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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@0.et>. Solutios to previously stated problems ca be see at < Solutios to the problems stated i this issue should be posted before March 5, : Proposed by Keeth Korbi, New York, NY Ratioalize ad simplify the fractio x + 4 x06x x if x = : Proposed by Nicusor Zlota, Traia Vuia Techical College, Focsai, Romaia If x > 0, the [x] [x] 4 + [x] + {x} + {x} 4 {x} 4 + [x] + {x} 4, where [.] ad 4 4 {.} respectively deote the iteger part ad the fractioal part of x : Proposed by Titu Zvoaru, Comăesti, Romaia ad Neculai Staciu, George Emil Palade School, Buzău, Romaia Prove that there are ifiitely may positive itegers a, b such that 8a b 6a b = : Proposed by Oleh Fayshtey, Leipzig, Germay Let a, b, c be the side-legths, α, β, γ the agles, ad R, r the radii respectively of the circumcircle ad icircle of a triagle. Show that a cosβ γ + b cosγ α + c cosα β b + c cos α + c + a cos β + a + b cos γ = 6Rr. 54: 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. Prove that FF+ = is a irratioal umber ad determie it *.

2 The asterisk idicates that either the author of the problem or the editor are aware of a closed form for the irratioal umber. 54: Proposed by Ovidiu Furdui ad Alia Sîtămăria, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia Fid all differetiable fuctios f : 0, 0,, with f =, such that f =, x > 0. x fx Solutios 5409: Proposed by Keeth Korbi, New York, NY Give isosceles trapezoid ABCD with AB < CD, ad with diagoal AC = AB + CD. Fid the perimeter of the trapezoid if ABC has iradius ad if ACD has iradius 5. Solutio by Michael N. Fried, Be-Gurio Uiversity, Beer-Sheva, Israel Let AB = x, AC = AD = c, AC = K so that, sice AC = AB + CD, we ca write DC = K x. The key observatio is that if the triagle ABC is reflected ad trasposed so that BC coicides with AD, the resultig figure AEDC is a equilateral triagle. This is so because: The trapezoid is isosceles, so that EDA = π ADC, ad, therefore, EDC is a straight lie By the give, AB + DC = ED + DC = x + K x = K, ad, therefore, EA = AC = CE = K.

3 With the geometry of the situatio i mid, oe ca ow easily see that sice the diameter of O is 4 ad the diameter of O is 70, the legth of the side of the equilateral triagle i.e. the diagoal of the origial trapezoid caot be less tha 94 uits. This will be importat later. Now, sice the twice the area of a triagle is the product of its iradius ad its perimeter, we fid that twice the area of the triagle AED is c + K + x ad twice the area of the triagle ADC is 5c + K + K x. O the other had, sice we have observed that EAC is equilateral, twice the areas of these triagles are also, respectively, Kx ad KK x. Thus, we ca write the followig two equatios: c + K + x = Kx 4 c + K x = KK x 70 Usig the law of cosies i the triagle AED ad the fact that agle < AED = π, we have a third equatio: c = K + x Kx Thus, we have three quadratic equatios i three ukows, c, K, ad x. We will show that this ca be reduced to a sigle quadratic equatio i K, from which we will be able to fid x ad c. To make the algebra easier to write out, let us use the followig otatios: q = 4 70 p = Q = qk P = pk The reaso for the latter two will become clear i a momet. Eiatig c from equatios ad, we fid that x = KpK above: x = KP P + Q Kp+q or usig our otatio Note, ca be writte as K + c = qk x = Qx similarly, ca be writte K + c = P K x, so substitutig 4 ito i this form, we fid, after some easy maipulatios: c = K P Q P + Q 5 P + Q O the other had, substitutig 4 ito ad rearragig terms we obtai: K c = P + Q + P P P + Q 6 P + Q Squarig 5, equatig it with 6, ad cacelig K P +Q, we obtai: P Q P + Q = P + Q + P P P + Q Ad after some simplificatio, we have the equatio: P QP Q P + Q + = 0 4

