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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@3.et>. Solutios to previously stated problems ca be see at < Solutios to the problems stated i this issue should be posted before Jue 5, 53: Proposed by Keeth Korbi, New York, NY A covex cyclic quadrilateral with iteger legth sides is such that its area divided by its perimeter equals. Fid the maximum possible perimeter. 53: Proposed by Tom Moore, Bridgewater State Uiversity, Bridgewater, MA If is a eve perfect umber, > 6, ad φ is the Euler phi-fuctio, the show that φ is a fourth power of a iteger. Fid ifiitely may itegers such that φ is a fourth power. 533: Proposed by Agel Plaza, Uiversidad de Las Palmas de Gra Caaria, Spai Let a, b, c, d be positive real umbers. Prove that a + b + c + d + a b + b c + c d + d a + /. 53: Proposed by Michael Brozisky, Cetral Islip, NY Determie whether or ot there exist ozero costats a ad b such that the coic whose polar equatio is a r = siθ b cosθ has a ratioal eccetricity. 535: Proposed by José Luis Díaz-Barrero, Barceloa Tech, Barceloa, Spai Let x be a positive real umber. Prove that [x] x + {x} + [x]{x} 3x + {x} x + [x], where [x] is the greatest iteger fuctio ad {x} is the fractioal part of the real umber. I.e., {x} = x [x].

2 536: Proposed by Ovidiu Furdui, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia Calculate: l x + x x x dx. Solutios 583: Proposed by Keeth Korbi, New York, NY Fid the sides of two differet isosceles triagles that both have perimeter 6 ad area 8. Solutio by Elsie M. Campbell, Dioe T. Bailey, ad Charles Dimiie, Agelo State Uiversity, Sa Agelo, TX To begi, we will let the isosceles triagle be desigated with sides a, a, x ad height h. With give perimeter 6, ad, usig the Pythagorea Theorem ad, x = 6 a x = 8 a, x h + = a h + 8 a = a Thus, with give area 8,, ad 3, Usig Mupad, the solutios are h = 9 a 8. 6 a9 a 8 = 8 8 a = a 8 a 3 5 a + 6, a 53, 985 =. a = , 65, Usig, a = does ot yield a triagle with perimeter 6. Hece, usig, whe a = , x = , ad whe a = 65, x = 3. Therefore, the isosceles triagles are , , ad 65, 65, 3.

3 With some persistece, these solutios ca be verified to yield a isosceles triagle with perimeter 6 ad area 8. Solutio by Arkady Alt, Sa Jose, CA Let b be legth of the lateral sides ad a be half of legth of the base. The { a + b = 6 a b a = 8 We have a 8 a = { { a + b = 8 a b a = < a 8/ a 8 a = a 8 a = 6 9 a 3 8a + =. { b = 8 a a 8 a = ad the equatio Sice a 3 8a + = a 6 a 9a 78 ad the quadratic equatio a 9a 78 = have oly oe positive root a = the we obtai two differet isosceles triagles with side-legths b, a, b = 65, 3, 65,, , Solutio 3 by Kee-Wai Lau, Hog Kog, Chia Let the sides of the isosceles triagles be a, a, 6 a. By Hero s formula for the area of a triagle we obtai 8 a a 8 =, or or Hece a = 65, a a 8 5 =, a 65a 75a = So the sides of the isosceles triagles are 65, 65, 3 ad, , Solutio by Bria D. Beasley, Presbyteria College, Clito, SC Ay such triagle has sides with legths x, x, ad 6 x, where 8/ < x < 8. Hero s formula the implies which i tur is equivalet to 8 = 88 x x 8, x 3 5x + 6x = x 65x 75x =. We fid three real solutios to this equatio, amely x = 65 ad x = 75 ± 7 77/; however, oe of these yields x 9.3, which is outside the ecessary domai. Hece we obtai two triagles, correspodig to x = 65 ad x 5.68: 65, 65, 3; , , , 5.68, 7.6.

