2003 AMC 10B Problem 17

1. The sphere and the cone have the same diameter, so they have the same radius. Call it $r$, and call the cone's height $h$.
2. The frozen ice cream is a sphere, with volume $\frac{4}{3}\pi r^3$.
3. When it melts it takes up only $75\% = \frac{3}{4}$ of that:
$$\frac{3}{4} \times \frac{4}{3}\pi r^3 = \pi r^3.$$
(The $\frac{3}{4}$ and $\frac{4}{3}$ cancel.)
4. This melted ice cream exactly fills the cone, whose volume is $\frac{1}{3}\pi r^2 h$. So
$$\pi r^3 = \frac{1}{3}\pi r^2 h.$$
5. Divide both sides by $\pi r^2$: $r = \frac{h}{3}$, so $h = 3r$.
6. The height is $3$ times the radius, a ratio of $3:1$.

The answer is $\boxed{\textbf{(B)}\ 3:1}$.

Both volume formulas contain $\pi r^2$, so after you set them equal and cancel, only a simple link between $h$ and $r$ is left. The problem is set up so the $\frac{3}{4}$ from melting cancels the $\frac{4}{3}$ in the sphere formula. The key is to apply the $75\%$ before comparing.

Same diameter means the sphere and the cone share the radius $r$. Let the cone's height be $h$.

Frozen volume: $\frac{4}{3}\pi r^3$. After melting it shrinks to $75\%$ of that:
$$\frac{3}{4} \cdot \frac{4}{3}\pi r^3 = \pi r^3.$$

This exactly fills the cone, whose capacity is $\frac{1}{3}\pi r^2 h$. Equating,
$$\pi r^3 = \frac{1}{3}\pi r^2 h \quad\Longrightarrow\quad h = 3r.$$

So $h : r = 3 : 1$, and the answer is $\boxed{\textbf{(B)}\ 3:1}$.