Test de Departajare pentru MofM 2014 (Bucureşti) Enunţuri & Soluţii


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1 Test de Departajare petru MofM 04 Bucureşti Euţuri & Soluţii Problem. Give + distict real umbers i the iterval [0,], prove there exist two of them a b, such that ab a b < Solutio. Idex the umbers 0 a 0 < a < < a. If a 0 = 0 we re doe; if ot, a k+ a k a k+ a k = a a 0 a k+ a k <, so there will exist 0 k such that a k+ a k a k+ a k < by a averagig argumet. Alterative Solutio. L. Ploscaru Idexăm umerele 0 a 0 < a < < a. Di pricipiul cutiei, va exista u idice 0 k astfel îcât a k+ a k. Dar avem a k+ a k = a k+ a k a k+ a k +a k+ a k > a k+ a k a k+ a k cu iegalitate strictă, căci a k+ a k, deci a k+ a k a k+ a k <. Remark. The mai idea is to otice that cubes of the variables are of good use, via the equivalet forms aba b = a b a b or a b = a ba b + ab. Problem. What is the miimum umber m of edges of K the complete graph o 4 vertices that ca be colored red, such that ay K 4 subgraph cotais a red K? For example, m4 =. Solutio. The aswer is i fact quite easy to get. Assume the edge ab is ot red. The the fact that amog ay {a,b, x, y} has to exist a red triagle forces x y to be red, ad moreover, either ax, ay to be red or bx,by to be red. That meas K {a,b} = K is red. Let A be the set of vertices x such that ax is red, ad B be the set of vertices y such that by is red; it follows A B = K \ {a,b}. If we could take x A \ B ad y B \ A, the {a,b, x, y} would be a cotradictio, so say B \ A =, thus A = K \ {a,b}, therefore K {b} = K is red. That is eough, so m = /. Alterative Solutio. Cosider the largest red clique K r cotaied i such a graph G. If r, for ay v G K r there must exist some f v K r such that the edge v f v is ot red. Now, if there exist v, w G K r such that f v f w, the the subgraph iduced by {v, w, f v, f w} caot cotai a red triagle. O the other had, havig f v = ω costat for all v G K r, makes that i order for the subgraph iduced by {v, w,ω,u}, for arbitrary v, w G K r ad u G ω, to cotai a red triagle, we eed vw to be red; but the K = G ω is a red clique. Alterative Solutio. A. Măgălie Demostrăm că î orice astfel de graf va exista măcar u vârf de grad roşu cel puţi. Cum m4 =, rezultă pri iducţie că m + m + = / + = /, cu modelul miim descris mai sus. Problem. Let 0 < p Q be fixed real umbers, ad let a,b, x ad y be positive real ax p ay Q umbers, such that. Determie the maximum value of a + bx + y, ad bx Q by Q the cases of equality. SGALL S LEMMA
2 Solutio. Let us ormalize, by takig λ = y x, µ = b, m = mi{λ,µ}, M = max{λ,µ}, a p p = p ax ad Q = Q m Q, ad dividig all iequatios by ax, to get ax M Q. mm Q We thus eed to maximize + m + M. We claim the maximum is p +Q. If m <, the + m + M + mm < + M + Q p +Q. If m, the m M 0, so m + M + mm, thus + m + M + mm + mm p +Q. Equality is reached if ad oly if p =, Q = M ad m =. Goig back to the origial variables, the above meas a + bx + y p +Q, with equality occurig if ad oly if p = ax, Q = by ad y = x or b = a. Problem 4. Say that a odegeerate triagle is fuy if it satisfies the coditio that the altitude, media, ad agle bisector draw from oe of the vertices partitio the triagle ito 4 ooverlappig triagles whose areas form i some order a 4term arithmetic sequece. Oe of these 4 triagles is allowed to be degeerate. Fid, with proof, all fuy triagles. MATH PRIZE FOR GIRLS 