HMMT February

Let the centers of $\omega_{1}$ and $\omega_{2}$ be $O_{1}$ and $O_{2}$. Let $DE$ intersect $\omega_{1}$ again at $Q$, and let $DF$ intersect $\omega_{2}$ again at $R$. Note that since $\omega_{1}$ and $\omega_{2}$ must be tangent to $AD$ at the same point (by equal tangents), so $AD$ must be the radical axis of $\omega_{1}$ and $\omega_{2}$, so $RQEF$ is cyclic. Thus, we have $$\angle O_{1}QR=\angle EQR-\angle O_{1}QE=180^{\circ}-\angle EFD-\angle O_{1}EQ=90^{\circ}$$ Thus, we have $QR$ is tangent to $\omega_{1}$, and similarly it must be tangent to $\omega_{2}$ as well. Now, note that by Monge's theorem on $\omega, \omega_{1}$, and $\omega_{2}$, we have that $P$ must be the intersection of the external tangents of $\omega_{1}$ and $\omega_{2}$. Since $RQ$ is an external tangent, we have $P, Q$, and $R$ are collinear. Thus, by power of a point, we have $PE \cdot PF=PR \cdot PQ$. Note that $PR=10 \sqrt{30}$ and $PQ=12 \sqrt{30}$. Thus, we have $PE \cdot PF=3600$.