AT-OEMO-REG 2002

Find the smallest natural number $x> 0$ so that all following fractions are simplified $\frac{3x+9}{8},\frac{3x+10}{9},\frac{3x+11}{10},...,\frac{3x+49}{48}$ , i.e. numerators and denominators are relatively prime.

Solve the following system of equations over the real numbers: $2x_1 = x_5 ^2 - 23$ $4x_2 = x_1 ^2 + 7$ $6x_3 = x_2 ^2 + 14$ $8x_4 = x_3 ^2 + 23$ $10x_5 = x_4 ^2 + 34$

In the convex $ABCDEF$ (has all interior angles less than $180^o$ ) with the perimeter $s$ the triangles $ACE$ and $BDF$ have perimeters $u$ and $v$ respectively. a) Show the inequalities $\frac{1}{2} \le \frac{s}{u+v}\le 1$ b) Check whether $1$ is replaced by a smaller number or $1/2$ by a larger number can the inequality remains valid for all convex hexagons.

Let $a_0, a_1, ..., a_{2002}$ be real numbers. a) Show that the smallest of the values $a_k (1-a_{2002-k})$ ( $0 \le k \le 2002$ ) the following applies: it is smaller or equal to $1/4$ . b) Does this statement always apply to the smallest of the values $a_k (1-a_{2003-k})$ ( $1 \le k \le 2002$ ) ? c) Show for positive real numbers $a_0, a_1, ..., a_{2002}$ : the smallest of the values $a_k (1-a_{2003-k})$ ( $1 \le k \le 2002$ ) is less than or equal to $1/4$ .