4 But sice Qx = K + c > 0 ad P K x = K + c > 0, either P or Q ca be 0, so, we have: P Q P + Q + = 0 At this poit, we ca substitute pk = P ad qk = Q, to obtai the followig quadratic equatio i K: pqk p + qk + 6 = 0 7 Substitutig p = 70 ad q = 4 ad solvig, we obtai two solutios: K = or K = As we oted above, K caot be less tha 94, so we have oly K = 80. Usig 4 ad 5 to fid x ad c, we fid the that the perimeter of ABCD= c + K = Solutio by David E. Maes, SUNY College at Oeota, Oeota, NY Let x = AB, y = CD ad z = AD = BC. The the perimeter of the trapezoid ABCD is x + y + z = 6 whe x = 7, y = 6 ad z = 7. Deote the area of polygo X by [X]. The, by Ptolemy s theorem, AC = xy + z. Therefore, x + y = xy + z. Solvig for z, we get z = x + xy + y. The height h of the trapezoid, accordig to the Pythagorea theorem, is give by Therefore, ad h = z y x = x + y. [ABC] = x x + y [ACD] = y x + y. Let r deote the iradius of triagle T. The r s = [T ] where s is the semiperimeter of T. For each of the triagles ACD ad ABC, this formula reduces to 5 x + y + z = yx + y, 4 6x + y + z = xx + y, 4 respectively. Multiplyig the first equatio by x, the secod by y ad the subtractig the secod equatio from the first yields the followig upo simplificatio: Sice z = x + xy + y, we have z 5 x 6y = 6y xy 5 x. x + xy + y 5 x 6y = 6y xy 5 x. Squarig both sides of this equatio ad the simplifyig it, oe obtais the equatio 6y 49xy 945x = 0. 4

Regardig this equatio as a quadratic i y, oe obtais the followig roots y = 49x ± 49x + 46945x 7 49x ± 759x =.

Thus, our solutios are parametrized by x ad the problem ow is to fid the values of x that satisfy the two equatios for the iradius. To that ed suppose 6 x + 6 7 x + 7 7 x = 4 x x + 6 7 x. The x = 7.

5 Regardig this equatio as a quadratic i y, oe obtais the followig roots y = 49x ± 49x x 7 49x ± 759x =. 7 Sice y > 0 we disregard the egative root so that Moreover, z = x + xy + y = y = x = 6 7 x. x x x = 7 7 x. Thus, our solutios are parametrized by x ad the problem ow is to fid the values of x that satisfy the two equatios for the iradius. To that ed suppose 6 x x x = 4 x x x. The x = 7. Similarly, the equatio 5 x + y + z = 4 yx + y yields x = 7. Hece, y = 6 7 x = 6 ad z = 7 7 x = 7 so that the perimeter of the trapezoid ABCD is x + y + z = 6. Solutio by Nikos Kalapodis, Patras, Greece Let P be the itersectio of diagoals AC ad BD. Sice trapezoid ABCD is isosceles, the triagles ABC ad BAD, as well as, the triagles ACD ad BDC are cogruet, SAS criterio. 5

