2003 AMC 10B Problems/Problem 17

Problem

An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies $75\%$ of the volume of the frozen ice cream. What is the ratio of the cone's height to its radius? (Note: a cone with radius $r$ and height $h$ has volume $\pi r^2 h / 3$ and a sphere with radius $r$ has volume $4 \pi r^3 / 3$.)

$\textbf{(A) } 2:1 \qquad\textbf{(B) } 3:1 \qquad\textbf{(C) } 4:1 \qquad\textbf{(D) } 16:3 \qquad\textbf{(E) } 6:1$

Solution

Let $r$ be the radius of both the cone and the sphere of ice cream, and let $h$ be the height of the cone. The cone can hold a maximum of $\pi r^2h/3$ melted ice cream. The volume of the frozen ice cream is $4\pi r^3/3.$ The volume of the melted ice cream is $75 \%$ of that, which is $\pi r^3.$ Since the melted ice cream fills the cone exactly,

\begin{align*}\pi r^3&=\pi r^2h/3\\ r&=h/3\\ 3&=h/r\end{align*}

Then the ratio of the cone's height to its radius is $h:r=\boxed{\textbf{(B)}\ 3:1}$

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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