a 2 +b 2 +c 2 ab+bc+ca.

Transcription

1 All Problems o the Prize Exams Sprig 205 The source for each problem is listed below whe available; but eve whe the source is give, the formulatio of the problem may have bee chaged. Solutios for the problems preseted here were obtaied without cosultig sources for these solutios eve whe available, ad additioal iformatio o how to solve these problems might be obtaied by cosultig the origial sources. The Juior Prize Exam was ot give this year. ) SENIOR ) Let p be a prime umber, ad let r be the remaider whe p is divided by 30. Show that r is also prime or r =. Source: Problem 8, Középiskolai Matematikai Lapok, Vol. IV, No. 952), p. 24. See Solutio: We may assume that p > 30; otherwise, we would have r = p, ad p is a prime. We have r = p 30q for some iteger q; further 0 r < 30. The prime divisors of 30 are 2, 3, ad 5; sice oe of these is a divisor of p, they caot be a divisor of r either. Thus, uless r =, the smallest prime divisor of r is 7. Uless r = or r is this prime itself, we must have r 7 2 = 49. Sice r < 30, this is ot possible; so, ideed, r = or else is a prime. 2) SENIOR 2) Give real umbers a, b, ad c, show that a 2 +b 2 +c 2 ab+bc+ca. Source: Problem 3, Középiskolai Matematikai és Fizikai Lapok, Vol. I, No ), p. 40. See Solutio: We have 0 a b) 2 +b c) 2 +c a) 2 = 2a 2 +b 2 +c 2 ab bc ca), whece the assertio follows. Clearly, equality holds oly i case a = b = c. 3) SENIOR 3) Give six cosecutive itegers, show that there is oe amog them that is relatively prime to all others. Two itegers are called relatively prime if their greatest commo divisor is.) Source: Problem 2, first category, roud 2 for 0th grades, Daiel Aray Mathematics Competitio, 960. See Source: Solutio: The differece of ay two amog six cosecutive itegers is at most 5. Sice the greatest commo divisor of two umbers also divides their differece, the oly prime factors that the greatest commo divisor of these two itegers ca have are 2, 3, ad 5. So, i order to fid a umber amog the six that is relatively prime to the others, we oly eed to fid oe umber that is ot divisible by 2, 3, ad 5. There is oe amog the six umbers, say, that is divisible by 6. The oe of the umbers, 5, +, ad +5 is divisible by 2 or 3, ad least two of them is amog the six cosecutive All computer processig for this mauscript was doe uder Debia Liux. The Perl programmig laguage was istrumetal i collatig the problems. AMS-TEX was used for typesettig.

2 umbers. Amog these two, oly oe ca be divisible by 5, sice the differece oly of +5 ad 5 amog them is divisible by 5, ad ot both of these umbers are amog the six cosecutive umbers. Thus, there will be oe umber amog, 5, +, ad +5 that is ot divisible by 5 ad is amog the six cosecutive umbers. This umber will be relatively prime to all the others. 4) SENIOR 4) Let be a positive iteger, ad let a, a 2,..., a be real umbers. Write fx) = a k sikx. Assume that fx) x for all x > 0. Prove that ka k. Source: Based o Problem 4, APICS Atlatic Provices Coucil o the Scieces, Caada) Mathematics Cotest 993. See Solutio: We have sicx lim = c x 0 x for ay real c. Hece fx) lim x 0 x = ka k. I view of the iequality fx) x, the absolute value of the right-had side must be less tha or equal to, establishig the result. 5) SENIOR 5) Let a 0 for all ad assume that a k k=+ for. Show that a k is coverget ad its sum is less tha 2ea, where e is the base of the atural logarithm. Source: Problem 6, Schweitzer Miklós Emlékversey Miklós Schweitzer Memorial Competitio), Hugary 958. See Solutio: We claim that we have a k ) 2 a k a k=0 +2 k ) for all 0. This is easy to show by iductio o. Ideed, for = 0 this just says a a, sice i this case the product o the right-had side is empty, ad the value of the empty product is by covetio. So let ad the assume that ) is true for this. By the assumptio of the problem, we have 2 2 a k 2 + k=2 + a k,

