2006 AMC 12A Problems/Problem 13

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Problem

2006 AMC 12A Problem 13.gif

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 $a$. We find that the radius of the largest circle is $4-a$ and the radius of the second largest circle is $3-a$. Thus, $4-a+3-a=5\iff a=1$. The radii of the other circles are $3$ and $2$. The sum of their areas is $\pi+9\pi+4\pi=14\pi\iff\mathrm{(E)}$

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 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

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