2006 AMC 12A Problem 13

Below is the professionally curated solution for Problem 13 of the 2006 AMC 12A, 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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Concepts:tangent circles system of equations circle area

Difficulty rating: 1330

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?

$12\pi$

$\dfrac{25\pi}{2}$

$13\pi$

$\dfrac{27\pi}{2}$

$14\pi$

Solution:

If $r, s, t$ are the radii at the vertices, then $r + s = 3,\ r + t = 4,\ s + t = 5.$ Adding all three gives $r + s + t = 6,$ so $r = 1,\ s = 2,\ t = 3.$

The sum of the areas is $\pi(1^2 + 2^2 + 3^2) = 14\pi.$

Thus, the correct answer is E.

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