2004 AMC 12B Problem 19

Question

A truncated cone has horizontal bases with radii

18

and

2.

A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

6

4\sqrt{5}

9

10

6\sqrt{3}

Take the cross-section through the axis: an isosceles trapezoid with an inscribed circle

Tangent lengths give a slant side of

18 + 2 = 20;

drop a height to form a right triangle

The axial cross-section is a trapezoid

ABCD

with parallel sides

2

and

18

and an inscribed circle (a great circle of the sphere). By equal tangent lengths from

B

and

C,

the slant side

BC = 18 + 2 = 20.

Dropping a perpendicular from

C

to the bottom base gives a right triangle with horizontal leg

18 - 2 = 16,

so the height is

\sqrt{20^2 - 16^2} = 12.

The sphere’s radius is half the height,

6.

Thus, the correct answer is A.

Problem 19 in other years