6 It follows that the triagles P AB ad P CD are isosceles with P A = P B ad P C = P D. Furthermore, sice they are similar cogruet agles we have P A AB = P C CD = P A + P C AB + CD = AC =. Thus, P A = AB ad P C = CD. AB + CD From ad we coclude that triagles P AB ad P CD are equilateral. Let p = AB = P A = P B, q = CD = P C = P D, r = BC = AD, t = p + q + r ad h the height of the trapezoid. The we have p q = ph qh = [ABC] p + q + r p + t = = or pq = tq 5p [ACD] 5p + q + r 5q + t Sice the trapezoid is isosceles, it is cyclic, so by Ptolemy s Theorem we have pq + r = p + q 4 or pq = tp + q r 5 By ad 5 we obtai 58p + q = r 6 Fially, applyig the well-kow formula r = s a ta A i triagles ACD ad BAC we have p + q + r p + q + r = 5 = r r = q p, i.e. q p = Solvig the system of equatios 4, 6 ad 7 we fid p = 7, q = 6, ad r = 7. Therefore the perimeter of trapezoid is p + q + r = 6. Also solved by Ed Gray, Highlad Beach, FL; Kee-Wai Lau, Hog Kog, Chia; Albert Stadler, Herrliberg, Switzerlad; Malik Sheykhov studet at the Frace-Azerbaija Uiversity i Azerbaija ad Talma Residli studet at Azerbaija Medical Uiversity i Baku, Azerbaija; David Stoe ad Joh Hawkis, Georgia Souther Uiversity, Statesboro, GA, ad the proposer 540: Proposed by Arkady Alt, Sa Jose, CA For the give itegers a, a, a fid the largest value of the iteger semiperimeter of a triagle with iteger side legths t, t, t satisfyig the iequalities t i a i, i =,,. Solutio by Kee-Wai Lau, Hog Kog, Chia Without loss of geerality, we assume that a a a. Let Let L = Maximum t,t,t S T = {,,..., a }, T = {,,..., a }, T = {,,..., a } T = {t, t, t : t T, t T, t T } ad S = T {t, t t : t, t, t are the side legths of a triagle}. t + t + t. We show that L = a + a + a, if a + a > a a + a, if a + a a. Case : a + a > a 6

7 We have a, a, a S ad clearly L = a + a + a. Case : a + a a We have a + a, a, a S so that L a + a. If t, t, t T ad t > a + a, the t, t, t / S. If t, t, t T the t < a + a, the t + t + t < a + a + a + a = a + a. Hece, L = a + a i this case. This completes the solutio. Solutio by proposer Let s = t + t + t. Sice t i < s, i =,, the by the triagle iequality our problem becomes the followig: Fid the maximum of s for which there are positive iteger umbers t, t, t satisfyig t i mi{a i, s, i =,,, t + t + t = s. First ote that s, t i, i =,,. Ideed, sice t i s the s t i, i =,, ad therefore t = s t t = s t + s t. Cyclicly we obtia t, t Hece, s 6 s. Sice t = s t t, t mi{a, s }, the s t t mi{a, s } max{s t a, s + t } t s t, ad therefore, we obtai the iequality for t, amely that max{s t a, s + t, } t mi{s t, a, s } with the coditios of solvability beig: s t a s s + a t s t a a s a a t s + t a s + a t s t t s Sice s s, the together with t mi{a, s } gives us the bouds for t max{s + a, s a a, s + a, } t = mi{a, s }. Sice a i, i =,, the s + a s, s + a s ad the solvability coditio for becomes a + a + a s + a a s a + a, s a a a s, s + a a s a + a, s a a s s +a + a. { } Thus, s a + a + a = mi, a + a, a + a, a + a is the largest iteger value of the semiperimeter. Solutio by Ed Gray, Highlad Beach, FL We cosider several special cases: a If a = a = a = k, we ca equate t i = a i for each i. The perimeter is the 6k ad the semiperimeter is k. 7

