1987 AMC 12 Problem 22

Question

A ball was floating in a lake when the lake froze. The ball was removed (without breaking the ice), leaving a hole

24

cm across at the top and

8

cm deep. What was the radius of the ball (in centimeters)?

8

12

13

8\sqrt3

6\sqrt6

Take a vertical cross-section through the center of the circular hole

The half-chord is

12,

while the center-to-ice distance is

r-8

In a vertical cross-section, half the

24

-cm hole is a chord segment of length

12.

If the sphere radius is

r,

the distance from its center to the ice plane is

r-8.

The resulting right triangle gives

(r-8)^2+12^2=r^2.

Solving yields

16r=208,

so

r=13.

Thus the correct answer is C.

Problem 22 in other years