2001 AMC 10 Problem 21

1. Three-dimensional shapes are hard to draw, so cut both solids straight down through their shared center line and look at the flat slice.
2. The cone's slice is a triangle with a base of $10$ (the cone's diameter) and a height of $12$.
3. Call the cylinder's radius $r$. Its diameter is $2r$, and its height equals its diameter, so its height is also $2r$. Its slice is a $2r$ by $2r$ square sitting on the base of the triangle, with its top two corners touching the slanted sides.
4. Look above the square. There is a small triangle with its base along the top of the square, which is $2r$ wide. Its height is what is left of $12$, which is $12 - 2r$.
5. The small triangle's base is parallel to the big triangle's base, so the two triangles have the same shape. Their base divided by height must be the same:
$$\frac{2r}{12 - 2r} = \frac{10}{12}.$$
6. Cross-multiply: $12 \times 2r = 10 \times (12 - 2r)$, so $24r = 120 - 20r$.
7. Then $44r = 120$, and $r = \frac{120}{44} = \frac{30}{11}$.
8. Check: $2r = \frac{60}{11} \approx 5.45$ and $12 - 2r \approx 6.55$; the ratio is about $0.83$, which matches $\frac{10}{12}$.

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

When two round solids share the same center line, a slice through that line shows everything you need. Here "cylinder inside a cone" becomes "square inside a triangle." A line across a triangle, parallel to its base, always cuts off a smaller triangle of the same shape, and that gives the equation.

Use coordinates. Put the center of the cone's base at $(0, 0)$ and the tip at $(0, 12)$. The right slanted side goes from $(5, 0)$ to $(0, 12)$. It drops $12$ over a run of $5$, so its equation is $y = 12 - \frac{12}{5}x$. The square's top right corner is at $(r, 2r)$ and lies on this line, so $2r = 12 - \frac{12}{5}r$. Multiply by $5$: $10r = 60 - 12r$, so $22r = 60$ and $r = \frac{30}{11}$.