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Problem 813

AMC 10/12, early questions

Geometry Difficulty 3.7 Multiple choice CEMC Fermat · Canada · 2024

A cylinder contains some water. A solid cone with the same height
and half the radius of the cylinder is submerged into the water until
the circular face of the cone lies flat on the circular base of the
cylinder, as shown.

Once this is done, the depth of the water is half of the height of
the cylinder. If the cone is then removed, the depth of the water will
be what fraction of the height of the cylinder?

(The volume of a cylinder with radius $r$ and height $h$ is $\pi r^2h$ and the volume of a cone with radius $r$ and height $h$ is $\frac{1}{3}\pi r^2h$ .)

Pick one

Official solution

Suppose that the radius of the cylinder is $r$ and the height of the cylinder is $h$ . This means that the volume of the cylinder is $\pi r^2 h$ ; the volume of half of the cylinder is $\$\frac{1}{2}\pi$ r^2
h. Also, the radius of the cone is

Figure for this problem $\frac{1}{2}r$ and the height of the cone is

Figure for this problem h. This means that the volume of the cone is

Figure for this problem $\frac{1}{3}\pi \left(\frac{1}{2}r\right)^2$
h $or$ $\frac{1}{12}\pi$ r^2 h$.

When the cone is divided into two pieces by a horizontal plane at half
of its height, the top portion of the cone is a cone with the same
proportions, but with dimensions $\frac{1}{2}$ of those of the larger cone. This means that the volume of the top portion is $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$ of that of the cone, which equals $\frac{1}{8} \cdot \frac{1}{12}\pi r^2 h$ or $\frac{1}{96}\pi r^2 h$ . To see this in another way, we note that this top portion of the cone has height $\frac{1}{2}h$ and should have radius $\$\frac{1}{2} \cdot \frac{1}{2}r$ (because the radius decreases proportionally to the height). This means that the volume of this portion is

Figure for this problem $\frac{1}{3}\pi \left(\frac{1}{4}r\right)^2 \cdot \frac{1}{2}h$ $which is again$ $\frac{1}{96}\pi$ r^2 h. Using this information, the bottom portion of the cone has volume

Figure for this problem $\frac{7}{8} \cdot \frac{1}{12} \pi$ r^2 h =
$\frac{7}{96} \pi$ r^2 h. Now, when the cone is in the cylinder and the cylinder is filled with water to half of its height, the volume of the bottom half of the cylinder is filled with the bottom portion of the cone and with the water. Therefore, the volume of water is the difference between half of the volume of the cylinder and the volume of the bottom portion of the cone, or

Figure for this problem $\frac{1}{2}\pi$ r^2 h - $\frac{7}{96}\pi$
r^2 h = $\frac{48}{96}\pi$ r^2 h - $\frac{7}{96}\pi$ r^2 h =
$\frac{41}{96}\pi$ r^2 h. When the cone is removed, the water then occupies a cylinder with radius

Figure for this problem r $and volume$ $\frac{41}{96}\pi$ r^2 h. If the depth of the water in this configuration is

Figure for this problem d $, then$ $\pi$ r^2 d = $\frac{41}{96}\pi$ r^2 h $and so$ d = $\frac{41}{96}h$ , which means that the depth of the water is

Figure for this problem $\frac{41}{96}\$$
of the height of the cylinder.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.