1980 AMC 12 Problem 19

Question

Let

C_1,

C_2,

and

C_3

be three parallel chords of a circle on the same side of the center. The distance between

C_1

and

C_2

is the same as the distance between

C_2

and

C_3.

The lengths of the chords are

20,

16,

and

8.

The radius of the circle is

12

4\sqrt7

\frac{5\sqrt{65}}3

\frac{5\sqrt{22}}2

not uniquely determined by the given information

A perpendicular from the center bisects each chord

Let the nearest chord be distance

u

from the center and the common spacing be

v

Let

r

be the radius,

u

the distance to the

20

-unit chord, and

v

the common spacing. Then

\begin{aligned} r^2&=u^2+10^2,\\ r^2&=(u+v)^2+8^2,\\ r^2&=(u+2v)^2+4^2. \end{aligned}

Consecutive subtraction gives

2uv+v^2=36

and

2uv+3v^2=48,

so

v^2=6

and

u=\frac{15}{\sqrt6}.

Hence

r^2=u^2+100=\frac{275}{2},

and

r=\frac{5\sqrt{22}}{2}.

Therefore, the correct answer is D.

Problem 19 in other years