2011 AMC 12A Problems/Problem 17

Problem

Circles with radii $1$ , $2$ , and $3$ are mutually externally tangent. What is the area of the triangle determine by the points of tangency?

$\textbf{(A)}\ \frac{3}{5} \qquad \textbf{(B)}\ \frac{4}{5} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{6}{5} \qquad \textbf{(E)}\ \frac{4}{3}$

The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.

The 3 triangles determined by one center and the two points of tangency that particular circle has with the other two are, by Law of Sines,

$\frac{1}{2} \dot 1 \dot 1 \dot 1 = \frac{1}{2}$

$\frac{1}{2} \dot 2 \dot 2 \dot \frac{4}{5} = \frac{8}{5}$ (Error compiling LaTeX. Unknown error_msg)

$\frac{1}{2} \dot 3 \dot 3 \dot \frac{3}{5} = \frac{27}{10}$ (Error compiling LaTeX. Unknown error_msg)

which add up to $4.8$ . Thus the area we're looking for is $6 - 4.8 = 1.2 = \frac{6}{5} \Rightarrow \boxed{D}$ .

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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