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Berkeley Math Circle Monthly Contest 8

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with $\angle ACB = 90^{\circ}$ . The inscribed circle of $\triangle ABC$ touches sides $AC$ and $BC$ at $D$ and $E$ , respectively. On the circumscribed circle of $\triangle ABC$ , the midpoints of minor arcs $AC$ and $BC$ are respectively $P$ and $Q$ . Prove that $D$ , $E$ , $P$ , and $Q$ are all collinear.

Solution

Solution:

Let $M$ be the midpoint of $AC$ , let $O$ be the circumcenter of $\triangle ABC$ , and let $F$ be the point where the incircle touches $AB$ .

Note that $\triangle CDE$ is a right isosceles triangle and therefore $\angle CDE = 45^{\circ}$ . Also, $\angle PMD$ is right since $OM \perp AC$ and $OM$ passes through $P$ . Let us prove that $\triangle PMD$ is right isosceles:
$PM = OP - OM = \frac{AB - BC}{2} = \frac{AF + FB - BE - EC}{2} = \frac{AD - DC}{2} = \frac{AC - 2DC}{2} = MC - DC = MD.$
Therefore $\angle PDM = 45^{\circ}$ and
$\angle PDM + \angle MDE = \angle PDM + 180^{\circ} - \angle CDE = 45^{\circ} + 180^{\circ} - 45^{\circ} = 180^{\circ}.$
This proves that $P$ lies on line $DE$ . Similarly, we can prove that $Q$ lies on line $DE$ .

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