Problem (Geometry):
In right triangle $A B C$ with right angle $C$, line $\ell$ is drawn through $C$ and is parallel to $\overline{A B}$. Points $P$ and $Q$ lie on $\overline{A B}$ with $P$ between $A$ and $Q$, and points $R$ and $S$ lie on $\ell$ with $C$ between $R$ and $S$ such that $P Q R S$ is a square. Let $\overline{P S}$ intersect $\overline{A C}$ in $X$, and let $\overline{Q R}$ intersect $\overline{B C}$ in $Y$. The inradius of triangle $A B C$ is 10 , and the area of square $P Q R S$ is 576 . Compute the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$.

Solution:
Note that in right triangle $A B C$ with right angle $C$, the inradius $r$ is equal to $\frac{a+b-c}{2}$, where $a=B C, b=A C$, and $c=A B$, because the inradius equals the distance from the vertex of the right angle $C$ to (either) point of tangency along $\overline{A C}$ or $\overline{B C}$. Thus the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$ is equal to one-half the difference between the sum of the lengths of the legs of these triangles and the sum of the lengths of the hypotenuses of these triangles. Let $t$ be the side length of square $P Q R S$. Then the sum of the lengths of the legs of triangles $A X P, C X S, C Y R$, and $B Y Q$ is

$$
\begin{aligned}
& A P+P X+X S+S C+C R+R Y+Y Q+Q B \\
= & A P+P S+S R+R Q+Q B \\
= & A P+t+t+t+Q B \\
= & A B-P Q+3 t \\
= & c-t+3 t \\
= & c+2 t .
\end{aligned}
$$



The sum of the lengths of the hypotenuses of triangles $A X P, C X S, C Y R$, and $B Y Q$ is $A X+X C+C Y+Y B=A C+C B=b+a$. Hence the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$ is $\frac{c+2 t-(a+b)}{2}=t-r$. Thus the desired sum equals $\sqrt{576}-10=24-10=\mathbf{1 4}$.

Final answer: 14