2012 AMC 12B Problem 15

Attempt Problem 15 of the 2012 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 2012 AMC 12B solutions, or check the answer key.

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15.

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?

18\dfrac{1}{8}

14\dfrac{1}{4}

1010\dfrac{\sqrt{10}}{10}

56\dfrac{\sqrt{5}}{6}

105\dfrac{\sqrt{10}}{5}

Answer: C
Concepts:conevolumePythagorean Theorem
Difficulty rating: 1800
Solution:

Each sector forms a cone with slant height 12.12. The smaller sector's arc length is 1203602π12=8π,\tfrac{120}{360}\cdot2\pi\cdot12=8\pi, so its base radius is 44 and its height is 12242=82.\sqrt{12^2-4^2}=8\sqrt2.

The larger sector (central angle 240240^\circ) has arc length 16π,16\pi, base radius 8,8, and height 12282=45.\sqrt{12^2-8^2}=4\sqrt5.

The ratio of volumes is 13π428213π8245=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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