JBMO 2002

The triangle $ABC$ has $CA = CB$ . $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$ ). $D$ is the foot of the perpendicular from $C$ to $PB$ . Show that $PA + PB = 2 \cdot PD$ .

Two circles with centers $O_{1}$ and $O_{2}$ meet at two points $A$ and $B$ such that the centers of the circles are on opposite sides of the line $AB$ . The lines $BO_{1}$ and $BO_{2}$ meet their respective circles again at $B_{1}$ and $B_{2}$ . Let $M$ be the midpoint of $B_{1}B_{2}$ . Let $M_{1}$ , $M_{2}$ be points on the circles of centers $O_{1}$ and $O_{2}$ respectively, such that $\angle AO_{1}M_{1}= \angle AO_{2}M_{2}$ , and $B_{1}$ lies on the minor arc $AM_{1}$ while $B$ lies on the minor arc $AM_{2}$ . Show that $\angle MM_{1}B = \angle MM_{2}B$ .

Ciprus

Find all positive integers which have exactly 16 positive divisors $1 = d_1 < d_2 < \ldots < d_{16} =n$ such that the divisor $d_k$ , where $k = d_5$ , equals $(d_2 + d_4) d_6$ .

Prove that for all positive real numbers $a,b,c$ the following inequality takes place $$\frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2} .$$ Laurentiu Panaitopol, Romania