4 Questio. I geeral, if we seek all isosceles triagles of the form x, x, P x that have perimeter P ad area A, the we obtai the equatio 6P x 3 P x + 8P 3 x P + 6A =. The give values P = 6 ad A = 8 produce exactly two such triagles. For what values of P ad A would we fid o triagles, oe triagle, two triagles, or three triagles? Also solved by Bruo Salgueiro Faego, Viveriro, Spai; Ed Gray, Highlad Beach, FL; Paul M. Harms, North Newto, KS; Jahageer Kholdi ad Farideh Firoozbakht, Uiversity of Isfaha, Khasar, Ira; David E. Maes, SUNY College at Oeota, Oeota, NY; Agel Plaza, Uiversidad de Las Palmas, de Gra Caaria, Spai; Michael Thew, Studet, St. George s School, Spokae, WA; Albert Stadler, Herrliberg, Switzerlad, ad the proposer. 58: Proposed by Tom Moore, Bridgewater State Uiversity, Bridgewater, MA Prove: a mod 8,, b mod 53,, c mod 968, 3, d mod 389,. Solutio by David E. Maes, SUNY College at Oeota, Oeota, NY Note the followig cogrueces:. Therefore, 3 mod, 3 3 mod 7, 3 9 mod mod 37, 3 8 mod mod, mod 7, mod 9, mod 37 3, mod 63. Recall the elemetary property of cogrueces : if a b mod m ad a b mod ad gcdm, =, the a b mod m Therefore, a sice gcd, 7 =, it follows from ad that mod 8,

5 b sice gcd9, 8 =, it follows from a ad 3 that mod 53 = 9 8, c sice gcd37, 53 =, it follows from b ad that mod 968 3, d sice gcd63, 968 =, it follows from c ad 5 that mod 389. This completes the solutio. Solutio by Albert Stadler, Herrliberg, Switzerlad We have 8= 7, 53 = 7 9, 968 = , 389 = , 3 3 mod 8, 3 9 mod 9, 3 7 mod 37, 3 8 mod 63. Statemet a is true fore =, statemet b is true for =, statemet c is true for = 3, statemet d is true for =. The geeral statmet the follows by iductio: If 3 3 mod a where a, 3 = the moda. Also solved by Arkady Alt, Sa Jose, CA; Dioe T. Bailey, Elsie M. Campbell, ad Charles Dimiie, Agelo State Uiversity, Sa Agelo, TX; Bria D. Beasley, Presbyteria College, Clito, SC; D.M. Bătietu Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu, Geroge Emil Palade Geeral School, Buzău, Romaia ad Titu Zvoaru, Comăesti, Romaia; Bruo Salgueiro Faego, Viveiro, Spai; Ed Gray, Highlad Beach, FL; Paul M. Harms, North Newto, KS; Jahageer Kholdi ad Farideh Firoozbakht, Uiversity of Isfaha, Khasar, Ira; Keeth Korbi, New York, NY; Kee-Wai Lau, Hog Kog, Chia; David Stoe ad Joh Hawkis, Georgia Souther Uiversity, Statesboro, GA, ad the proposer. 585: Proposed by D.M. Bătietu Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu, Geroge Emil Palade Geeral School, Buzu, Romaia Let {a }, ad {b } be positive sequeces of real umbers with a b + + a = a R + ad = b R +. b For x R, calculate a si x cos x cos x + b + b. Solutio by Arkady Alt, Sa Jose, CA Sice the a a + a = a, the by the Stolz Theorem b + +! b! = b + + b = b + + b + = = b. + b b = a. Also ote that 5

6 b By the Multiplicative Stolz Theorem b +! +! + b+ + +! +! Let c = Sice + b + b =! c cos x l =. c cos x = e, b! b! Hece, c cos x = +! +. = b yields b! = b. + = b, = the c =, ad, therefore, l c cos x cos x l c = cos x l c = cos x l l c cos x + b +. ccos x b = l c cos x = Sice + b + = b + + b b b +, the + b+ + + b = b b e = e +! ad, therefore, c cos x = cos x. Ad sice a cos x si x + b + cos x b = a cos si x x cos x cos x b! + b +! the b a cos x si x + b + cos x b = a si x b cos x e cos x c cos x = a si x b cos x e cos x cos x. Solutio by Perfetti Paolo, Departmet of Mathematics, Tor Vergata Uiversity, Rome, Italy a cos si x x + b + b x cos = a si x si x b cos x + b + b cos x 6

7 = a si x b cos x + b + b cos x = a si x b cos x + b + l b + b + b cos x cos x l + b + b cos x. By Cesaro-Stolz, ad Now b / / = b b + + = b + + = b e a = a + a. b+ cos x + b cos x = b+ cos x + + cos x cos x cos x + cos x cos x b = bcos x e cos x ecos x b cos x =. Moreover, b+ cos x + b cos x = b+ cos x b+ cos x + b cos x = b+ cos x cos x b cos x cos x + cos x + cos x cos x b+ + = b cos x ecos x b cos x = ecos x The it is thus a si x bcos x e cos x cos x. Solutio 3 by Albert Stadler, Herrliberg, Switzerlad By assumptio a + a = a + o, b + = be o, as. So, b a = a + a j a j = a+o, b =!b b j =!b e o = e b e o, as. j b j j= 7 j=

