2005 AMC 12B Problems/Problem 14

Problem

This problem has not been edited in. If you know this problem, please help us out by adding it. 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]$

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2005 AMC 12B Problems/Problem 14

Problem

Solution

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