2011 AMC 12A Problems/Problem 17

Revision as of 16:19, 22 September 2013 by Armalite46 (talk | contribs) (Solution)

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 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,

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

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

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

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

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png