2004 AMC 12B Problem 19

Attempt Problem 19 of the 2004 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12B solutions, or check the answer key.

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19.

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?

66

454\sqrt{5}

99

1010

636\sqrt{3}

Answer: A
Concepts:conespheretangent line
Difficulty rating: 1870
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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