VN-TST 1996

In the plane we are given $3 \cdot n$ points ( $n>$ 1) no three collinear, and the distance between any two of them is $\leq 1$ . Prove that we can construct $n$ pairwise disjoint triangles such that: The vertex set of these triangles are exactly the given 3n points and the sum of the area of these triangles $< 1/2$ .

For each positive integer $n$ , let $f(n)$ be the maximal natural number such that: $2^{f(n)}$ divides $\sum^{\left\lfloor \frac{n - 1}{2}\right\rfloor}_{i=0} \binom{n}{2 \cdot i + 1} 3^i$ . Find all $n$ such that $f(n) = 1996.$

Find the minimum value of the expression:

$$f(a,b,c)= (a+b)^4+(b+c)^4+(c+a)^4 - \frac{4}{7} \cdot (a^4+b^4+c^4).$$

Given 3 non-collinear points $A,B,C$ . For each point $M$ in the plane ( $ABC$ ) let $M_1$ be the point symmetric to $M$ with respect to $AB$ , $M_2$ be the point symmetric to $M_1$ with respect to $BC$ and $M'$ be the point symmetric to $M_2$ with respect to $AC$ . Find all points $M$ such that $MM'$ obtains its minimum. Let this minimum value be $d$ . Prove that $d$ does not depend on the order of the axes of symmetry we chose (we have 3 available axes, that is $BC$ , $CA$ , $AB$ . In the first part the order of axes we chose $AB$ , $BC$ , $CA$ , and the second part of the problem states that the value $d$ doesn't depend on this order).

There are some people in a meeting; each doesn't know at least 56 others, and for any pair, there exist a third one who knows both of them. Can the number of people be 65?

Find all reals $a$ such that the sequence $\{x(n)\}$ , $n=0,1,2, \ldots$ that satisfy: $x(0)=1996$ and $x_{n+1} = \frac{a}{1+x(n)^2}$ for any natural number $n$ has a limit as n goes to infinity.