10. Triangle $A B C$ has side lengths $A B=65, B C=33$, and $A C=56$. Find the radius of the circle tangent to sides $A C$ and $B C$ and to the circumcircle of triangle $A B C$.
Answer: 24
Solution: Let $\Gamma$ be the circumcircle of triangle $A B C$, and let $E$ be the center of the circle tangent to $\Gamma$ and the sides $A C$ and $B C$. Notice that $\angle C=90^{\circ}$ because $33^{2}+56^{2}=65^{2}$. Let $D$ be the second intersection of line $C E$ with $\Gamma$, so that $D$ is the midpoint of the arc $A B$ away from $C$. Because $\angle B C D=45^{\circ}$, one can easily calculate $C D=89 \sqrt{2} / 2$. The power of $E$ with respect to $\Gamma$ is both $r(65-r)$ and $r \sqrt{2} \cdot(89 \sqrt{2} / 2-r \sqrt{2})=r(89-2 r)$, so $r=89-65=24$.