SOLUTIONS - Springer

1 SOLUTIONSM ethods of the contrary, namely that 2+ 3+ 5=r, whereris a rational the equality 2+ 3=r 5 to obtain 5+2 6=r2+5 2r 5. It followsthat 2 6+2r 5 is itself rational. Squaring again, we find that 24+20r2+8r 30is rational, and hence 30 is rational, too. Pythagoras method for proving that 2isirrational can now be applied to show that this is not true. Write 30=mnin lowestterms; then transform this intom2=30n2. It follows thatmis divisible by 2 and because2(m2)2=15n2it follows thatnis divisible by 2 as well. So the fraction was not in lowestterms, a contradiction.

2 We conclude that the initial assumption was false, and therefore 2+ 3+ 5 is that such numbers do exist, and let us look at their prime factorizations. Forprimespgreater than 7, at most one of the numbers can be divisible byp, and the partitioncannot exist. Thus the prime factors of the given numbers can be only 2, 3, 5, and now look at repeated prime factors. Because the difference between two numbersdivisible by 4 is at least 4, at most three of the nine numbers are divisible by 4. Also, atmost one is divisible by 9, at most one by 25, and at most one by 49.

3 Eliminating these atmost 3+1+1+1=6 numbers, we are left with at least three numbers among the ninethat do not contain repeated prime factors. They are among the divisors of 2 3 5 7,and so among the numbers2,3,5,6,7,10,14,15,21,30,35,42,70 ,105, the difference between the largest and the smallest of these three numbersis at most 9, none of them can be greater than 21. We have to look at the sequence1,2,3,..,29. Any subsequence of consecutive integers of length 9 that has a termgreater than 10 contains a prime number greater than or equal to 11, which is from 1,2.

4 ,10 we cannot select nine consecutive numbers with the requiredproperty. This contradicts our assumption, and the problem is of example 22,32,52,..,432, where we considered the squares of the first 14 primenumbers, shows thatn that there exista1,a2,..,a16, pairwise relatively prime integers greater than1 and less than 2005, none of which is a prime. Letqkbe the least prime number in thefactorization ofak,k=1,2,..,16. Letqibe the maximum ofq1,q2,..,q15. Thenqi p16=47. Becauseaiis not a prime,aiqiis divisible by a prime number greater thanor equal toqi.

5 Henceai q2i=472>2005, a contradiction. We conclude thatn= by contradiction, we assume that none of the colors has the desired there exist distancesr g bsuch thatris not attained by red points,gby greenpoints, andbby blue points (for these inequalities to hold we might have to permute thecolors).Consider a sphere of radiusrcentered at a red point. Its surface has green and bluepoints only. Sinceg, b r, the surface of the sphere must contain both green and bluepoints. ChooseMa green point on the sphere. There exist two pointsPandQon thesphere such thatMP=MQ=gandPQ=b.

6 So on the one hand, eitherPorQisgreen, or elsePandQare both blue. Then either there exist two green points at distanceg, namelyMandP,orQ, or there exist two blue points at distanceb. This contradictsthe initial assumption. The conclusion follows.(German mathematical olympiad , 1985) by contradiction, let us assume that the area of the overlap of any two sur-faces is less than19. In this case, ifS1,S2,..,Sndenote the nine surfaces, then the areaofS1 S2is greater than 1+89, the area ofS1 S2 S3is greater than 1+89+79,..,and the area ofS1 S2 S9is greater than1+89+79+ +19=459=5,a contradiction.

7 Hence the conclusion.(L. Panaitopol, D. Serb anescu,Probleme de Teoria Numerelor si Combinatoricapentru Juniori(Problems in Number Theory and Combinatorics for Juniors), GIL, 2003) that such anfexists. We focus on some particular values of the variable. Letf(0)=aandf(5)=b,a, b {1,2,3},a =b. Because|5 2|=3,|2 0|=2, wehavef(2) =a, b,sof(2)is the remaining number, sayc. Finally, because|3 0|=3,|3 5|=2, we must havef(3)=c. Therefore,f(2)=f(3). Translating the argumentto an arbitrary numberxinstead of 0, we obtainf(x+2)=f(x+3), and this violates the condition from the definition.

8 It follows that such a function doesnot by contradiction, let us assume that such a function exists. Setf(3)= the inequality 23<32, we obtainMethods of Proof32533=f(2)3=f(23)<f(32)=f(3)2=k2,hencek>5. Similarly, using 33<25, we obtaink3=f(3)3=f(33)<f(25)=f(2)5=35=243<343= implies thatk<7, and consequentlykcan be equal only to 6. Thus we shouldhavef(2)=3 andf(3)=6. The monotonicity offimplies that 2u<3vif and onlyif 3u<6v,u, vbeing positive integers. Taking logarithms this means thatvu>log23ifand only ifvu>log36. Since rationals are dense, it follows that log23=log36.

9 Thiscan be written as log23=1log23+1, and so log23 is the positive solution of the quadraticequationx2 x 1=0, which is the golden ratio1+ 52. The equality21+ 52=3translates to 21+ 5=9. But this would imply65536=25 <25(1+ 5)=95= have reached a contradiction, which proves that the functionfcannot exist.( Venkatachala,Functional Equations: A Problem Solving Approach, PrismBooks PVT Ltd., Bangalore, 2002) constant functionf(x)=k, wherekis a positive integer, is the only possiblesolution. That any such function satisfies the given condition is easy to suppose there exists a nonconstant solutionf.

10 There must exist two positiveintegersaandbsuch thatf(a)<f(b). This implies that(a+b)f(a) < af(b)+bf ( a ) <(a+b)f (b), which by the given condition is equivalent to(a+b)f(a)<(a+b)f (a2+b2)<(a+b)f (b). We can divide bya+b>0 to find thatf(a)<f(a2+b2)<f(b).Thus between any two different values offwe can insert another. But this cannot go onforever, sinceftakes only integer values. The contradiction shows that such a functioncannot exist. Thus constant functions are the only SOLUTIONS .(Canadian mathematical olympiad , 2002) thatA,B, andasatisfyA B=[0,1],A B= ,B=A+a.