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2003 AMC 10B Problem 17

2003 AMC 10B Problem 17AMC 10 Step by Step

Name: Date: September 13, 2026

Solid Geometry · difficulty 3 of 5 · about 2 min · position P16-20

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 $4 \pi r^3 / 3$ .)

A)
$2:1$
B)
$3:1$
C)
$4:1$
D)
$16:3$
E)
$6:1$

Answer

Key insight

Melted volume is 3/4 of 4/3 pi r^3 = pi r^3, and setting this equal to the cone's pi r^2 h / 3 gives h = 3r.

Solution

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

Why this works

Both volumes carry the same factor $\pi r^2$ , so once the melting factor is applied the equation reduces to a linear relation between $h$ and $r$ . The $\frac{3}{4}$ and $\frac{4}{3}$ cancel neatly by design; the whole problem is careful bookkeeping of the given formulas.

The trap

Forgetting the 75 percent factor and equating 4/3 pi r^3 with the cone volume, which gives 4:1.

Common mistakes

Forgetting the $75\%$ factor and equating $\frac{4}{3}\pi r^3$ with the cone volume, which gives $4:1$ .
Applying the $75\%$ the wrong way (dividing by $0.75$ instead of multiplying), which gives $\frac{16}{3}$ , choice (D).

Techniques

Set up the equation/formula and compute; no special trick needed