2006 AMC 12A Problems/Problem 13

Revision as of 19:12, 4 November 2006 by Lotrgreengrapes7926 (talk | contribs) (Solution)

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

Invalid username
Login to AoPS