Difference between revisions of "2005 AMC 12B Problems/Problem 14"
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== Problem == | == Problem == | ||
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A circle having center <math>(0,k)</math>, with <math>k>6</math>, is tangent to the lines <math>y=x</math>, <math>y=-x</math> and <math>y=6</math>. What is the radius of this circle? | A circle having center <math>(0,k)</math>, with <math>k>6</math>, is tangent to the lines <math>y=x</math>, <math>y=-x</math> and <math>y=6</math>. What is the radius of this circle? | ||
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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>. | + | 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 \boxed{\mathrm{E}}</math>. |
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== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2005|ab=B|num-b=13|num-a=15}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 20:12, 24 December 2020
Problem
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, 
.
![[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]](http://latex.artofproblemsolving.com/9/2/1/9214d60dfddf137d3b4e216cf2abcdd4d76c8e90.png)
See also
| 2005 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
