2001 AMC 10 Problem 21

Slice both solids with a plane containing the shared axis. The cone appears as an isosceles triangle with base $10$ and height $12$. The cylinder, whose height equals its diameter $2r$, appears as a $2r \times 2r$ square standing on the base of the triangle with its two upper corners on the slanted sides.

Above the square is a small triangle with base $2r$ (the top of the square) and height $12 - 2r$. Its base is parallel to the big triangle's base, so the two triangles are similar and their base-to-height ratios agree:
$$\frac{2r}{12 - 2r} = \frac{10}{12} = \frac{5}{6}.$$
Cross-multiplying: $12r = 60 - 10r$, so $22r = 60$ and $r = \frac{30}{11}$.

The answer is $\boxed{\textbf{(B)}\ \frac{30}{11}}$.

Solids of revolution sharing an axis are fully described by their axial cross-section, which turns "cylinder inscribed in a cone" into "rectangle inscribed in a triangle." The similar triangle cut off at the top is the standard tool for that 2-D picture. Draw the section first; the 3-D words are only a wrapper.

Coordinates: put the center of the base at the origin with the apex at $(0, 12)$. The right slanted edge is the line through $(5, 0)$ and $(0, 12)$, i.e. $y = 12 - \frac{12}{5}x$. The cylinder's top-right corner $(r, 2r)$ lies on it: $2r = 12 - \frac{12}{5}r$, giving $\frac{22}{5}r = 12$ and $r = \frac{30}{11}$.

• Using r instead of 2r for the cylinder's height, which corrupts the similarity ratio.
• Comparing the small triangle's half-base $r$ with the full base $10$ (instead of the half-base $5$), which yields a wrong ratio.