2021 AMC 10B Problem 10

P1-10~2 minPrint

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

An inverted cone with base radius $12 \text{cm}$ and height $18\text{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of $24\text{cm}$ . What is the height in centimeters of the water in the cylinder?

Show solution without answering

Key insight

Pouring keeps the amount of water the same. Set the cone's volume, (1/3)π · 12² · 18, equal to π · 24² · h, so h = 1.5.

Solution

Pouring water from one container to another does not change how much water there is. So the volume of the cone equals the volume of the water in the cylinder.

1. Volume of the cone. A cone's volume is one third of base area times height:
$\tfrac{1}{3} \times \pi \times 12^2 \times 18 = \tfrac13 \times 144 \times 18 \times \pi = 864\pi.$
2. Volume of the water in the cylinder. It is a short cylinder with radius $24$ and unknown height $h$ :
$\pi \times 24^2 \times h = 576\pi h.$
3. Set them equal. $576\pi h = 864\pi$ , so
$h = \frac{864}{576} = \frac{3}{2} = 1.5.$

The answer is $\boxed{\textbf{(A)}\ 1.5}$ .

Why this works

"Poured into" always means "same volume." You can also see the answer with ratios: the cylinder's radius is $2$ times the cone's, so its base is $4$ times as big, and the cone only holds one third of a cylinder with its own base. So the water height is $18 \div 4 \div 3 = 1.5$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The volume of water equals the cone's volume:
$V = \frac{1}{3}\pi (12)^2 (18) = \frac{1}{3}\pi \cdot 144 \cdot 18 = 864\pi.$

In the cylinder this water forms a cylinder of radius $24$ and unknown height $h$ :
$\pi (24)^2 h = 576\pi h = 864\pi \quad\Longrightarrow\quad h = \frac{864}{576} = \frac{3}{2}.$

The answer is $\boxed{\textbf{(A)}\ 1.5}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Geometry, Chapter 15: Curved Surfaces