10. Let $A B C$ be a triangle with $A B=13, B C=14$, and $C A=15$. Let $\Gamma$ be the circumcircle of $A B C$, let $O$ be its circumcenter, and let $M$ be the midpoint of minor arc $\widehat{B C}$. Circle $\omega_{1}$ is internally tangent to $\Gamma$ at $A$, and circle $\omega_{2}$, centered at $M$, is externally tangent to $\omega_{1}$ at a point $T$. Ray $A T$ meets segment $B C$ at point $S$, such that $B S-C S=4 / 15$. Find the radius of $\omega_{2}$.
Answer: $1235 / 108$ Let $N$ be the midpoint of $B C$. Notice that $B S-C S=\frac{4}{15}$ means that $N S=\frac{2}{15}$. let lines $M N$ and $A S$ meet at $P$, and let $D$ be the foot of the altitude from $A$ to $B C$. Then $B D=5$ and $A D=12$, so $D N=2$ and $D S=\frac{32}{15}$. Thus $N P=A D \frac{S N}{S D}=12 \frac{2 / 15}{32 / 15}=\frac{3}{4}$. Now $O B=R=\frac{a b c}{4 A}=\frac{(13)(14)(15)}{4(84)}=\frac{65}{8}$, so $O N=\sqrt{O B^{2}-B N^{2}}=\sqrt{\left(\frac{65}{8}\right)^{2}-7^{2}}=\frac{33}{8}$. Thus $O P=\frac{27}{8}$ and $P M=O M-O P=\frac{19}{4}$. By Monge's theorem, the exsimilicenter of $\omega_{1}$ and $\Gamma$ (which is $A$ ), the insimilicenter of $\omega_{1}$ and $\omega_{2}$ (which is $T$ ), and the insimilicenter of $\omega_{2}$ and $\Gamma$ (call this $P^{\prime}$ ) are collinear. But notice that this means $P^{\prime}=O M \cap A T=P$. From this we get
$$
\frac{\text { radius of } \omega_{2}}{R}=\frac{M P}{O P}=\frac{38}{27}
$$

Thus the radius of $\omega_{2}$ is $\frac{65}{8} \cdot \frac{38}{27}=\frac{1235}{108}$.