Problem 813

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 $\frac{1}{2}r$ and the height of the cone
is $h$.

This means that the volume of the cone is $\frac{1}{3}\pi \left(\frac{1}{2}r\right)^2 h$ 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 $\frac{1}{3}\pi \left(\frac{1}{4}r\right)^2 \cdot \frac{1}{2}h$ which is again $\frac{1}{96}\pi r^2 h$.

Using this information, the bottom portion of the cone has volume $\frac{7}{8} \cdot \frac{1}{12} \pi r^2 h = \frac{7}{96} \pi r^2 h$.

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 $\frac{1}{2}\pi r^2 h - \frac{7}{96}\pi r^2 h = \frac{48}{96}\pi r^2 h - \frac{7}{96}\pi r^2 h = \frac{41}{96}\pi r^2 h$.

When the cone is removed, the water then occupies a cylinder with radius
$r$ and volume $\frac{41}{96}\pi r^2 h$.

If the depth of the water in this configuration is $d$, 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 $\frac{41}{96}$
of the height of the cylinder.