Harvard-MIT Mathematics Tournament

Let $C$ be the fourth vertex of parallelogram $APCQ$. The midpoint $M$ of $\overline{PQ}$ is the intersection of the diagonals of this parallelogram. Because $M$ has equal power with respect to the two circles $\omega_{1}$ and $\omega_{2}$, it lies on $\overleftrightarrow{AB}$, the circles' radical axis. Therefore, $C$ lies on $\overleftrightarrow{AB}$ as well.

Using a series of parallel lines and inscribed arcs, we have:

$$\angle APC = \angle APQ + \angle CPQ = \angle APQ + \angle PQA = \angle ABP + \angle QBA = \angle PBQ = \angle XPB$$

where the last equality follows from the fact that $PX \parallel QB$.

We also know that $\angle BXP = 180^{\circ} - \angle PAB = \angle CAP$, so triangles $BXP$ and $CAP$ are similar. By the spiral similarity theorem, triangles $BPC$ and $XPA$ are similar, too.

By analogous reasoning, triangles $BQC$ and $YQA$ are similar. Then we have:

$$\frac{AX}{AY} = \frac{AX / BC}{AY / BC} = \frac{AP / CP}{AQ / CQ} = \frac{AP^{2}}{AQ^{2}}$$

where the last equality holds because $APCQ$ is a parallelogram. Using the Law of Sines, the last expression equals $\frac{\sin^{2} 15^{\circ}}{\sin^{2} 30^{\circ}} = 2 - \sqrt{3}$.