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. Let's label the center of the smaller circle (which is also the center of the larger circle) as $D$ . Let's also label the point where the smaller circle intersects $BC$ as $E$ . So $ABC$ is similar to $DEC$ . Since $DE$ is the radius of the smaller circle, call the length $x$ and since $DC$ is the radius of the bigger circle, call that length $3x$ . The diameter, $AC$ is $6x$ . So,

$\frac{AB}{AC} = \frac {DE}{DC} \Rightarrow \frac{12}{6x} = \frac{x}{3x}$

$\fbox{B}$

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 • 26 • 27 • 28 • 29 • 30
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1992 AHSME Problems/Problem 11

Problem

Solution (Similarity)

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