# Difference between revisions of "1992 AHSME Problems/Problem 11"

## Problem

$[asy] draw(circle((0,0),18),black+linewidth(.75)); draw(circle((0,0),6),black+linewidth(.75)); draw((-18,0)--(18,0)--(-14,8*sqrt(2))--cycle,black+linewidth(.75)); dot((-18,0));dot((18,0));dot((-14,8*sqrt(2))); MP("A",(-18,0),W);MP("C",(18,0),E);MP("B",(-14,8*sqrt(2)),W); [/asy]$

The ratio of the radii of two concentric circles is $1:3$. If $\overline{AC}$ is a diameter of the larger circle, $\overline{BC}$ is a chord of the larger circle that is tangent to the smaller circle, and $AB=12$, then the radius of the larger circle is

$\text{(A) } 13\quad \text{(B) } 18\quad \text{(C) } 21\quad \text{(D) } 24\quad \text{(E) } 26$

## Solution (Similarity)

We are given that $BC$ is tangent to the smaller circle. Using that, we know where the circle intersects $BC$, it creates a right triangle. We can also point out that since $AC$ is the diameter of the bigger circle and triangle $ABC$ is inscribed the semi-circle, that angle $B$ is a right angle. Therefore, we have $2$ similar triangles.

$\fbox{B}$