2005 AMC 12B Problems/Problem 14

Problem

A circle having center $(0,k)$, with $k>6$, is tangent to the lines $y=x$, $y=-x$ and $y=6$. What is the radius of this circle?

$\mathrm{(A)}\ 6\sqrt{2}-6 \qquad \mathrm{(B)}\ 6 \qquad \mathrm{(C)}\ 6\sqrt{2} \qquad \mathrm{(D)}\ 12 \qquad \mathrm{(E)}\ 6+6\sqrt{2}$

Solution

Let $R$ be the radius of the circle. Draw the two radii that meet the points of tangency to the lines $y = \pm x$. 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 $R$ and the diagonal is $k = R+6$. The diagonal of a square is $\sqrt{2}$ times the side length. Therefore, $R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow E$.

[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]

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
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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