AMC12 2006 12A, Problem 13

The vertices of a $3-4-5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?

Solution: Let the radius of the smallest circle be $r_A$, the radius of the second largest circle be $r_B$, and the radius of the largest circle be $r_C$. \[r_A + r_B = 3\] \[r_A + r_C = 4\] \[r_ B + r_C = 5\] Adding up all these equations and then dividing both sides by 2, we get, \[r_A + r_B + r_C = 6\] Then, we get $r_A = 1$, $r_B = 2$, and $r_C = 3$ Then we get $1^2 \pi + 2^2 \pi + 3^2 \pi = 14 \pi \iff$ $\fbox{14\pi}$.

Answer: 14\pi