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2004 AMC 12B Problem 19

Problem 19 of 25HarderGeometry

A truncated cone has horizontal bases with radii 1818 and 2.2. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

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Solution

The axial cross-section is a trapezoid ABCDABCD with parallel sides 22 and 1818 and an inscribed circle (a great circle of the sphere). By equal tangent lengths from BB and C,C, the slant side BC=18+2=20.BC = 18 + 2 = 20. Dropping a perpendicular from CC to the bottom base gives a right triangle with horizontal leg 182=16,18 - 2 = 16, so the height is 202162=12.\sqrt{20^2 - 16^2} = 12. The sphere’s radius is half the height, 6.6. Thus, the correct answer is A.

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Concepts: cone · sphere · tangent line

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.