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2004 AMC 12B Problems/Problem 19

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Problem

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?

\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 4\sqrt{5}\qquad\mathrm{(C)}\ 9\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ 6\sqrt{3}

Solution

Consider a trapezoidal (label it ABCD as follows) cross-section of the truncate cone along a diameter of the bases:

import olympiad;size(220);defaultpen(0.7);pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2),...

Above, E,F, and G are points of tangency. By the Two Tangent Theorem, BF = BE = 18 and CF = CG = 2, so BC = 20. We draw H such that it is the foot of the altitude \overline{HD} to \overline{AB}:

import olympiad;size(450);defaultpen(0.7);pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2),...

By the Pythagorean Theorem, r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = \boxed{6} \Rightarrow \mathrm{(A)}.

See also

2004 AMC 12B (ProblemsResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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