Difference between revisions of "2006 AMC 12A Problems/Problem 13"
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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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| − | <math>\ | ||
== See also == | == See also == | ||
* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] | ||
Revision as of 18:12, 4 November 2006
Problem
The vertices of a 
right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?
Solution
Let the radius of the smallest circle be 
. We find that the radius of the largest circle is 
and the radius of the second largest circle is 
. Thus, 
. The radii of the other circles are 
and 
. The sum of their areas is 
