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

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[[Image:2006 AMC 12A Problem 13.gif]]
 
[[Image:2006 AMC 12A Problem 13.gif]]
  
The vertices of a <math>3-4-5</math> right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?
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The [[vertex|vertices]] of a <math>3-4-5</math> [[right triangle]] are the centers of three mutually externally tangent [[circle]]s, as shown. What is the sum of the areas of the three circles?
  
 
<math> \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</math>
 
<math> \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</math>
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== Solution ==
 
== Solution ==
  
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>
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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>
  
 
== See also ==
 
== See also ==
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
*[[2006 AMC 12A Problems/Problem 12|Previous Problem]]
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*[[2006 AMC 12A Problems/Problem 14|Next Problem]]
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{{AMC box|year=2006|n=12A|num-b=13|num-a=15}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 19:45, 31 January 2007

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


{{{header}}}
Preceded by
Problem 13
AMC 12A
2006
Followed by
Problem 15
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