CONO-SUR 1992

Find a positive integrer number $n$ such that, if yor put a number $2$ on the left and a number $1$ on the right, the new number is equal to $33n$ .

Let $P$ be a point outside the circle $C$ . Find two points $Q$ and $R$ on the circle, such that $P,Q$ and $R$ are collinear and $Q$ is the midpopint of the segmenet $PR$ . (Discuss the number of solutions).

Consider the set $S$ of $100$ numbers: $1; \frac{1}{2}; \frac{1}{3}; ... ; \frac{1}{100}$ . Any two numbers, $a$ and $b$ , are eliminated in $S$ , and the number $a+b+ab$ is added. Now, there are $99$ numbers on $S$ . After doing this operation $99$ times, there's only $1$ number on $S$ . What values can this number take?

Prove that there aren't any positive integrer numbers $x,y,z$ such that $x^2+y^2=3z^2$ .

In a $\triangle {ABC}$ , consider a point $E$ in $BC$ such that $AE \perp BC$ . Prove that $AE=\frac{bc}{2r}$ , where $r$ is the radio of the circle circumscripte, $b=AC$ and $c=AB$ .

Consider a $m*n$ board. On each box there's a non-negative integrer number assigned. An operation consists on choosing any two boxes with $1$ side in common, and add to this $2$ numbers the same integrer number (it can be negative), so that both results are non-negatives. What conditions must be satisfied initially on the assignment of the boxes, in order to have, after some operations, the number $0$ on every box?.