutful to utlze the dervatve whch s: 4 f x 6x 5 78x + x 39 3/ 3 3x We note that 5 f f , so the functon s at x and x At x, t s decreasng and at x t s ncreasng Therefore, there s a pont x wth < x < and f x The functon s ncreasng at x, where the dervatve s greater than, so f x >, the functon s greater than It would be good f the functon stays postve for x > Note the second dervatve s: 7 f x 3x 4 56x + 7/4 3x / At x, f /8 7465, and clearly ncreases as x ncreases We conclude there can be no real roots greater than Now we look at the stuaton where x, whch s a root f 55 Therefore, values of fx for x slghtly less than must be postve If x < /3, we note the radcal term becomes negatve and complex terms wll be ntroduced, negatng the exstence of real roots We need to consder the regon /3 < x < The value of f/3 /3 6 6/ /3 39/3 + 64/79 68/ /9 39/3 + 64/79 > Also, f /3 /43 Ths s unexpectedly postve, whch means the functon rses from 64/79 at x /3 as x ncreases, then there must be a pont x such that /3 < x < and f x After x > x, the dervatve turns negatve and the functon descends to at x Therefore, there can be no other real roots other than x and x Soluton 3 by Anna V Tomova, Varna, Bulgara The decson area s: 3x x 3, x6 6x x 39x + We let t + 3x t x, and after substtuton we obtan 3 t + t + 6t 8 54t t 4 79t 3 + t + 64

12 Lookng for low-valued postve nteger roots to the above equaton so that we can use the factor theorem, we see that t and t allows us to rewrte the equaton as t t t + 3t 9 + 9t 8 + 5t + 75t t t t t + 453t + 3 Because all of the coeffcents n the thrd factor are postve, we see that there are no other x t + x postve roots So, 3 x t + x 3 Soluton 4 by Bran D Beasley, Presbyteran College, Clnton, SC We note that x and x satsfy the gven equaton, and we show that those are the only real roots Squarng both sdes of the equaton and factorng yelds x x fx, where fx x + 3x 9 + 7x 8 37x 7 5x 6 49x x 4 5x 3 + 8x 37x + 54 Snce we must have x /3 n the orgnal equaton, t suffces to show that fx for each x /3 We wrte fx gx + hx, where and gx x x x x x hx x x x x Then h x x x x Snce h x > on,, h s ncreasng on, Also, h /3 >, so we have h x > for each x /3 Thus h s ncreasng on [/3, wth h/3 >, so hx > for each x /3 as needed Hence fx > for each x /3 Also solved by Bruno Salguero Fanego, Vvero, Span; Kee-Wa Lau, Hong Kong, Chna; Perfett Paolo, Department of Mathematcs, Tor Vergata Unversty, Rome Italy; Davd Stone and John Hawkns, Georga Southern Unversty, Statesboro, GA, and the proposer 545: Proposed by Ovdu Furdu, Techncal Unversty of Clu-Napoca, Clu-Napoca, Romana Let k be a postve nteger Calculate where a denotes the floor the nteger part of a x k y x k dxdy, Soluton by Bran Brade, Chrstopher Newport Unversty, Newport News, VA Reverse the order of ntegraton, and then wrte x y k dx dy xk x x k dy dx y xk x k y x k dy dx + n x/n x/n+ x k dy dx y xk

13 For x y, x/y, whle for x/n + y x/n, x/y n, so x k y x k dx dy n k n x/n x/n+ dy dx xk Now, so x/n x/n+ x k dy dx k + k + k + x k y x k dx dy k + k + x k x k+ n k+ x k+ n + k+ dx x dx n k+ n + k+ n k+ n + k+ n n, n k n k+ n + k+ n k n k n k+ By the bnomal theorem, It follows that n k n k n k n k n k n k+ k k n k k + k k + k n k, n +, and n n k n k n k+ k + k n n + k + k ζ +, where ζx denotes the Remann zeta functon Fnally, x k y x k dx dy k + k + k ζ + Soluton by Ángel Plaza, Unversty of Las Palmas de Gran Canara, Span For any postve nteger m, let us assume that m x y < m +, whch t s equvalent to 3

14 x m + y < x The proposed ntegral, say I, becomes m I m k m m k m m k x k x m x m+ + x k k + k + m k + m m k + m dy dx ] x m x m+ dx m k+ m + k+ m k m + k+ m m + k k m + k+ from where, I k k + ζ + k + Soluton 3 by Perfett Paolo, Department of Mathematcs, Tor Vergata Unversty, Rome Italy We change varables x/y t, x u and the ntegral becomes dt t du t k+ t k u + The frst ntegral s zero so we get dt du t k+ t k u dt du t k+ t k u q+ q k q q k + k + k + dt t k+ n k + lm q k n q k+ q + k+ q q q + k q + q + k q k k q q + k+ + k + k q + q q k + k + q k q + q }{{} telescope k q k k k k k q + k q + ζ + 4 k + k k ζ +

15 Also solved by Bruno Salguero Fanego, Vvero, Span; Ed Gray, Hghland Beach, FL; Kee-Wa Lau, Hong Kong, Chna; Mot Levy, Rehovot, Israel; Albert Stadler, Herrlberg, Swtzerland; Davd Stone and John Hawkns, Georga Southern Unversty, Statesboro, GA, and the proposer End Notes Sacht Msra of Delh, Inda should have been credted wth havng solved 544, and Davd Stone and John Hawkns of Georga Southern Unversty, Statesboro, GA should have been credted for havng solved 5444 Once agan, mea culpa 5