Then color each point with the same colour... I think there is something missing... Pierre.

could you please post the complete test? if possible also in spanish just in case there some doubts about it.

Pierre, I think you misread.

Pierre: If we color each point with the same color then all the rectangles will be monocromatic. Please ask me if you have any doubt about the statement mszew: Do you speak spanish? The problem in spanish: Probar que es posible colorear los puntos latice del plano tal que no exista ningun rectangulo (con lados paralelos a los ejes coordenados y vertices en 4 puntos latice) con area igual a una potencia de 2 que tenga sus 4 vertices del mismo color

yes, I do! Could you be so kind to post the other problems?...

I swear I didn't see the 'not' before monochromatic Okay, so I have to give an answer : Let's color $ (x,y)$ red iff $ x+y \neq 1 \mod[3]$ and blue otherwise. In the following $ [..]$ means area. - Let $ B$ be a rectangle with all vertices blue, sides parallel to the grid lines. Let $ (a,b)$ and $ (c,d)$ be two of its vertices. Then $ a+b = a+d = 1 \mod[3]$ so that $ b-d = 0 \mod [3]$. Thus $ [B]$, which is a multiple of $ b-d$, is divisible by $ 3$, and so cannot be a power of $ 2$. - Let $ R$be a rectangle with all vertices red and sides parallel to the grid lines. Let $ X(a,b),Y(a,d),Z(c,d),T(c,b)$ be its vertices, with $ c>a$ and $ d>b$. Let $ x=a+b$. Assume, for a contradiction, that $ [R] = (c-a)(d-b)$ is a power of $ 2$. It follows that $ c=a+2^p$ and $ d=b+2^q$ for some non-negative integers $ p,q$. Moreover, since the four vertices are red, we have : $ x \neq 1 \mod[3]$ and $ x+(-1)^p \neq 1 \mod[3]$ and $ x+(-1)^q \neq 1 \mod[3]$ and $ x+(-1)^p + (-1)^q \neq 1 \mod[3]$. Thus, modulo $ 3$, we have $ \{x,x+(-1)^p \} = \{0,2 \} = \{x,x+(-1)^q \}$. It follows that $ x + (-1)^p = x + (-1)^q \mod [3]$. Therefore, $ p,q$ have the same parity. But, this forces $ x,x+(-1)^p,x+(-1)^p+(-1)^q$ to be pairwise distinct modulo $ 3$. In particular, one of them must be $ 1 \mod [3]$, a contradiction. Hence result. Pierre.

Oh my god!! I can't believe that. Yesterday in the olympiad, I think the problem in this way too (construct a coloration) and in 1½ hours I tried with mod 2^k, I wrote the co-ordinates in binary and I taked their 2-adic value and more and every failed. Thanks you very much Pierre. mszew: the first five problems are easy and it's posible too that they aren't original. However I will to post them here. First day: 1- We place the numbers from 1 to 81 in a 9x9 board. Prove that exist k in {1,2,...,9} so that the product of the numbers in the k-th column is diferent to the product of the numbers in the k-th row. 2- Let H a regular hexagon and let P a point in the plane of H. Let V(P) the sum of the distances from P to the vertices of H and let L(P) the sum of the distances from P to the edges of H. a)Find all points P so that L(P) is minimun b)Find all points P so that V(P) is minimun 3-A boy write three times the number n in a blackboard. He can make the followen move: He flock one of the numbers and write in its place the sum of the two others less 1. After several moves one of the three numbers in the blackboard is 900. Find all the posible values of n Second day: 1- a^2+a+b^2≥a^4+a^3+b^4 a,b≥0 Proof that (1-a^4)/a^2≥ (b^2-1)/b 2- Let ABC an acute-angle triangle. Let R a rectangle with vertices in the edges of ABC. Let O de center of R. a)Find the locus of all the points O b)Decide if there is a poit that is the center of three of this rectangles. 3- Prove that we can color the lattice points in the plane with two color so that every rectangle with vertices in the lattice points and edges parallels to the co-ordinate axis that have area 2^n is not monocromatic. mszew: Where are you from?

Prove that we can color the lattice points in the plane with two color so that every rectangle with vertices in the lattice points and edges parallels to the co-ordinate axis that have area $2^n$ is not monocromatic.

rest problems