8 We have used a weak form of Stirlig s formula, amely! = e +o as. We coclude a si x cos x cos x + b + b = = si x a + o si x + be +o cos x be +o cos x = = si x+cos x a + o si x be +o cos x = a + o si x be +o cos x cos x = a + o si x be +o cos x cos x + O + cos x = + O a si x b cos x e cos x cos x as. Commet by Bruo Salgueiro Faego, Viveiro, Spai A more geeral questio ca be see i problem 75 from the joural Mathproblems, available at < d=> see page ad solved at < d=7>see pages 6-8> Also solved by Kee-Wai Lau, Hog Kog, Chia, ad the proposer. 586: Proposed by Michael Brozisky, Cetral Islip, NY I Cartesialad, where immortal ats live, a at is assiged a specific equilateral triagle EF G ad three distict positive umbers < a < b < c. The at s job is to fid a uique poit P x, y such that the distaces from P to the vertices E, F ad G of his triagle are proportioate to a : b : c respectively. Some ats are eterally doomed to a impossible search. Fid a relatioship betwee a, b ad c that guaratees evetual success; i.e., that such a uique poit P actually exists. Solutio by David E. Maes, SUNY College at Oeota, Oeota, NY Let s be the legth of the side of EF G ad suppose we are give three distict positive itegers < a < b < c such that a + b > c, b + c > a ad c + a > b. Recall the followig: the symmetric equatio 3 x + y + z + w = x + y + z + w relates the size of a equilateral triagle ABC to the distaces of a poit from its three vertices. Substitutig a, b ad c for x, y ad z respectively ad solvig for w the gives the triagle s side say w = s ad the existece of a poit P. By Pompeiu s Theorem, if P is a arbitrary poit a equilateral triagle ABC, the there exists a triagle with 8

9 sides of legth P A, P B, P C. Moreover, the theorem remais valid for ay poit P i the plae of triagle ABC ad that the triagle is degeerate if ad oly if P lies o the circumcircle of ABC. Therefore, a + b > c, b + c > a ad c + a > b. Fially usig a dilatio trasformatio from ABC to EF G with a dilatio factor of s, it follows s s that there exists a poit P = P whose s distaces from the three vertices are s s s P E = a s, P F = b s ad P G = c s. Hece, P E a = P F = P G = s b c s so that the distacs from P to the vertices E, F ad G are proportioate to a : b : c respectively. Solutio by Michael Fried, Be Gurio Uiversity, Beer-Sheva, Israel Sice this is Cartesialad, we might as well place the equilateral triagle i the Cartesia plae ad give the vertices coveiet coordiates, say, E =,, F =,, ad G =, 3 see figure below. Let us set α = b/c = P E/P F, β = a/c = P G/P F, ad γ = a/b = P G/P E. The the locus of poits P with P E/P F = α is the Apolloius circle: α x + y x + + y = Similarly, the locus of poits P with P G/P E = γ is the Apolloius circle: γ x + + y x + y 3 = The coditio that the system of equatios, α x + y x + + y = γ x + + y x + y 3 = 9

10 has a solutio, that is, that the two Apolloius circles have a itersectio is after some messy but routie algebra is: = 6 [ γ γ + α + α γ α + 3α ] After some further maipulatio, this come dow to the iequality: α γ α + α γ α From which we have the coditio: α γ + α Or goig back to the defiitio α = b/c, γ = a/b, we have: So that, c a b b + c b b c a b + c Sice a, b, c are positive umbers, ad sice this must be true o matter which Apolloius circle ratio we begi with, we have the triagle-like iequalities: a b + c b a + c c a + b Oe should ote that if the circles α x + y x + + y = ad γ x + + y x + y 3 = itersect, they will geerally itersect i two poits P ad P, where both P G/P F ad P G/P F = a/c, ad a sigle Apolloius circle with respect to G ad F will pass through these poits. Observe too, the three circles the have the same radical axis, amely, P P see figure below.

