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 determined by the points of tangency?

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

Solution

$[asy] unitsize(.5cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(3,0), C=(0,4); dot (A); dot (B); dot (C); draw(A--B); draw(A--C); draw(B--C); draw(Circle(A,1)); draw(Circle(B,2)); draw(Circle(C,3)); [/asy]$

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

The areas of the three triangles determined by the center and the two points of tangency of each circle are, by Law of Sines,

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

$\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{4}{5} = \frac{8}{5}$

$\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}$

which add up to $4.8$ . The area we're looking for is the large 3-4-5 triangle minus the three smaller triangles, or $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$ .

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

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2011 AMC 12A Problems/Problem 17

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