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Switzerland — Final Round

Geometry Difficulty 5.8 AIME, harder Prove it Switzerland

Problem:

Let $ABC$ be an acute triangle with $AB = AC$ and let $D$ be a point on the side $BC$ . The circle with centre $D$ passing through $C$ intersects the circumcircle of $ABD$ in $P$ and $Q$ , where $Q$ is the point closer to $B$ . The line $BQ$ intersects $AD$ in $X$ and $AC$ in $Y$ . Prove that $PDXY$ is cyclic.

Solution

Solution:

We first claim that $P \in AC$ . Indeed, let $P'$ be the second intersection between the circle centered at $D$ and $AC$ . Then
$\angle ABD = \angle ABC = \angle ACB = \angle P'CD = 180^\circ - \angle AP'D$
so that $ABDP'$ is cyclic. This implies that $P = P'$ , in particular $P \in AC$ .

As $DP = DQ$ , the arcs $DP$ and $DQ$ subtend angles of same measure on the circle $(APDQB)$ , so that $\angle QBD = \angle PAD$ . Hence, $\triangle DBX$ and $\triangle DAC$ are similar (alternatively $ABXC$ is cyclic), implying that $\angle DXB = \angle DCA$ .

This concludes the problem, as we wanted to prove that
$\angle DPY = \angle DPC = \angle DCP = \angle DCA = \angle DXB = 180^\circ - \angle DXY$

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