Difference between revisions of "2006 AMC 12A Problems/Problem 13"

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== Solution ==
 
== Solution ==
  
{{solution}}
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Let the radius of the smallest circle be <math> a </math>. We find that the radius of the largest circle is <math>4-a</math> and the radius of the second largest circle is <math>3-a</math>. Thus, <math>4-a+3-a=5\iff a=1</math>. The radii of the other circles are <math>3</math> and <math>2</math>. The sum of their areas is <math>\pi+9\pi+4\pi=14\pi\iff\mathrm{(E)}</math>
 
 
<math>\rm{(E)}\,14\pi</math>
 
  
 
== See also ==
 
== See also ==
  
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]

Revision as of 19:12, 4 November 2006

Problem

2006 AMC 12A Problem 13.gif

The vertices of a $3-4-5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?

$\mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ }  14\pi$

Solution

Let the radius of the smallest circle be $a$. We find that the radius of the largest circle is $4-a$ and the radius of the second largest circle is $3-a$. Thus, $4-a+3-a=5\iff a=1$. The radii of the other circles are $3$ and $2$. The sum of their areas is $\pi+9\pi+4\pi=14\pi\iff\mathrm{(E)}$

See also

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