2006 AMC 12A Problem 13

Attempt Problem 13 of the 2006 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12A solutions, or check the answer key.

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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 these circles?

12\pi

\dfrac{25\pi}{2}

13\pi

\dfrac{27\pi}{2}

14\pi

The radii at two adjacent vertices sum to the side length between them

Solve

r + s = 3,\ r + t = 4,\ s + t = 5

r, s, t

are the radii at the vertices, then

r + s = 3,\ r + t = 4,\ s + t = 5.

Adding all three gives

r + s + t = 6,

r = 1,\ s = 2,\ t = 3.

The sum of the areas is

\pi(1^2 + 2^2 + 3^2) = 14\pi.

Thus, the correct answer is E.

Problem 13 in other years