8 b Suppose a = a = a = k +. We ote that a + a = 4k + ad a = k. We defie t = a, t = a ad t = a. c Suppose that a = a ad a is larger tha either oe. I this case we set t = a ad t = a. It does t matter if a, a are both eve or both odd, t + t is eve. We ow have to avoid a potetial problem. It must be true that t + t t. Therefore, sice if a is large, we eed to defie t = a x, where x is the iteger which is the smallest such that a x is eve ad t + t > t. Sice t + t + t is eve, the semiperimeter is itegral. d Suppose that a = a ad a is smaller tha either oe, i this case set t = a, t = a, so that t + t is eve. If a =, we let t =. If a >, but odd, we we set t = a. The t + t + t equals the perimeter which is eve ad with a iteger semiperimeter, ad the triagle iequalities hold. e Fially, we have the geeral case: a < a < a. We set t = a, t = a. If t + t is eve we eed t to be eve. If a is very far so that a + a < a, we let t = a x, where x is the smallest iteger which simultaeously makes t + t + t eve ad t + t > t. If t + t is odd, we employ a similar calculatio. Solutio 4 by Paul M. Harms, North Newto, KS Suppose a a a. The largest perimeter would be a + a + a where t i = a i, i =,, provided that we have a triagle, i.e., a + a > a. If a + a > a, ad the perimeter is a eve iteger, the the largest value of a iteger semiperimeter is a + +a. If the perimeter is a odd iteger, the a must be at least ad we could use sides t = a, t = a ad t = a. The largest iteger semiperimeter for this case is a + a + a. Now cosider the case where a + a a. A triagle with a maximum perimeter is whe t = a, a = t, ad t = a + a. Here t > a, a ad the perimeter is the odd iteger a + a. To get the largest iteger semiperimeter we could use t = a, t = a ad t = a + a which has a + a as the largest iteger semiperimeter. Also solved by Jeremiah Bartz ad Timothy Prescott, Uiversity of North Dakota, Grad Forks, ND; David Stoe ad Joh Hawkis, Georgia Souther Uiversity, Statesboro, GA 54: Proposed by D.M. Bătietu-Giurgiu, Matei Basarab Natioal College, Bucharest, Romaia ad Neculai Staciu, George Emil Palade Geeral School, Buzău, Romaia Let a, b be real valued positive sequeces with a = b = a R + If a a = b R ad b a = c R compute a + + +! b! Note: R + meas the positive real umbers without zero. 8.

9 Solutio by Bria Bradie, Christopher Newport Uiversity, Newport News, VA By Stirlig s approximatio, so It the follows that! e! π +/ e, ad + +! +. e a + + +! b! + a + e b e = e [ + a + a b a + a] ad a + + +! b! = e [ + a + a b a + a] = b c + a. e Solutio : by Moti Levy, Rehovot, Israel. +! + a+! b = = =! +! + + +! + + +! a + a + +! + a!! + a + a + a + a b a + a +! +!. b a! a b a + a +! +!! So we are challeged with two its: ad +! +!. We will show that both its equal to e. We begi by statig the well-kow asymptotic expasio of the Gamma fuctio: For positive iteger, e x x x πx e Γ x + + x, x.! π +,. Usig + + ad π, we get e! +,, 9

10 or which implies We coclude that! e +,,! = e. +! +! =!! e +! + +!! Now back to the origial it + ; +! +! +! + +! + +! +! + + +! +! e +! +! + e + = + + e. +! +! = e.! = e b e c + a e = a + b c. e + a + a b a + a +! +! Also solved by Arkady Alt, Sa Jose, CA; Paul M. Harms, North Newto, KS; Paolo Perfetti, Departmet of Mathematics, Tor Vergata, Rome, Italy, ad the proposers. 54: Proposed by Michal Kremzer, Gliwice, Silesia, Polad Give positive iteger M. Fid a cotiuous, o-costat fuctio f : R R such that f fx = f [x], for all real x, ad for which the maximum value of fx is M. Note: [x] is the greatest iteger fuctio. Solutio by Tommy Dreyfus, Tel Aviv Uiversity, Israel Let fx = 0 except for M < x < M +, where fx = M M M x +. 0

