2012 AMC 10B Problem 17

1. Picture rolling a paper sector into a cone. The curved edge of the sector wraps around to become the circle at the bottom of the cone. The straight edges (length $12$) become the slanted side of the cone, running from the tip to the bottom circle.
2. The two sectors have angles $120^\circ$ and $360^\circ - 120^\circ = 240^\circ$.
3. The full circle of radius $12$ has circumference $2\pi \cdot 12 = 24\pi$.
- The $120^\circ$ sector is $\frac{120}{360} = \frac13$ of it, so its curved edge is $8\pi$.
- The $240^\circ$ sector is $\frac23$ of it, so its curved edge is $16\pi$.
4. The curved edge becomes the base circle, so the base radius is the edge length divided by $2\pi$:
- small cone: $r = 8\pi \div 2\pi = 4$;
- large cone: $r = 16\pi \div 2\pi = 8$.
5. The height, the base radius and the slant side $12$ form a right triangle. By the Pythagorean theorem:
- small cone: $h = \sqrt{12^2 - 4^2} = \sqrt{128} = 8\sqrt2$;
- large cone: $h = \sqrt{12^2 - 8^2} = \sqrt{80} = 4\sqrt5$.
6. A cone's volume is $\frac13 \pi r^2 h$. The $\frac13\pi$ is the same for both, so it cancels in the ratio:
$$\frac{4^2 \cdot 8\sqrt2}{8^2 \cdot 4\sqrt5} = \frac{128\sqrt2}{256\sqrt5} = \frac{\sqrt2}{2\sqrt5}.$$
7. Multiply top and bottom by $\sqrt5$ to clean it up: $\dfrac{\sqrt{10}}{2 \cdot 5} = \dfrac{\sqrt{10}}{10}$.

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

A cone made from a sector keeps two lengths: the sector's radius becomes the slant side, and the sector's arc becomes the base circle. With those two facts and the Pythagorean theorem you can find the radius and height of any rolled-up cone. Here the two cones share the same slant side, so the wider cone is shorter; that is why you have to compute both volumes rather than guess from the angles.