Question

A circle centered at $A$ with a radius of $1$ and a circle centered at $B$ with a radius of $4$ are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is

$\dfrac{1}{3}$

$\dfrac{2}{5}$

$\dfrac{5}{12}$

$\dfrac{4}{9}$

$\dfrac{1}{2}$

Solution:

When two mutually tangent circles of radii $r$ and $s$ both rest on a line, the distance between their points of tangency is $2\sqrt{rs}.$

The big circles' contact points are $2\sqrt{1 \cdot 4} = 4$ apart. Placing the small circle of radius $x$ between them, its two tangent distances add up: $2\sqrt{1 \cdot x} + 2\sqrt{4 \cdot x} = 4.$

Then $6\sqrt{x} = 4,$ so $\sqrt{x} = \dfrac{2}{3}$ and $x = \dfrac{4}{9}.$

Thus, the correct answer is D.

2001 AMC 12 Problem 18

Solution:

Problem 18 in Other Years