The vertices of a $5$-$12$-$13$ right triangle are the centers of three mutually externally tangent circles. What is the sum of the areas of the three circles?
Let the radii of the circles centered at $A,B,C$ be $r_a,r_b,r_c$. Since the circles are mutually externally tangent, the distance between two centers is the sum of their radii: $r_a+r_b=5$, $r_a+r_c=12$, $r_b+r_c=13$.
Adding, $2(r_a+r_b+r_c)=30$, so $r_a+r_b+r_c=15$. Then $r_c=15-5=10$, $r_b=15-12=3$, $r_a=15-13=2$.
The sum of the areas is $(2^2+3^2+10^2)\pi=\boxed{113\pi}$.
