2001 AMC 12 Problem 18

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}

Answer: D

Concepts:tangent circles Pythagorean Theorem

Difficulty rating: 1820

Small hint:

For two circles of radii

r

and

s

both tangent to a line and to each other, their contact points on the line are

2\sqrt{rs}

apart

Big hint:

The contact point of the small circle splits the segment:

2\sqrt{x} + 2\sqrt{4x} = 2\sqrt{4}

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.