$\angle PBF = \angle ACF$ , $\triangle FPB \sim \triangle FQC$. By Gliding principle $\triangle FNM$ is also similar to them. So $\angle FNM = \angle FPB = 90^\circ$. BTW, is there a page for this contest on AoPS?

$K $ is projection of point $F $ on $AB $ and $AC $. $P, Q $ and $K $ colinear, by Simson line. $PAQF $-cyclic. $\angle FPQ=\angle FAQ=\angle FBC$, $\angle BFC=\angle BAC=\angle PFQ$ $\implies$ $\triangle PFQ \sim \triangle BFC $ and $FN $, $FM $ are median of $\triangle PFQ$, $\triangle BFC $ , respectively. Since, $\angle FNK= \angle FMK $, so $FNMK $-cyclic. Therefore, $\angle FNM=\angle FKM=90^{\circ}$

zuss77 wrote: ...BTW, is there a page for this contest on AoPS? zuss77, yes there is. I will soon add more editions from previous years to AOPS. It's under National and Regional Contests -> Bulgaria Contests -> Int'l Fest. of Young Mathematicians, Sozopol. The competition is held in Bulgaria annually and it is also open for international participation.

Pinko wrote: Point $F$ lies on the circumscribed circle around $\Delta ABC$, $P$ and $Q$ are projections of point $F$ on $AB$ and $AC$ respectively. Prove that, if $M$ and $N$ are the middle points of $BC$ and $PQ$ respectively, then $MN$ is perpendicular to $FN$. Baltic Way 2007