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2012 AMC 12B Problem 15

Problem 15 of 25IntermediateGeometry

Jesse cuts a circular paper disk of radius 1212 along two radii to form two sectors, the smaller having a central angle of 120120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?

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Solution

Each sector forms a cone with slant height 12.12. The smaller sector’s arc length is 120360⋅2π⋅12=8π,\tfrac{120}{360}\cdot2\pi\cdot12=8\pi, so its base radius is 44 and its height is 122−42=82.\sqrt{12^2-4^2}=8\sqrt2. The larger sector (central angle 240∘240^\circ) has arc length 16π,16\pi, base radius 8,8, and height 122−82=45.\sqrt{12^2-8^2}=4\sqrt5. The ratio of volumes is 13π⋅42⋅8213π⋅82⋅45=1010.\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}. Thus, the correct answer is C.

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Concepts: cone · volume · Pythagorean Theorem

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