1981 AMC 12 Problem 23

Question

Equilateral

\triangle ABC

is inscribed in a circle. A second circle is tangent internally to the circumcircle at

T

and tangent to sides

AB

and

AC

at points

P

and

Q.

If side

BC

has length

12,

then segment

PQ

has length

6

6\sqrt3

8

8\sqrt3

9

Use symmetry to place both circle centers and

T

on the altitude from

A

Relate the smaller circle’s radius to the circumradius using the

30^\circ

half-angle at

A

The circumradius of the equilateral triangle is

R=\frac{12}{\sqrt3}=4\sqrt3.

If the smaller radius is

r,

its center lies on the altitude and is

2r

from

A,

because its distance to either side is

r

and the half-angle is

30^\circ.

Internal tangency at the bottom gives

2r=2R-r,

so

r=\frac{2R}{3}=\frac{8\sqrt3}{3}.

The tangency points on the two sides are separated by

PQ=r\sqrt3=8.

Therefore, the correct answer is C.

Problem 23 in other years