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

m (→Problem) |
(→Solution) |
||

Line 9: | Line 9: | ||

== 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> | |

− | |||

− | <math>\ | ||

== See also == | == See also == | ||

* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] |

## Revision as of 19:12, 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