10. Triangle $A B C$ has incenter $I$. Let $D$ be the foot of the perpendicular from $A$ to side $B C$. Let $X$ be a point such that segment $A X$ is a diameter of the circumcircle of triangle $A B C$. Given that $I D=2$, $I A=3$, and $I X=4$, compute the inradius of triangle $A B C$.
Answer:
$\frac{11}{12}$
Solution: Let $R$ and $r$ be the circumradius and inradius of $A B C$, let $A I$ meet the circumcircle of $A B C$ again at $M$, and let $J$ be the $A$-excenter. We can show that $\triangle A I D \sim \triangle A X J$ (e.g. by $\sqrt{b c}$ inversion), and since $M$ is the midpoint of $I J$ and $\angle A M X=90^{\circ}, I X=X J$. Thus, we have $\frac{2 R}{I X}=\frac{X A}{X J}=\frac{I A}{I D}$, so $R=\frac{I X \cdot I A}{2 I D}=3$. But we also know $R^{2}-2 R r=I O^{2}=\frac{2 X I^{2}+2 A I^{2}-A X^{2}}{4}$. Thus, we have
$$
r=\frac{1}{2 R}\left(R^{2}-\frac{2 I X^{2}+2 I A^{2}-4 R^{2}}{4}\right)=\frac{11}{12} .
$$