2021 AMC 10A Spring Problem 12

Question

Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are

3

cm and

6

cm. Into each cone is dropped a spherical marble of radius

1

cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

1:1

47:43

2:1

40:13

4:1

Answer: E

Concepts:cone similarity volume

Difficulty rating: 1660

Small hint:

Equal liquid volumes relate the two initial liquid heights

Big hint:

The marble adds the same displaced volume in each cone, so the final cone-below-surface volumes are still equal

Video solution:

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Written solution:

Let the initial liquid heights in the narrow and wide cones be

h_1

and

h_2.

Since the liquid volumes are equal,

\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,

so

h_1=4h_2.

After the identical marbles are dropped in, each cone must contain the same final volume below the liquid surface: the original liquid volume plus the volume of one marble. If the new liquid-surface radii are

3x

and

6y,

similarity gives new heights

h_1x

and

h_2y.

Thus

\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y.

Using

h_1=4h_2,

this simplifies to

x^3=y^3,

so

x=y.

The rise ratio is therefore

\begin{aligned} h_1(x-1):h_2(y-1) &= h_1:h_2 \\ &= 4:1. \end{aligned}

Thus, E is the correct answer.