Difference between revisions of "2011 AMC 12A Problems/Problem 17"

Revision as of 21:48, 11 February 2011

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,

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

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

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

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

2011 AMC 12A (Problems • Answer Key • Resources)
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

@@ Line 20: / Line 20: @@
 <math>\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}</math>
-which add up to <math>4.8</math>. Thus the area we're looking for is <math>6 - 4.8 = 1.2 = \frac{6}{5} \Rightarrow \boxed{D}</math>.
+which add up to <math>4.8</math>. Thus the area we're looking for is <math>6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}</math>.
 == See also ==
 {{AMC12 box|year=2011|num-b=16|num-a=18|ab=A}}