Question

The straight line $AB$ is divided at $C$ so that $\overline{AC}=3\overline{CB}.$ Circles are described on $\overline{AC}$ and $\overline{CB}$ as diameters and a common tangent meets $\overline{AB}$ produced at $D.$ Then $\overline{BD}$ equals:

the diameter of the smaller circle

the radius of the smaller circle

the radius of the larger circle

$\overline{CB}\sqrt3$

the difference of the two radii

Answer: B

Concepts:tangent circles tangent line homothety

Difficulty rating: 1700

Small Hint:

The point $D$ is the external center of similitude of the two circles

Big Hint:

Let $\overline{CB}=u$ and compare the distances from $D$ to the two centers in the ratio $3:1$

Solution:

Let $\overline{CB}=u,$ so $\overline{AC}=3u$ and $\overline{AB}=4u.$ Measured from $A,$ the circle centers are at $\frac{3u}{2}$ and $\frac{7u}{2},$ and their radii are in the ratio $3:1.$ If $AD=z,$ the common external tangent makes $D$ the external center of similitude, so $\frac{z-\frac32u}{z-\frac72u}=3.$ Hence $z=\frac{9u}{2},$ and $BD=z-4u=\frac u2,$ which is the radius of the smaller circle.

Thus, the correct answer is B.

1954 AMC 12 Problem 26

Answer: B

Difficulty rating: 1700

Small Hint:

Big Hint:

Solution:

Problem 26 in Other Years