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

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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.
  
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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,
+
The areas of the three triangles determined by the center and the two points of tangency of each circle are, by Law of Sines,
  
 
<math>\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}</math>
 
<math>\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}</math>
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<math>\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}</math>
 
<math>\frac{1}{2} \cdot 3 \cdot 3 \cdot \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>.
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which add up to <math>4.8</math>. The area we're looking for is the large 3-4-5 triangle minus the three smaller triangles, or <math>6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 15:09, 20 January 2014

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 $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$.

See also

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

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