again a romanian problem, taken from the isl 2003. boy, we used a french problem last year in our tst-s, and now you guys are paying us back by using two romanian problems in this year's tst-s? :P so nice ..

Note that, at least one of the contestants ('Raf' on mathlinks) has solved it with the improved bound of 124 instead of 96 (as noticed by Darij) and a very nice and easy solution. I'll give it if he doesn't. Pierre.

I belive that raf should have valsed through this imo tst, since I already solved 4 questions in less than 20 minutes.

He seems to have a made a minor error in the pb 3 but he will probably score around 40-42/42. He said he spent many times to find his error because he cannot believe to have solve the problem so easily with 124 instead of 96. Maybe, it was the most difficult part for him , everyone has its own problems.... Pierre.

Valentin, Raphal went to the IMO last year, do you remember him by any chance?

well I remember the whole french team, cuz I kept seeing them on the airports on my way on and back (bucharest -> charles de gaule -> tokio and back) but I couldn't pin point him by name

So, my solution for the second part of the problem (first part is very easy) Let ABC be an equilateral triangle with A,B end C in differents disks. Let O1 be the center of the disk wich contain A. O2 th center of the disk wich contain B and O3 the center of the disk wich contain C. Lemma: There isn't an equilateral triangle PQR with P,Q and R at integers points. Proof: We can assume that P(0,0) Let be Q(u,v) and R(u',v') the coordinates of Q and R. If 2 divides u,v,u' and v' points Q'(u/2,v/2) and R'(u'/2,v'/2) are such that PQ'R' is equilateral too. So we can assume that u is odd for example. Then you look at the two cases of v modulo 2, you get a contradiction with: u^2+v^2=u'^2+v'^2=(u-u')^2+(v-v')^2 So the lemma is true. Now by triangle inequality: O1O2-O1A-O2B=<AB=<O1O2+O1A+O2B Let a=Ab=AC=BC Moreover we have O1O2-O1A-O2B>=O1O2-2/1000 and O1O2+O1A+O2B=<O1O2+2/1000 So we get a-2/1000=<O1O2=<a+2/1000 We get with the sames arguments: a-2/1000=<O1O2,O2O3,O1O3=<a+2/1000 So O1O2^2,O2O3^2 and O1O3^2 are all in [(a-2/1000)^2,(a+2/1000)^2] But at least two of O1O2^2,O2O3^2 and O1O3^2 are different because of the lemma and there are all integers. But the size of the interval [(a-2/1000)^2,(a+2/1000)^2] is 8a/1000 and contain at least two integers differents so we have 8a/1000>=1 so a>=125.

i remember a long hair boy of the romanian team and I think there is a boy of your team at Louis le Grand now in France, no?

yes Manolescu is following classe preparatoire and you probably remember our dear Kappa which you will probably meet again in Greece

This problem was used in one of the Swedish correspondence tests. For the first part there were two different official solutions: one using Kroenecker's Approximation Theorem with complex numbers, and the other using Pigeonhole together with some other stuff. Wheras the second part, the prove was by contradiction. It is quite strange how the same question can appear on almost every Team TST in Europe

... and I have used the "well-known theorem of Dedekind" (yeah, it is the Kronecker density theorem, but not for complex numbers)... Darij