VN-TST 1991

1.) In the plane let us consider a set $S$ consisting of $n \geq 3$ distinct points satisfying the following three conditions:

I. The distance between any two points $\in S$ is not greater than 1. II. For every point $A \in S$ , there are exactly two "neighbor" points, i.e. two points $X, Y \in S$ for which $AX = AY = 1$ . III. For arbitrary two points $A, B \in S$ , let $A', A''$ be the two neighbors of $A, B', B''$ the two neighbors of $B$ , then $A'AA'' = B'BB''$ .

Is there such a set $S$ if $n = 1991$ ? If $n = 2000$ ? Explain your answer.

For a positive integer $n>2$ , let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be a sequence of $n$ positive reals which is either non-decreasing (this means, we have $a_{1}\leq a_{2}\leq \ldots \leq a_{n}$ ) or non-increasing (this means, we have $a_{1}\geq a_{2}\geq \ldots \geq a_{n}$ ), and which satisfies $a_{1}\neq a_{n}$ . Let $x$ and $y$ be positive reals satisfying $\frac{x}{y}\geq \frac{a_{1}-a_{2}}{a_{1}-a_{n}}$ . Show that: $$\frac{a_{1}}{a_{2}\cdot x+a_{3}\cdot y}+\frac{a_{2}}{a_{3}\cdot x+a_{4}\cdot y}+\ldots+\frac{a_{n-1}}{a_{n}\cdot x+a_{1}\cdot y}+\frac{a_{n}}{a_{1}\cdot x+a_{2}\cdot y}\geq \frac{n}{x+y}.$$

Let $\{x\}$ be a sequence of positive reals $x_1, x_2, \ldots, x_n$ , defined by: $x_1 = 1, x_2 = 9, x_3=9, x_4=1$ . And for $n \geq 1$ we have:

$$x_{n+4} = \sqrt[4]{x_{n} \cdot x_{n+1} \cdot x_{n+2} \cdot x_{n+3}}.$$

Show that this sequence has a finite limit. Determine this limit.

Let $T$ be an arbitrary tetrahedron satisfying the following conditions:

I. Each its side has length not greater than 1, II. Each of its faces is a right triangle.

Let $s(T) = S^2_{ABC} + S^2_{BCD} + S^2_{CDA} + S^2_{DAB}$ . Find the maximal possible value of $s(T)$ .

For every natural number $n$ we define $f(n)$ by the following rule: $f(1) = 1$ and for $n>1$ then $f(n) = 1 + a_1 \cdot p_1 + \ldots + a_k \cdot p_k$ , where $n = p_1^{a_1} \cdots p_k^{a_k}$ is the canonical prime factorisation of $n$ ( $p_1, \ldots, p_k$ are distinct primes and $a_1, \ldots, a_k$ are positive integers). For every positive integer $s$ , let $f_s(n) = f(f(\ldots f(n))\ldots)$ , where on the right hand side there are exactly $s$ symbols $f$ . Show that for every given natural number $a$ , there is a natural number $s_0$ such that for all $s > s_0$ , the sum $f_s(a) + f_{s-1}(a)$ does not depend on $s$ .

Let a set $X$ be given which consists of $2 \cdot n$ distinct real numbers ( $n \geq 3$ ). Consider a set $K$ consisting of some pairs $(x, y)$ of distinct numbers $x, y \in X$ , satisfying the two conditions:

I. If $(x, y) \in K$ then $(y, x) \not \in K$ . II. Every number $x \in X$ belongs to at most 19 pairs of $K$ .

Show that we can divide the set $X$ into 5 non-empty disjoint sets $X_1, X_2, X_3, X_4, X_5$ in such a way that for each $i = 1, 2, 3, 4, 5$ the number of pairs $(x, y) \in K$ where $x, y$ both belong to $X_i$ is not greater than $3 \cdot n$ .