# Difference between revisions of "2005 AMC 12A Problems/Problem 16"

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− | + | Set <math>s =1</math> so that we only have to find <math>r</math>. Draw the segment between the center of the third circle and the large circle; this has length <math>r+1</math>. We then draw the [[radius]] of the large circle that is perpendicular to the [[x-axis]], and draw the perpendicular from this radius to the center of the third circle. This gives us a [[right triangle]] with legs <math>r-3,r-1</math> and [[hypotenuse]] <math>r+1</math>. The [[Pythagorean Theorem]] yields: | |

<div style="text-align:center;"><math>(r-3)^2 + (r-1)^2 = (r+1)^2</math><br /><math>r^2 - 10r + 9 = 0</math><br /><math>r = 1, 9</math></div> | <div style="text-align:center;"><math>(r-3)^2 + (r-1)^2 = (r+1)^2</math><br /><math>r^2 - 10r + 9 = 0</math><br /><math>r = 1, 9</math></div> | ||

− | Quite obviously <math>r > | + | Quite obviously <math>r > 1</math>, so <math>r = 9 \boxed{(D)}</math>. |

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## Latest revision as of 01:56, 10 January 2021

## Problem

Three circles of radius are drawn in the first quadrant of the -plane. The first circle is tangent to both axes, the second is tangent to the first circle and the -axis, and the third is tangent to the first circle and the -axis. A circle of radius is tangent to both axes and to the second and third circles. What is ?

## Solution

### Solution 1

Set so that we only have to find . Draw the segment between the center of the third circle and the large circle; this has length . We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legs and hypotenuse . The Pythagorean Theorem yields:

Quite obviously , so .