20. 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 $P Q$ than $B$. Let points $R$ and $S$ lie along rays $P A$ and $Q A$, respectively, so that $P Q=A R=A S$ and $R$ and $S$ are on opposite sides of $A$ as $P$ and $Q$. Let $O$ be the circumcenter of triangle $A S R$, and let $C$ and $D$ be the midpoints of major arcs $A P$ and $A Q$, respectively. If $\angle A P Q$ is 45 degrees and $\angle A Q P$ is 30 degrees, determine $\angle C O D$ in degrees.
Answer: 142.5
We use directed angles throughout the solution.
Let $T$ denote the point such that $\angle T C D=1 / 2 \angle A P Q$ and $\angle T D C=1 / 2 \angle A Q P$. We claim that $T$ is the circumcenter of triangle $S A R$.
Since $C P=C A, Q P=R A$, and $\angle C P Q=\angle C P A+\angle A P Q=\angle C P A+\angle A C P=\angle C A R$, we have $\triangle C P Q \cong \triangle C A R$. By spiral similarity, we have $\triangle C P A \sim \triangle C Q R$.
Let $T^{\prime}$ denote the reflection of $T$ across $C D$. Since $\angle T C T^{\prime}=\angle A P Q=\angle A C P$, we have $\triangle T C T^{\prime} \sim$ $\triangle A C P \sim \triangle R C Q$. Again, by spiral similarity centered at $C$, we have $\triangle C T R \sim \triangle C T^{\prime} Q$. But $C T=C T^{\prime}$, so $\triangle C T R \cong \triangle C T^{\prime} Q$ and $T R=T^{\prime} Q$. Similarly, $\triangle D T T^{\prime} \sim \triangle D A Q$, and spiral similarity centered at $D$ shows that $\triangle D T A \cong \triangle D T^{\prime} Q$. Thus $T A=T^{\prime} Q=T R$.
We similarly have $T A=T^{\prime} P=T S$, so $T$ is indeed the circumcenter. Therefore, we have $\angle C O D=$ $\angle C T D=180^{\circ}-\frac{45^{\circ}}{2}-\frac{30^{\circ}}{2}=142.5^{\circ}$.