Auswahlwettbewerb zur Internationalen Mathematik-Olympiade

Solution:

We denote the perpendicular bisector of the segment $PQ$ by $m$ and the circle $QPCF$ by $k$. Because $DP \perp BC$ and $DQ \perp AC$, $D$ also lies on $k$, and $CD$ is in fact a diameter of $k$. The segments $QE$ and $PE$ are symmetric with respect to $m$; at the same time $m$ is also an axis of symmetry of $k$. Therefore the chords $CQ$ and $FP$ are symmetric with respect to $m$, hence so are $C$ and $F$ (by assumption $P, E$ and $F$ are collinear!). Thus the perpendicular bisector of $CF$ coincides with $m$; it follows that $m$ passes through the circumcenter $O$ of $ABC$.
Now we consider the midpoint $M$ of the segment $AB$. Because $CM \perp DM$, $M$ also lies

on $k$. Because $\angle ACM=\angle MCB$, it follows from the inscribed angle theorem that the chords $MP$ and $MQ$ of $k$ are equal in length. This shows, however, that $m$ passes through $M$. Since both $O$ and $M$ lie on $m$ and on $CM$, it follows that $O=M$ and hence $\angle ACB=90^{\circ}$.

Solution:

As carried out in the first solution, we establish that the points $P, C, F, Q, D$ and the midpoint $M$ of $AB$ lie on the circle $k$ with diameter $CD$, and that the perpendicular bisector $m$ of $PQ$ passes through $M$.
Let $G$ be the intersection point of $BF$ and $CM$, and let us denote $\alpha=\angle FBA$. Because $E$ lies on $m$, because $P, E$ and $F$ are collinear,

and because $CFAB$ and $CFQP$ are cyclic quadrilaterals, it follows that $\angle PQC=\angle PQE=\angle EPQ=\angle FPQ=\angle FCQ=\angle FCA=\angle FBA=\alpha$.
Furthermore $\angle FEM=\angle FEQ+\angle QEM=2\alpha+(90^{\circ}-\alpha)=90^{\circ}+\alpha$ holds. Likewise, by exterior angles, $\angle FGM=90^{\circ}+\alpha$ holds; therefore $MGEF$ is a cyclic quadrilateral.
It follows that $\angle CGE=\angle MFE=\angle MFP=\angle MCP$. Hence $GE \parallel BC$, and thus we obtain $\angle EAF=\angle CAF=\angle CBF=\angle EGF$, so that $GEFA$ is likewise a cyclic quadrilateral. From this it follows that $\angle ACB=\angle AFB=\angle AFG=180^{\circ}-\angle GMA=90^{\circ}$, which was to be shown.