2002 AMC 12A Problem 18

Question

Let

C_1

and

C_2

be circles defined by

(x - 10)^2 + y^2 = 36

and

(x + 15)^2 + y^2 = 81,

respectively. What is the length of the shortest line segment

\overline{PQ}

that is tangent to

C_1

at

P

and to

C_2

at

Q?

15

18

20

21

24

The shortest such segment is the internal tangent; it crosses

\overline{AB}

between the centers

The crossing point divides

\overline{AB}

in the ratio of the radii

6 : 9;

use similar right triangles

The centers are

A = (10, 0)

and

B = (-15, 0),

with radii

6

and

9,

so

AB = 25.

The shortest tangent is the internal one, meeting

\overline{AB}

at a point

D

that splits it in the ratio

6 : 9,

giving

D = (0, 0).

The right triangles

APD

and

BQD

are similar with ratio

2 : 3.

Then

PD = \sqrt{10^2 - 6^2} = 8

and

QD = \sqrt{15^2 - 9^2} = 12,

so

PQ = 8 + 12 = 20.

Thus, the correct answer is C.

Problem 18 in other years