HMMT November

Solution 1: Denote the second circle by $\gamma$. Let $T$ and $r$ be the center and radius of $\gamma$, respectively, and let $X$ and $H$ be the tangency points of $\gamma$ with $\omega$ and $\overline{P Q}$, respectively. Let $O$ be the center of $\omega$, and let $M$ be the midpoint of $\overline{Q R}$. Note that $Q M=M R=\frac{1}{2} Q R=3$, so $\triangle P M Q$ and $\triangle P M R$ are 3-4-5 triangles. Since $\triangle Q H T \sim \triangle Q M P$ and $H T=r$, we get $Q T=\frac{5}{4} r$. Then $T M=Q M-Q T=3-\frac{5}{4} r$. By the extended law of sines, the circumradius of $\triangle P Q R$ is $O P=\frac{P R}{2 \sin \angle P Q R}=\frac{5}{2(4 / 5)}=\frac{25}{8}$, so $O M=M P-O P=4-\frac{25}{8}=\frac{7}{8}$. Also, we have $O T=O X-X T=\frac{25}{8}-r$. Therefore, by the Pythagorean theorem, $$\left(3-\frac{5}{4} r\right)^{2}+\left(\frac{7}{8}\right)^{2}=\left(\frac{25}{8}-r\right)^{2}$$ This simplifies to $\frac{9}{16} r^{2}-\frac{5}{4} r=0$, so $r=\frac{5}{4} \cdot \frac{16}{9}=\frac{20}{9}$.