11 The f is cotiuous, attais its maximum at f M + = M, ad ffx = f[x] = 0 for all x. Solutio by Albert Stadler, Herrliberg Switzerlad { M si πx, if x < 0 or x > M Let fx = 0, if 0 x M. fx is cotiuous ad o-costat. I additio f = 0 for all itegers. 0 fx M ad the maximum M is assumed. f [x] = 0 for all real x sice [x] is a iteger. ffx = 0 for all real x, sice 0 x M ad fy = 0 for 0 y M. Solutio by Moti Levy, Rehovot, Israel Let f : R R be defied as follows M is positive iteger: { [ ] M x f x = M+ si πx, for M + x M + 0, otherwise. The fuctio f x is cotiuous ad its maximum value over R is M. Clearly sice si π [x] = 0, f [x] = 0. [ ] By its defiitio 0 f x M. Hece, f f x = 0, sice fx M+ = 0. f f x = { M [ ] fx M+ si πf x = 0, for M + x M + 0, otherwise. We coclude that f x is cotiuous ad o-costat fuctio with maximum value M, which satisfies ffx = f[x] = 0. Solutio 4 by The Ashlad Uiversity Udergraduate Problem Solvig Group, Ashlad, OH The followig fuctio satisfies the give coditios; x + M if M x M / fx = x + if M / < x M M otherwise. We ca easliy check that f is cotiuous by otig that: fm = M, f M = M, ad fm = M. We ow show f satisfies f fx = f [x]. Whe x M, [x] M ad ffx = fm = M = f[x]. Whe x M, [x] M ad ffx = fm = M = f[x]. Fially, whe M < x < M, [x] = M ad M Thus, ffx = M = f [x]. fx < M.

12 Also solved by Michael N. Fried, Be-Gurio Uiversity, Beer-Sheva, Israel, ad the proposer. 54: Proposed by José Luis Díaz-Barrero, Barceloa Tech, Barceloa, Spai Compute i j + i + j + ij. Solutio by Bria Bradie, Christopher Newport Uiversity, Newport News, VA i j + i + j + ij = = = = i j x i/ + j/ + x + y dy dx + y + x x 0 dx + x = x 4 + x = Solutio by Kee-Wai Lau, Hog Kog, Chia Sice + i + j + ij = + i + j, it is easy to check that i j + i + j + ij = + i= + i i= + i Now 0 < i= i= + i + i = = + i=, we have + i j i= 0 dx =, ad from + x = 0, so by, we obtai + i + i + j + ij =. Also solved by Arkady Alt, Sa Jose, CA; Bruo Salgueiro Faego, Viveiro, Spai; Moti Levy, Rehovot, Israel; Paolo Perfetti, Departmet of Mathematics, Tor Vergata, Rome, Italy; Albert Stadler, Herrliberg, Switzerlad, ad the proposer.

13 544: Proposed by Ovidiu Furdui, Techical Uiversity of Cluj-Napoca, Cluj-Napoca, Romaia Let A, B M C be such that 05AB 06BA = 07I. Prove that Here, C is the set of complex umbers. AB BA = O. Solutio by Athoy J. Bevelacqua, Uiversity of North Dakota, Grad Forks, ND Recall that the characteristic polyomial of a x matrix M is p M x = detm xi. A easy calculatio shows that p M x = x trm x + detm where detm is the determiate of M ad trm is its trace. By the Cayley-Hamilto Theorem we have p M M = 0. We first ote that AB ad BA have the same characteristic polyomial px because detab = detba ad trab = trba. We give 05AB 06BA = 07I. Addig AB to both sides of this equatio yields Takig the determiat of this we fid 06AB BA = AB + 07I. 06 detab BA = detab + 07I = p 07. Similarly addig BA to both sides of the origial equatio ad takig the determiat yields 05 detab BA = 05 detab BA = p 07. Thus ad so detab BA = detab BA = 05 detab BA Sice trab BA=trAB trba = 0 we see that the characteristic polyomial of AB BA is x. Thus, AB BA = 0. Essetially the same argumet would establish the followig mild geeralizatio: Let A, B M K were K is a field. Let s, t K with s ± ad t 0. The AB sba = ti implies AB BA = 0. Solutio by Albert Stadler, Herrliberg, Switzerlad AB ad BA have the same eigevalues, sice detxi AB = detxi BA. Ideed, whe A is osigular this result follows from the fact that AB ad BA are similar: BA = A ABA. For the case where both A ad B are sigular, oe may remark that the desired idetity is a equality betwee polyomials i x ad the coefficiets of the matrices. Thus, to prove this equality, it suffices to prove that it is veified o a o-empty ope subset for the usual topology, or, more geerally, for the Zariski topology of the space of all the coefficiets. As the o-sigular matrices form such a ope subset of the space of all matrices, this proves the result.

