2006 AMC 12A Problems/Problem 13

Problem

Original 2006 12A 13.png

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 (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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