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Harvard-MIT Mathematics Tournament

Geometry Difficulty 6.0 AIME, harder Prove it United States

Problem:

Let $\omega_{1}$ and $\omega_{2}$ be two circles that intersect at points $A$ and $B$ . Let line $l$ be tangent to $\omega_{1}$ at $P$ and to $\omega_{2}$ at $Q$ so that $A$ is closer to $PQ$ than $B$ . Let points $R$ and $S$ lie along rays $PA$ and $QA$ , respectively, so that $PQ = AR = AS$ and $R$ and $S$ are on opposite sides of $A$ as $P$ and $Q$ . Let $O$ be the circumcenter of triangle $ASR$ , and let $C$ and $D$ be the midpoints of major arcs $AP$ and $AQ$ , respectively. If $\angle APQ$ is $45$ degrees and $\angle AQP$ is $30$ degrees, determine $\angle COD$ in degrees.

Solution

We use directed angles throughout the solution.
Let $T$ denote the point such that $\angle TCD = \frac{1}{2} \angle APQ$ and $\angle TDC = \frac{1}{2} \angle AQP$ . We claim that $T$ is the circumcenter of triangle $SAR$ .
Since $CP = CA$ , $QP = RA$ , and $\angle CPQ = \angle CPA + \angle APQ = \angle CPA + \angle ACP = \angle CAR$ , we have $\triangle CPQ \cong \triangle CAR$ . By spiral similarity, we have $\triangle CPA \sim \triangle CQR$ .
Let $T'$ denote the reflection of $T$ across $CD$ . Since $\angle TCT' = \angle APQ = \angle ACP$ , we have $\triangle TCT' \sim \triangle ACP \sim \triangle RCQ$ . Again, by spiral similarity centered at $C$ , we have $\triangle CTR \sim \triangle CT'Q$ . But $CT = CT'$ , so $\triangle CTR \cong \triangle CT'Q$ and $TR = T'Q$ . Similarly, $\triangle DTT' \sim \triangle DAQ$ , and spiral similarity centered at $D$ shows that $\triangle DTA \cong \triangle DT'Q$ . Thus $TA = T'Q = TR$ .
We similarly have $TA = T'P = TS$ , so $T$ is indeed the circumcenter. Therefore, we have $\angle COD = \angle CTD = 180^\circ - \frac{45^\circ}{2} - \frac{30^\circ}{2} = 142.5^\circ$ .

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