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Berkeley Math Circle Monthly Contest 4

Geometry Difficulty 4.8 AIME Prove it United States

Problem:
Let $AB$ and $CD$ be two nonperpendicular diameters of a circle centered at $O$ , and let $Q$ be the reflection of $D$ about $AB$ . The tangent at $B$ meets $AC$ at $P$ , and $DP$ meets the circle again at $E$ . Prove that lines $AE$ , $BP$ , and $CQ$ are concurrent.

Solution

Let $X$ be the intersection of $CQ$ and $BP$ . We first note that $CQ \parallel AB$ since
$\angle CQA = \angle CBA = \angle BAD = \angle BAQ$
Note that $\triangle CPX \sim \triangle CDA$ since the angles at $X$ and $A$ are right and
$\angle CPX = 90 - \angle CAB = \angle CBA = \angle BDA \text{.}$
So $\triangle CPX \sim \triangle CDA$ ; rearranging the known facts
$\frac{CP}{CX} = \frac{CD}{CA} \quad \text{and} \quad \angle PCX = \angle DCA$
yields $\triangle CPD \sim \triangle CXA$ . In particular, $\angle CAX = \angle CDP = \angle CAE$ , so $A, E, X$ are collinear, as desired.

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