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Geometry Difficulty 3.7 AMC 10/12 Find the answer Canada

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 rr and height hh is $π\$\pi
r^2h and the volume of a cone with radius randheight and height his is 13π\frac{1}{3}\pi r^2h$.)

Pick one

Solution

Suppose that the radius of the cylinder is rr and the height of the cylinder is hh.

This means that the volume of the cylinder is πr2h\pi r^2 h; the volume of half of the
cylinder is $12π\$\frac{1}{2}\pi r^2
h$.

Also, the radius of the cone is 12r\frac{1}{2}r and the height of the cone
is hh.

This means that the volume of the cone is $13π(12r)2\$\frac{1}{3}\pi \left(\frac{1}{2}r\right)^2
hor or 112π\frac{1}{12}\pi r^2
h$.

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 12\frac{1}{2} of those of the larger
cone.

This means that the volume of the top portion is (12)3=18\left(\frac{1}{2}\right)^3 = \frac{1}{8}
of that of the cone, which equals 18⋅112πr2h\frac{1}{8} \cdot \frac{1}{12}\pi r^2 h
or 196πr2h\frac{1}{96}\pi r^2 h.

To see this in another way, we note that this top portion of the cone
has height 12h\frac{1}{2}h and should
have radius $12⋅12r$\$\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 $13π(14r)2⋅12h\$\frac{1}{3}\pi \left(\frac{1}{4}r\right)^2 \cdot \frac{1}{2}hwhichisagain which is again 196π\frac{1}{96}\pi r^2 h$.

Using this information, the bottom portion of the cone has volume $78⋅112π\$\frac{7}{8} \cdot \frac{1}{12} \pi r^2 h =
796π\frac{7}{96} \pi r^2 h$.

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 $12π\$\frac{1}{2}\pi r^2 h - 796π\frac{7}{96}\pi
r^2 h = 4896π\frac{48}{96}\pi r^2 h - 796π\frac{7}{96}\pi r^2 h =
4196π\frac{41}{96}\pi r^2 h$.

When the cone is removed, the water then occupies a cylinder with radius
rr and volume 4196πr2h\frac{41}{96}\pi r^2 h.

If the depth of the water in this configuration is dd, then $π\$\pi
r^2 d = 4196π\frac{41}{96}\pi r^2 handso and so d = 4196h$,\frac{41}{96}h\$, which means that the
depth of the water is 4196\frac{41}{96}
of the height of the cylinder.

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