1959 AMC 12 Problem 41

Question

On the same side of a straight line three circles are drawn as follows: a circle with a radius of

4

inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is:

24

20

18

16

12

Answer: D

Concepts:tangent circles distance formula

Difficulty rating: 1590

Small hint:

The small circle lies symmetrically between the two equal circles

Big hint:

If an equal circle has radius

R,

compare the center distance

\sqrt{R^2+(R-4)^2}

with

R+4

Solution:

Let the equal circles have radius

R.

Their centers are

2R

apart, so the center of the radius-

4

circle lies midway between them. The horizontal and vertical separations between its center and either large center are

R

and

R-4.

Tangency gives

R^2+(R-4)^2=(R+4)^2.

Simplifying yields

R^2-16R=0,

and positivity gives

R=16.

Thus, the correct answer is D.