11 . Ke Korbi, New York, NY Commets Give < a < b < c. If it is possible to costruct a triagle with sides a, b, c i which each of the agles is less tha, them there is a uique poit P.. Bruo Salgueiro Faego, Viveiro, Spai I the article by Oee Bottema O the distaces of a poit to the vertices of a triagle. joural Crux Mathematicorum, 98, 8, 6, it is proved amog other thigs the followig relatioship betwee the legths of the sides α = A A A 3, α = A 3 A A, α 3 = A A 3 A ad ay poit P i the plae of A A A 3 with distaces to the vertices d = P A, d = P A, d 3 = P A 3, the: a d + a d + a 3d 3 a a 3 cos α d d 3 a 3 a cos α d 3d a a cos α 3 d d a a a 3 cos α d a a a 3 cos α d a a a 3 cos α 3 d 3 + a a a 3 = called idetity 6 ad reciprocally. That is, that if d, d, d 3 are positive umbers satisfyig idetity 6 the there is a uique poit P such that P A = d, P A = d, P A 3 = d 3. This implies that idetity6 is the relatioship which solves a problem more geerally

12 tha the oe proposed. Note: I particular, if we suppose that A A A 3 is the equilateral triagle EF G of the statemet of the problem, with sides e = a = a = a 3 ad k is the costat of proportioality such that d = ka, d = kb, d 3 = kc the idetity 6, whe divided by e becomes k a + b + c + e k a b + a c + b c k a + b + c e + e =, which is the required relatioship i the origial statemet of the problem. O the other had, if we suppose that a poit P exists ad k is the costat of proportioality, such aht P E = ka, P F = kb, ad P G = kc, usig the idetity which appears i the editor s commet of SSM problem 5, or its equivalet, P E +P F +P G +EF = P E P F +P E P G +P F P G +P E EF +P F EF +P G EF, we obtai directly the relatioship which is required i the problem, that is, k a + b + c + e = k a b + a c + b c + k a + b + c e, which is also equivalet to equality i the published solutio # to 5. Also solved by Ed Gray, Highlad Beach, FL; Paul M. Harms, North Newto, KS, ad the proposer. 587: Proposed by José Luis Díaz-Barrero, Barceloa Tech, Barceloa, Spai Let u, v, w, x, y, z be complex umbers. Prove that Reux + vy + zw 3 u + v + w + 3 x + y + z. Solutio by Albert Stadler, Herrliberg, Switzerlad We ote that 3u 3 x = 3u 3u 3 x 3 x = 3 u + 3 x Reux. So, Reux 3 u + 3 x. Similarly, Revy 3 v + 3 y, ad Rezw 3 w + 3 z. The statemet follows by addig these iequalities. Solutio by David Dimiie ad Tatyaa Savchuk, Texas Istrumets, Ic., Dallas, TX We will prove the equivalet statemet 3 u + v + w Re ux + vy + zw + x + y + z. 3 Let u, u deote the real ad imagiary parts of u, respectively, ad similarly for v, w, x, y, z. The the right side of becomes

13 3 u + u + v + v + w + w u x u x + v y v y + w z w z which we rewrite as 3u u x + 3 x + 3u + u x + 3 x w w z + 3 z x + x + y + y + z + z, 3v v y + 3 y + Notig that 3a + ab + 3 b ad 3a ab + 3 b may be rewritte as 3a ad 3 b, respectively, becomes + 3w + w z + 3 z 3a + 3 b 3v + v y + 3 y. 3u 3u x + + 3v x + 3v y + + y w + 3w z + + z Sice 3 is a sum of squares of real umbers the expressio must be oegative, ad therefore holds. Solutio 3 by Paul M. Harms, North Newto, KS We kow that the real part of a fiite sum of complex umbers is less tha or equal to the modulus of the sum which is less tha or equal to the sum of the moduli. Also the modulus of a fiite product of complex umbers equals the product of the moduli. We have 3 u x + 3 v y + 3 w z. After squarig the three parts, movig terms ad dividig by 3, we ca obtai, u x + v y + z w 3 u + v + w + x + y + z. 3 From what was said ad show above, Reux + vy + zw u x + v y + z w 3 u + v + w + 3 x + y + z. Solutio by Kee-Wai Lau, Hog Kog, Chia We have 3 u + 3 x Reux 3 u + 3 x u x = 3 3 u x, ad similarly, 3 v + 3 y Revy, 3 z + 3 w Rezw. 3

