Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Segment $P Q$ is tangent to $\omega_{1}$ at $P$ and to $\omega_{2}$ at $Q$, and $A$ is closer to $P Q$ than $B$. Point $X$ is on $\omega_{1}$ such that $P X \| Q B$, and point $Y$ is on $\omega_{2}$ such that $Q Y \| P B$. Given that $\angle A P Q=30^{\circ}$ and $\angle P Q A=15^{\circ}$, find the ratio $A X / A Y$.
Let $C$ be the fourth vertex of parallelogram $A P C Q$. The midpoint $M$ of $\overline{P Q}$ is the intersection of the diagonals of this parallelogram. Because $M$ has equal power ${ }^{3}$ with respect to the two circles $\omega_{1}$ and $\omega_{2}$, it lies on $\overleftrightarrow{A B}$, the circles' radical axi: $\stackrel{4}{4}$. Therefore, $C$ lies on $\overleftrightarrow{A B}$ as well. Using a series of parallel lines and inscribed arcs, we have: \[ \angle A P C=\angle A P Q+\angle C P Q=\angle A P Q+\angle P Q A=\angle A B P+\angle Q B A=\angle P B Q=\angle X P B \] where the last equality follows from the fact that $P X \| Q B$. We also know that $\angle B X P=180^{\circ}-\angle P A B=\angle C A P$, so triangles $B X P$ and $C A P$ are similar. By the spiral similarity theorem ${ }^{5}$ triangles $B P C$ and $X P A$ are similar, too. By analogous reasoning, triangles $B Q C$ and $Y Q A$ are similar. Then we have: \[ \frac{A X}{A Y}=\frac{A X / B C}{A Y / B C}=\frac{A P / C P}{A Q / C Q}=\frac{A P^{2}}{A Q^{2}} \] where the last inequality holds because $A P C Q$ is a parallelogram. Using the Law of Sines, the last expression equals $\frac{\sin ^{2} 15^{\circ}}{\sin ^{2} 30^{\circ}}=2-\sqrt{3}$. $\fbox{2-\sqrt{3}}$.