3 ad so 2) By ) ad 2) we have +2 ) 2 a k + a k. a +2 k ) = +2 ) a +2 k ) +2 ) 2 a k the first iequality holds by ), ad the secod oe by 2). This completes the iductio, establishig ). If a = 0, the assertio of the problem immediately follows from ), so assume that a > 0. Writig logx for the atural logarithm of x ad usig the iequality log+x) < x, valid for all x > 0, by ) we have + log a k loga + log +2 k) k=0 < loga +log2+log + ) = loga +log2+log + ) for > 2, where the /2 i the secod lie cacels out the term of the sum for k =, which is ot eeded sice the first two terms of the secod sum i the first lie are explicitly writte out. Thus log a k loga +log2+log + ) As log+/2) /2, the iequality follows. a k < 2ea. 6) SENIOR 6) Let > 0 be a iteger. Cosider a polyomial i variables with real coefficiets. We kow that if every variable is ±, the value of the polyomial is positive or egative accordig as the umber of variables havig value is eve or odd. Prove that the degree of this polyomial is at least. Source: Problem 2, József Kürschák Mathematical Competitio, Hugary, 995. See Click o Hugary uder the headig Natioal ad Regioal competitios. Solutio: We may assume that the polyomial is symmetric i its variables. I fact, if the polyomial is Rx,x 2,...,x ), we ca take the polyomial 2 k + a k ; Qx,x 2,...,x ) = σ Rx σ),x σ2)...,x σ) ), 3

4 istead, where σ rus over all permutatios of {,2,...,} i.e., all oe-to-oe mappigs of this set oto itself). We may also assume that for ay i with i, x i occurs i each term of Qx,x 2,...,x ) with expoet 0 or. Ideed, i every term we ca replace x l i with is l is eve ad with x i if l is odd; this will ot chage the value of Qx,x 2,...,x ) for x i = ±. I this way, the degree of Qx,x 2,...,x ) will be at most. We are goig to show that there is a polyomial Px) such that ) P x i = Qx,x 2,...,x ) wheever x i = ± for all i with i, ad, further, the degree of P is ot higher tha that of Q. Such a polyomial P ca easily be costructed by recursio. Assume P 0 x) = 0 ad Q 0 x,x 2,...,x ) = Qx,x 2,...,x ). At the kth step, we will elimiate all terms of degree k from Q k ad itroduce a correspodig term of degree k ito P, while preservig the symmetry of the polyomial Q k. So, assume that P k x) ad Q k x,x 2,...,x ) have already bee costructed for some k with 0 k, ad the degree of Q k x,x 2,...,x ) is at most k; ad, further, the expoet of each occurrece of x i i this polyomial for i with i is 0 or. Let c k be the coefficiet of the term k x i i this polyomial for some costat c k possibly c k = 0), ad let P k+ x) = P k x)+c k x k. Cosider the polyomial ) k. Q k x,x 2,...,x ) c k x i I this polyomial, replace all occurreces of x l i with if l is eve, ad with x i if l is odd, ad call the resultig polyomial Q k+ x,x 2,...,x ). The degree of this polyomial is at most k zero beig the degree of a ozero costat polyomial, ad beig the degree of the idetically zero polyomial), ad ) P k x i +Q k x,x 2,...,x ) = P k+ x i )+Q k+ x,x 2,...,x ) wheever x i = ±. For k = + we will have Q + x,x 2,...,x ) = 0 ad ) P + x i = Qx,x 2,...,x ) wheever x i = ± i ). For k with 0 k, takig x i = for i k ad x i = for i > k, we therefore have P 2k) = ) k. This meas that Px) has at least sig chages, ad so it has at least real) zeros. Therefore, the degree of Px) is at least, completig the proof. 7) SENIOR 7) Prove that the equatio y = y 2 +x, y0) = 0 does ot have a solutio o the iterval 0, 3). Source: Ural State Uiversity- DMM Olympiad, 2005, Problem 3. Click o the last item i the middle colum, with the headig Udergraduate Competitios. 4

5 Solutio: Assumig yx) is such a solutio, we have O the other had, ad so yx) = 0 yt) ) 2 +t ) dt y y 2 +x =, y y 2 for x. + Itegratig this o the iterval [,x] for x > we obtai 0 tdt = x y t) yx) dy ) 2 dt = yt) + y 2 + = arctayx) arctay) dt = x. y) Notig that y) 0 ad so arctay) 0 ad arctayx) lim y arctay = π/2, this iequality implies that x π/2, ad so x + π/2 < 3. Hece there is o fuctio yx) with y0) = 0 that is cotiuous at 0 ad satisfies the differetial equatio y = y 2 + x o the iterval 0, 3). 5