Difference between revisions of "2005 AMC 12B Problems/Problem 14"
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== Problem == | == Problem == | ||
− | {{ | + | 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? |
+ | |||
+ | <math> | ||
+ | \mathrm{(A)}\ 6\sqrt{2}-6 \qquad | ||
+ | \mathrm{(B)}\ 6 \qquad | ||
+ | \mathrm{(C)}\ 6\sqrt{2} \qquad | ||
+ | \mathrm{(D)}\ 12 \qquad | ||
+ | \mathrm{(E)}\ 6+6\sqrt{2} | ||
+ | </math> | ||
+ | |||
== 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 \boxed{\mathrm{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 == | ||
− | + | {{AMC12 box|year=2005|ab=B|num-b=13|num-a=15}} | |
+ | {{MAA Notice}} |
Latest revision as of 21:12, 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, .
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 |
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