# Difference between revisions of "2006 AMC 12A Problems/Problem 13"

## Problem

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? $\mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ } 14\pi$

## 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 $1^2 \pi + 2^2 \pi + 3^2 \pi = 14 \pi \iff\mathrm{(E)}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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