0 Solutio. L. Ploscaru Să presupuem că cele trei ceviee pleacă di A, cu AB < AC ABC u poate evidet fi isoscel î A; di ipoteză se deduce şi că triughiul fuy u poate fi obtuzughic î B sau C.. Ordiea dreptelor este AB îălţimea bisectoarea mediaa AC se demostrează evetual uitâdue la picioarele lor pe BC. Ideea pricipală este să demostrăm că u triughi fuy ABC e dreptughic parateza di ipoteză face aluzie la această posibilitate; dacă u erau triughiuri fuy dreptughice, u îşi avea rostul. Să zicem că M este mijlocul lui BC ; atuci aria[abm] = aria[ac M], deci clar AC M e triughiul cu cea mai mare arie. Fie q, q + r, q + r, q + r ariile. Cele triughiuri mici îl partiţioează pe ABM, deci q + r = q + r, de ude q = 0, iar atuci sigurul fel î care di cele 5 drepte de mai sus pot coicide este AB BC, adică ABC este dreptughic î B î afară de cazul imposibil câd ABC este isoscel î A. Acum problema e aproape gata; luăm D piciorul bisectoarei, şi pri simpla formulă aria = baza îălţimea, vom obţie că {BD,DM, MC } = {x,x,x} petru u x real pozitiv. Evidet MC = x, iar atuci î fiecare ditre cele două cazuri aplicăm teorema bisectoarei ca să aflăm valoarea raportului AB/AC = cos A, şi am termiat. Obţiem A {arccos/5, arccos/ = π/} deci uul ditre triughiuri este cel de ughiuri 0,60,90, dar mai există u caz. Problem 5. For positive real umbers a,b,c with a +b +c, prove the iequality a + bc + b + ca + c + ab ad determie its cases of equality. Show that if a + b + c <, the iequality may hold o more. DAN SCHWARZ, variat of Italia Test Solutio. It is eough to cosider the case a + b + c =. Ideed, for k we have ka + kbkc a + bc et.al.
3 We the have + bc + b + c = 5 a, hece fuctio f : [0,] R give by f t = have by Jese s iequality a f a + f b + f c + b + c f a et.al. The the t 5 t = 0 is clearly covex, therefore we 5 t + bc a 5 a = f =. Thus the iequality is proved, with the obvious equality case whe a + b + c = ad a = b = c =. For a +b +c < the iequality will hold o more; just cosider 0 < a = b = c = k <, ad the LHS = k + k <. Alterative Solutio. Tryig the CauchySchwarz iequality, just for a + b + c = see to be eough a + bc + b + ca + c a + b + c + ab + bc + ca = + ab + bc + ca + ab + ab + bc + ca will ot work this time, sice the hopeful cotiuatio towards value would require 6 + 4ab + bc + ca 9 + ab + bc + ca, i.e. ab + bc + ca, which i fact it is precisely the other way aroud. If however we try a commo trick, ad write a + bc + b + ca + c + ab = a 4 a + a bc + b 4 b + b ca + c 4 c + c ab, the we ca cotiue by CauchySchwarz a 4 a + a bc + b 4 b + b ca + c 4 c + c ab a + b + c a + b + c + abca + b + c = 9 + abca + b + c. Now, i order to cotiue with, we eed abca + b + c, which holds true, sice a + b + c / abc = ad a + b + c a + b + c = ; the equality case follows as above. Alterative Solutio. C. Popescu The required iequality is a cosequece of the followig iequality a + bc a + b + c + a + b + c. To prove the latter, apply Jese s iequality to the covex fuctio t +t, t >, at t = bc, t = ca ad t = ab, with weights λ = a /a + b + c, λ = b /a + b + c ad λ = c /a + b + c, respectively, to obtai a a + b + c + bc + a bc = a +b +c ad get thereby a Now, + bc a + b + c a + b + c + abca + b + c. a + b + c a + b + c + abca + b + c, so abca + b + c a + b + c a + b + c a + b + c = a + b + c, a This eds the proof. + bc a + b + c a + b + c + a + b + c = a + b + c + a + b + c.
