# Difference between revisions of "2005 AMC 12B Problems/Problem 14"

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== Problem == | == Problem == | ||

{{problem}} | {{problem}} | ||

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== Solution == | == Solution == | ||

+ | Let <math>R</math> be the radius of the circle. Draw the two radii that meet the points of tangency to the lines <math>y = \pm x</math>. We can see that a square is formed by the origin, two tangency points, and the center of the circle. The side lengths of this square are <math>R</math> and the diagonal is <math>k = R+6</math>. The diagonal of a square is <math>\sqrt{2}</math> times the side length. Therefore, <math>R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow E</math>. | ||

+ | |||

+ | <center> | ||

+ | <asy> | ||

+ | real Xmin,Xmax,Ymin,Ymax; | ||

+ | real R = 6+6*sqrt(2); | ||

+ | Xmin = -16; Xmax = 16; Ymin = -4; Ymax = 40; | ||

+ | xaxis(Xmin,Xmax,Arrows); | ||

+ | yaxis(Ymin,Ymax,Arrows); | ||

+ | label("$x$",(Xmax+0.25,0),S); | ||

+ | label("$y$",(0,Ymax+0.25),E); | ||

+ | draw((Xmin,-Xmin)--(-Ymin,Ymin)); | ||

+ | draw((Xmax,Xmax)--(Ymin,Ymin)); | ||

+ | draw((Xmin,6)--(Xmax,6)); | ||

+ | dot((0,0)); dot((R/sqrt(2),R/sqrt(2))); dot((-R/sqrt(2),R/sqrt(2))); dot((0,R*sqrt(2))); | ||

+ | draw((0,0)--(R/sqrt(2),R/sqrt(2))--(0,R*sqrt(2))--(-R/sqrt(2),R/sqrt(2))--(0,0)); | ||

+ | draw(Circle((0,6+R),R)); | ||

+ | label("$R$",(0,6+R/2),(0,0)); | ||

+ | label("$R$",(-R/(2*sqrt(2)),3*R/(2*sqrt(2))),0.5*NW); | ||

+ | label("$R$",(R/(2*sqrt(2)),3*R/(2*sqrt(2))),0.5*NE); | ||

+ | label("$R$",(-R/(2*sqrt(2)),R/(2*sqrt(2))),0.5*SW); | ||

+ | label("$R$",(R/(2*sqrt(2)),R/(2*sqrt(2))),0.5*SE); | ||

+ | label("$6$",(0,3),(0,0)); | ||

+ | </asy> | ||

+ | </center> | ||

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

* [[2005 AMC 12B Problems]] | * [[2005 AMC 12B Problems]] |

## Revision as of 23:05, 9 June 2010

## Problem

*This problem has not been edited in. If you know this problem, please help us out by adding it.*
A circle having center , with , is tangent to the lines , and . What is the radius of this circle?

## Solution

Let be the radius of the circle. Draw the two radii that meet the points of tangency to the lines . We can see that a square is formed by the origin, two tangency points, and the center of the circle. The side lengths of this square are and the diagonal is . The diagonal of a square is times the side length. Therefore, .