2012 AMC 10B Problem 17

When a sector is rolled into a cone, its arc becomes the circumference of the base and its radius ($12$) becomes the slant height.

Smaller sector ($120^\circ$): arc length $\frac{120}{360} \cdot 2\pi \cdot 12 = 8\pi$, so the base radius is $r_1 = 4$. Height: $h_1 = \sqrt{12^2 - 4^2} = \sqrt{128} = 8\sqrt2$.

Larger sector ($240^\circ$): arc length $16\pi$, so $r_2 = 8$ and $h_2 = \sqrt{144 - 64} = \sqrt{80} = 4\sqrt5$.

The factor $\frac13\pi$ cancels in the ratio of volumes:
$$\frac{r_1^2 h_1}{r_2^2 h_2} = \frac{16 \cdot 8\sqrt2}{64 \cdot 4\sqrt5} = \frac{\sqrt2}{2\sqrt5} = \frac{\sqrt2\cdot\sqrt5}{2 \cdot 5} = \frac{\sqrt{10}}{10}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{\sqrt{10}}{10}}$.

A cone's net is a sector: arc $\leftrightarrow$ base circumference, sector radius $\leftrightarrow$ slant height. Those two facts plus the Pythagorean theorem give $r$ and $h$ for any rolled cone. Because both cones share slant height $12$, doubling the base radius does not double the height; the volumes must be computed separately before comparing.

• Taking the volume ratio to be the ratio of sector angles (1/2) or of base areas (1/4), forgetting that the two cones have different heights.
• Using $12$ as the height of each cone instead of the slant height, which gives the ratio $\frac14$.