14 The iequality of the problem follows by addig up the three iequalities above. Solutio 5 by Paolo Perfetti, Departmet of Mathematics, Tor Vergata Uiversity, Rome, Italy ad Reux + vy + zw ux + vy + zw = u x + v y + z w z w 3 z + 3 w is simply the AGM. Also solved by Arkady Alt, Sa Jose, CA; Bruo Salgueiro Faego, Viveiro, Spai; Ed Gray, Highlad Beach, FL ad the proposer. 588: Proposed by Ovidiu Furdui, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia Let a, b, c be real umbers. Fid the value of i= j= i + j + ai + bj + c. Solutio by Paolo Perfetti, Departmet of Mathematics, Tor Vergata Uiversity, Rome, Italy Aswer: l + Proof: We show that the it is idepedet o a, b, c allowig us to set a = b = c = for evaluatig it. If Q = [, ] [, ], the it becomes i,j= By writig the itegral as we have π/ π/ / si θ i + j = x i,j= ρ π/ ρ dρ dθ = π/ = i + j To show that the it is idepedet by a, b, c, we prove i,j= i + j + ai + bj + c = Q x + y dxdy. x + y dy dx ad passig to polar coordiates si θ dθ = l ta θ π/ = l +. π/ i,j= i + j + a i + b j + c for ay a, b, c, a, b, c. We itroduce a umber of positive costats C k, k =,,.... Sice i a a + j b b + c c C i + j ad i + j + ai + bj + c C i + j we have the boud

15 i + j + ai + bj + c = i + j + a i + b j + c a a + jb b + c c i + j + ai + bj + ci + j + a i + b j + c i + j + ai + bj + c + i + j + a i + b j + c C i + j i + j i + j i + j = C C i + j 3/ Thus i,j= i + j i + j 3/ j= i= ad it follows that for ay a, b, c, a, b, c i ij 3/ + i= j= j ij 3/ C 3/ i + j + ai + bj + c =. i + j + a i + b j + c i,j= I particular we ca take a = b = c = ad write i + j + ai + bj + c = i + j + ai + bj + c + i + j i + j The coclusio is that for ay a, b, c the it assumes the same value l +. Solutio by Ed Gray, Highlad Beach, FL Cosider the itegral x= y= x= y= dxdy x + y. Editor s commet: Ed used ituitio i movig from the double summatio to the double itegral by reasoig that the liear terms i the summatio would t cotribute much to the summatio for very large values of. His ituitio was right o target, as see i Paolo s solutio above. Ed evaluated the double itegral i the usual maer, by first itegratig the iside itegral with respect to x treatig y as a costat, ad the itegratig that aswer with respect to y, treatig x as a costat. x= dx = l x + y + x x + y 5 l y x=

16 = l + y + l y l + y + + l y = l + y + l + y + Ad ow we compute: y= y= l + y + dy y= l + y + dy. l + y + dy = y l + y + + l + y + y [ l + + Let s called this A. Ad evaluatig y= we obtai [ l ] + + Ad let s call this B. y ] [ + l + + l l + + l y + + dy = y l y + + y + l y + + y + l + [ + l + + l + ]. We ow evaluate A B. Doig this gives us l +. Solutio 3 by Kee-Wai Lau, Hog Kog, Chia We show that the it equal l +, idepedet of a, b, c. We first ote that i + j + ai + bj + c i + j i= j= i= j= y= y= ] = i= j= ai + bj + c i + j + ai + bj + c i + j i + j + ai + bj + c + i + j i= j= ai + bj + c i + j 3/ 6

17 = O i= j= ai + bj + c ij 3/ = O i / i= j 3/ j= + O i= i 3/ j / j= + O i= i 3/ j 3/ j= = O. The costats implied by O deped at most o a, b, ad c. It follows that the it of the problem i fact equals. Now the last it equals i + j i= j= i= j= i + j = which we are goig to evaluate. It is easy to check that ad d l y + x dy + y = x + y dydx x + y, d l x + x + dx + x l + x + l x = l + x + y l x. Hece dydx x + y = l + x + l x dx = l +, where we have used the fact that x + x l x =. This completes the solutio. Solutio by Aastasios Kotrois, Athes, Greece Let We have a = i= j= i + j + ai + bj + c. a + a = + i= + i ai + b + + c + j= + + j + a + + bj + c 7

18 + + a + b + + c = b + + c + d + But b = i= i + + ai + b + c = i= i/ + + ai/ + b/ + c/ = i= i/ + + O = i= i/ + + O x + dx = l + ad by symmetry, the same holds for c. Sice clearly d, by Cezàro Stolz i= j= i + j + ai + bj + c l +. Commet by Bruo Salgueiro Faego, Viveiro, Spai This problem ad its solutio appeared as challege exercise U i the joural Mathematical Reflectios. See: < /MR 9 Solutios.pdf >. Pages The required value is l +. Also solved by Arkady Alt, Sa Jose, CA; Albert Stadler, Herrliberg, Switzerlad, ad the proposer. Mea Culpa The ame of Michael Thew, a studet at St. George s School i Spokae, WA was iadvertetly omitted from the list of those who had solved 577 ad

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Problems Ted Eiseberg, Sectio Editor ********************************************************* This sectio of the Joural offers readers a opportuity to exchage iterestig mathematical problems ad solutios.