Competition problem — Geometry — Maths Olympiad Prep

Library / /6 of 25

Geometry Difficulty 5.5 AIME, harder Prove it Russia

A quadrilateral $ABCD$ is inscribed into a circle $\Gamma$ centered at $O$ . Its diagonals $AC$ and $BD$ are perpendicular to each other; let $P$ be their meeting point (the point $O$ lies inside the triangle $BPC$ ). A point $H$ is chosen on the segment $BO$ so that $\angle BHP = 90^\circ$ . The circumcircle $\omega$ of the triangle $PHD$ meets again the segment $PC$ at $Q$ . Prove that $AP = CQ$ .

Solution

Let $BT$ be a diameter of $\Omega$ ; then the points $P$ , $H$ , $T$ , and $D$ are concyclic (see Fig. 9). Thus $\angle PQD = 90^\circ$ , and the perpendicular bisectors to $PQ$ , $TD$ , and $AC$ coincide.

Consider the common perpendicular bisector $l$ to the segments $DT$ and $PQ$ . It passes through $O$ , and therefore is also the perpendicular bisector to $AC$ . Thus, the segments $AP$ and $CQ$ are symmetric with respect to $l$ and hence are equal.

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