Let $A,B,C$ and $D$ be points on a circle in this order. The chords $AC$ and $BD$ intersect in point $P$. The perpendicular to $AC$ through C and the perpendicular to $BD$ through $D$ intersect in point $Q$.
Prove that the lines $AB$ and $PQ$ are perpendicular.
Let $PQ\cap AB=X$. $\angle BAP=\angle PDC=\angle PQC$ and $\angle CPQ=\angle APX$; since $\angle CPQ+\angle PQC=90^\circ$, we are done.
Best regards,
sunken rock
Since $PCQD$ is cyclic with diameter $PQ$ we get $\measuredangle QPC=\measuredangle QDC=\alpha$. Furthermore since $ABCD$ is cyclic $\measuredangle BDC=\measuredangle CAB=90^{\circ}-\alpha$ $\Longrightarrow$ $\measuredangle CPQ+\measuredangle BAP=90^{\circ}$ hence $AB\perp PQ$
sunken rock wrote:
Let $PQ\cap AB=X$. $\angle BAP=\angle PDC=\angle PQC$ and $\angle CPQ=\angle APX$; since $\angle CPQ+\angle PQC=90^\circ$, we are done.
Best regards,
sunken rock
I suppose that's what most people had at the competition, as it is the easiest way. But there was one slight problem. Proving that $X$ exists. I believe there were no penalties given for people who did not do that, though.