14 Let x be a eigevalue of AB. The 0 = det xi AB = det xi BA I = det = det x I BA 05 x I AB 05 = 06 05x det I AB. 06 detxi AB is a quadratic polyomial i x, let s say detxi AB = ax + bx + c. the implies that ax + bx + c = x a + b coefficiets of the polyomials ad see that b = 404a, c = 07 a. + c. We compare the So the quadratic polyomial reads as ax + 404ax + 07 a = ax + 07 which shows that the characteristic polyomial of AB ad BA has -07 as a double zero, x is a eigevector of both AB ad BA correspodig to the eigevalue -07. Therefore 07 u there are umbers u ad v such that AB is similar to v ad BA is similar to. Therefore, AB BA 0 07 is thus similar to 0 u v = 0 0 0, which implies that AB BA = 0. Solutio by Michael N. Fried, Be-Gurio Uiversity, Beer-Sheva, Israel Let us write [A, B] for AB BA. Sice traceab = traceba, we have, as is well-kow, trace[a, B] = 0. Thus, keepig i mid that [A, B] is a matrix, the characteristic polyomial of [A, B] is x + det[a, B] = 0, so that if its eigevalues are λ ad λ, we have λ = λ = λ ad λ = det[a, B]. Moreover, sice every matrix satisfies its ow characteristic polyomial, [A, B] = det[a, B]I Therefore, [A, B] = 0, which is what we wat to show, if ad oly if λ = det[a, B] = 0, that is, if ad oly if λ = 0. We will show that, ideed, λ = 0 Cosider the give equatio pab p + BA = p + I. By addig BA or AB to both sides, we obtai, respectively: p[a, B] = p + I + BA 8 p + [A, B] = p + I + AB 9 4

15 Let λ be a eigevalue for [A, B] ad v the correspodig eigemvector. Thus, we have by 8: p[a, B]v = pλv = p + I + BAv Thus, BAv = pλ p + v so that, pλ p + is a eigevalue for BA. Sice λ is the other eigevalue of [A, B], we fid that pλ + p + is the secod eigevalue of BA. I the same way, usig equatio 9, we fid the eigevalues of AB to be p + λ p + ad p + λ + p + The determiat of ay matrix is of course equal to the product of the eigevalues. Moreover, detab = detba. Hece: pλ + p + pλ p + = p + λ + p + p + λ p + From which we have: p + p λ = 0 So that λ = 0, which is what we wished to prove. Solutio 4 by Bria D. Beasley, Presbyteria College, Clito, SC Assume A, [ B M ] C with AB [ ] + BA = + I for some positive iteger. a b e f Write A = ad B =. c d g h The AB BA = ki, where By hypothesis, we have: k = bg cf + af + bh be dfce + dg ag ch. ae + bg + ae + cf = + af + bh + be + df = 0 cf + dh + bg + dh = + ce + dg + ag + ch = 0. Thus af + bh be df = be + df ad ce + dg ag ch = ag + ch. Also, + / = af + bh/be + df = ce + dg/ag + ch, ad ae dh = + bg cf. Substitutig yields k = ae dhbg cf + + be + dfag + ch. 5

16 The + k = abeg acef bdgh + cdfh + + abeg + bceh + adfg + cdfh + = abeg + cdfh acef + bdgh + + adfg + bceh + = [ag + chbe + df adfg bceh] acef + bdgh + + adfg + bceh + = ag + chbe + df adfg + bceh acef + bdgh af + bh ce + dg = ag + chbe + df adfg + bceh acef + bdgh be + df ag + ch = af + bhce + dg adfg + bceh acef + bdgh = 0. Hece k = 0 as eeded. Also solved by Moti Levy, Rehovot, Israel, ad the proposer. Mea Culpa Paul M. Harms of North Newto, KS ad Jeremiah Bartz of Uiversity of North Dakota, Grad Forks, ND should have each bee credited with havig solved problem

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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.