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

(Solution)
(Solution)
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The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.
 
The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.
  
The 3 triangles determined by one vertex (one center) and two tangent points closest it
+
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,
 +
 
 +
<math>\frac{1}{2} \dot 1 \dot 1 \dot 1 = \frac{1}{2}</math>
 +
 
 +
<math>\frac{1}{2} \dot 2 \dot 2 \dot \frac{4}{5} = \frac{8}{5}</math>
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 +
<math>\frac{1}{2} \dot 3 \dot 3 \dot \frac{3}{5} = \frac{27}{10}</math>
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 +
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 ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=16|num-a=18|ab=A}}
 
{{AMC12 box|year=2011|num-b=16|num-a=18|ab=A}}

Revision as of 22:38, 11 February 2011

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?

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

Solution

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 $6 - 4.8 = 1.2 = \frac{6}{5} \Rightarrow \boxed{D}$.

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 12 Problems and Solutions