F on AB and G on CD satisfy AF F B = DG

Transcription

1 THE OLYMPIAD CORNER / 353 F o AB ad G o CD satisfy AF F B = DG GC. I fact, we shall see that if directed distaces are used the F ca be ay poit of the lie AB ad G the correspodig poit o DC. Let T = EH AB; the ET AB (sice l is parallel to AB ad perpedicular to EH). Because ABCD is cyclic, the triagles DEC ad AEB are oppositely similar. Because F ad G are correspodig poits i the similar triagles DEC ad AEB as are K ad T, we have EGK = EF T or, usig directed agles, KGE = EF T. () We ow set S = EF HK ad wat to prove that these lies are perpedicular at S. Because of the right agles at H ad K, the poits E, G, H, K lie o the circle whose diameter is EG, whece KHE = KGE (as directed agles). But SHE = KHE (because S KH) ad KGE = EF T (from ()), so it follows that SHE = EF T. Also, the vertically opposite agles at E are equal so that the triagles SHE ad T F E are similar. But F T E = 9, hece HSE = 9 ; that is, EF HK. OC9. Let be a positive iteger. If oe root of the quadratic equatio x ax + = is equal to + + +, prove that a 3. Origially questio 6 from the Kazahsta Natioal Olympiad, Grade 9. Solved by G. Apostolopoulos; Š. Arslaagić; M. Bataille; C. Curtis; M. Dicǎ; O. Geupel; L. Giugiuc; B. Ji ad E. T. H. Wag; N. Midttu; P. Perfetti; V. Pambuccia; G. Scărlătescu; D. Văcaru; ad T. Zvoaru. We give a solutio similar to those provided by most of the solvers. Let Sice we have s := s as + = a = s + s. Therefore, by the AM-GM iequality we get a. Copyright c Caadia Mathematical Society, 4

2 354/ THE OLYMPIAD CORNER To prove the other iequality we show first that s. The left had side iequality is immediate: s = =. We ext prove the right had side iequality by iductio. Whe =, it states that. Next, assume the statemet is true for some =, so Thus which completes the iductio. Now, sice s, we get which completes the proof a = s + s + + = = + Ä + ä = + = 3, A Taste Of Mathematics Aime-T-O les Mathématiques ATOM ATOM Volume I: Mathematical Olympiads Correspodece Program (995-96) by Edward J. Barbeau. This volume cotais the problems ad solutios from the Mathematical Olympiads Correspodece Program. This program has several purposes. It provides studets with practice at solvig ad writig up solutios to Olympiad-level problems, it helps to prepare studet for the Caadia Mathematical Olympiad ad it is a partial criterio for the selectio of the Caadia IMO team. There are curretly 3 boolets i the series. For iformatio o tiles i this series ad how to order, visit the ATOM page o the CMS website: Crux Mathematicorum, Vol. 39(8), October 3

3 SOLUTIONS / 38 Editor s commet. As is ofte the case with a trigoometry problem, there were several differet approaches exhibitig a variety of efficiecy ad recourse to other results. Kouba produced a solutio similar to the secod oe above ad Woo aalyzed the graph of y = ta x sec x to show that it lay above the lie y = π/3+x. Some solvers used the represetatio of the tagets ad cosies of the half agles of a triagle i terms of the sides, semi-perimeter, iradius ad area. Brow oted that the give iequality is equivalet to s Rr + 3r, while Lau reduced it to 3s (4R + r). Dicǎ proved this geeralizatio : Let A A... A be a covex go. The = ta A cos π = sec A [ : 335, 336] Proposed by G. Apostolopoulos. Let x, y, ad z be positive real umbers such that xyz = ad x 4 + y 4 + z 4 = 3. Determie all possible values of x4 + y 4 + z 4. Solved by A. Alt ; Š.Arslaagic ; D. Bailey, E. Campbell ad C. Dimiie ; M. Bataille ; C. Curtis ; R. Hess ; O. Kouba ; D. Kouais ; S. Maliić ( solutios) ; P. Perfetti ; A. Plaza ; C. M. Quag ; D. Smith ; D. R. Stoe ad J. Hawis ; I. Uchiha ; D. Văcaru ; T. Zvoaru ; ad the proposer. There was also a icorrect solutio. We give a solutio that is a composite of virtually all solutios received. By the AM-GM Iequality, we have 3 = x 4 + y 4 + z x 4 y 4 z 4 = 3. Thus, we must have the equality above, which implies that x 4 = y 4 = z 4 or x = y = z. Sice we ow that xyz =, it follows that x = y = z = ad so x 4 + y 4 + z 4 = [ : 335, 337] Proposed by M. Bataille. Let A A A 3 be a triagle with circumcetre O, icircle γ, icetre I, ad iradius r. For i =,, 3, let A i o side A ia i+ ad A i o side A i A i+ be such that A i A i OA i ad γ is the A i excircle of A i A i A i where A 4 = A, A 5 = A. Prove that (a) A A A A A 3A 4a a a 3 3 = (a + a + a 3 ) r (b) A A + A A + A 3A 3 = a + a + a 3 a a a 3 IK + 3a a a 3 a + a + a 3 Copyright c Caadia Mathematical Society, 4

4 SOLUTIONS / [ : 335, 337] Proposed by O. Furdui. Let f : [, ] R be a cotiuously differetiable fuctio ad let ã ã ã x = f + f + + f. Calculate lim (x + x ). Solved by A. Alt ; M. Bataille ; O. Kouba ; M. R. Moda ; P. Perfetti ; ad the proposer. There were five flawed solutios, three of which applied a ivalid coverse of the Stolz-Cesaro theorem. We preset solutios. Solutio by Omra Kouba. The required limit is equal to f(x)dx. We first ote that, if g : [, ] R is a cotiuously differetiable fuctio, the, usig itegratio by parts, we have that x ã ï g (x)dx = x ã ò g(x) = g() + g() g(x)dx. g(x)dx Apply this to the fuctio g(x) = f(( + x)/) for =,,,..., ad add the resultig equatios to obtai where x + f() + f() H (x) = f(x)dx = ã + x f. = x ã H (x)dx, Observe that, for each x [, ], H (x) is a Riema sum for the itegral f (t)dt ad H (x) sup [.] f. From the foregoig equatio ad its aalogue for +, we obtai that x + x f(x)dx = (H + (x) H (x)) x ã dx. As teds to ifiity, the itegrad o the right side teds poitwise ad boudedly to, so by the Lebesgue Domiated Coverget Theorem, we coclude that lim (x + x ) = f(x)dx. Copyright c Caadia Mathematical Society, 4

5 386/ SOLUTIONS Solutio by Paolo Perfetti. There exists ξ (( + ), ) for which x + x = ã f f + = = f (ξ ) ( + ) = = + = ãã + f() ã + f() f (ξ ) f ãã + = f ã + f() where ( + ) < ξ <. Note that f is uiformly cotiuous o [, ] ad that ξ ( + ) <. Therefore, for each ɛ >, whe is sufficietly large f (ξ ) f ã < ɛ for. Thus + Moreover = f (ξ ) f ãã < ɛ lim + = f Therefore, itegratig by parts, we fid that lim (x + x ) = f() ã = ( + ) xf (x)dx = ã ã ( + ) = ɛ. xf (x)dx. f(x)dx. Editor s commet. Maliić ad Ricardo oted that the proposer poses ad solves this problem i his boo Limits, Series, ad Fractioal Part Itegrals published by Spriger i 3. It is problem.3 o page 6 uder Miscellaeous Limits ; the solutio appears o page 5. Crux Mathematicorum, Vol. 39(8), October 3