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

## See also

 2006 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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

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