4 Remarks. Notice that the followig agai works immediately. a 4 + bc + b4 + ca + c4 + ab a + b + c + bc + ca + ab = 9 + ab + bc + ca. The origial Italia Test problem was to prove for a + b + c = the iequality + bc + + ca + + ab, much easier to hadle. A "brute force" solutio is also possible here, but more difficult to compute for the variat asked above. I fact a + b + c is both eeded, ad eough, for the Italia problem. Combiig the two, both holdig for a + b + c =, allows us to the claim that + a + bc + + b + ca + + c + ab. Problem 6. Fid the formula of the geeral term of a real umbers sequece x satisfyig { x = x + x = x x + 6 Solutio. It is clear the sequece is strictly icreasig. The x + x = x x + 6 = x + x x + + x x x + 6 allows us to write x + + x = x+ x So 4x + 5x = x + 6. Square it, write it for the ext idex, subtract the two ad factorize, i order to get 8x + x x + 5x + + x = 0, hece x + 5x + + x = 0. By the kow methods, the geeral solutio is x = α + β. Sice the sequece ca i fact be prologed to the left, to x 0 = 0, the coefficiets ca be determied to be α =, β =, so x = + +. Alterative Solutio. If we compute the first few terms ad "guess" this formula, it is a simple task to check it verifies the recurrece relatio, sice x + x = + = + +, x x + 6 = = + + The fact that the sequece is uiquely determied by its first term immediately follows from the fact that the mappig x x x + 6 is icreasig from to +, thus oetooe, or from the previously obtaied relatio 4x + = 5x + x + 6. There are merits i this approach, especially if oe has see i the past such relatios. Alterative Solutio. M. Iva Se vede că şirul este strict crescător, deci cu termei pozitivi. Deoarece fucţia f :, 0, dată de f x = x este bijectivă, x putem scrie uic x = a, cu a = şi a > petru toţi >. Relaţia de a recureţă coduce atuci la a + = a, deci a = petru orice. Pri urmare x =, care evidet verifică. Sfârşit. Soluţii origiale şi compilate de DAN SCHWARZ, 9 decembrie 0 4.
5 More developmets o Problem. Give + distict real umbers i the iterval [0,], prove there exist two of them a b, such that ab a b < Solutio. Idex the umbers 0 a 0 < a < < a. If a 0 = 0 we re doe; if ot, a k+ a k a k+ a k = a a 0 a k+ a k <, so there will exist 0 k such that a k+ a k a k+ a k < by a averagig argumet. We ca say more. We have a k+ a k a k+ a k = Hölder s iequality, so the boud becomes by the followig a a0 a a 0 a a 0, by +, LEMMA. For 0 x < y, the maximum of y x y x is +. Proof. Let f x, y = y x y x ; the d dy f x, y = y y x vaishes at y = x, outside 0,], so max f x, y = f x, = x<y x x. Accordigly, d dx f x, = x + x vaishes at x =, so max + f x, = 0 x< +. The equality case i Hölder forces a k = a 0 + k +, thus with a 0 = + it yields a k = k + + ad ideed a = + =. But the still the summads are ot all equal, so + is ot reachable. For that we would eed all summads equal, ad that + possibility has to be explored. However, we still improved to the existece of a,b with ab a b < +. A secod course of actio is to explore the maximum of y x y x = yxy x, with k x < y k +, ad apply it to the Alterate Solutio. LEMMA. For 0 a x < y b, the maximum of y x y x = yxy x is as show below. Proof. See as a fuctio i y, it is a parabola with apex at y = x/, thus havig it maximum at y = b. See ow as a fuctio i x, it is a parabola with both apex ad maximum at x = b/, amely the maximum will be b/4 if b/ a, but otherwise ab b a. A third course of actio is to maximize a commo value for each term usig a =. Alterative Approach. M. Băluă Putem vedea expresia a k+ a k a k+ a k, cu o miimă trasformare, drept sumă Riema sau Darboux. Avem, petru 0 a < b, ba < 4 b+a, şi 4 b+a b a < b a = Dar atuci ak+ a k x dx = a k+ a k a k+ a k < a a 0 x dx 0 b a x dx, căci ambele revi la b a > 0. 4 a k+ +a k a k+ a k < ak+ a k x dx, şi cum x dx =, vom găsi u k cu a k+a k a k+ a k < 5
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