A cone is filled with water. Two solid spheres are placed in the cone as shown in the diagram and water spills out. (The spheres are touching each other, each sphere touches the cone all of the way around, and the top of the top sphere is level with the top of the cone.) The larger sphere has radius twice that of the smaller sphere. If the volume of the water remaining in the cone is $2016 \pi$, what is the radius of the smaller sphere? (The volume of a sphere with radius $r$ is $\frac{4}{3} \pi r^{3}$. The volume of a cone with radius $r$ and height $h$ is $\frac{1}{3} \pi r^{2} h$.)
(A) $2 \sqrt{2}$
(B) 6
(C) 8
(D) $6 \sqrt{2}$
(E) $4 \sqrt[3]{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-065.jpg?height=352&width=198&top_left_y=1445&top_left_x=1343)
Let the radius of the smaller sphere be $r$.

Thus, the radius of the larger sphere is $2 r$.

We determine expressions for the height and radius of the cone in terms of $r$ and use these to help solve the problem.

By symmetry, the centres of the two spheres $(Q$ of the smaller sphere and $O$ of the larger sphere) lie on the line joining the centre of the circular top of the cone $(C)$ to the tip of the cone $(P)$.

Draw a vertical cross-section of the cone through the centre of the circular top of the cone and through the tip of the cone.

Each such cross-section will be an identical triangle.

Because the centres of the spheres lie on a line which is in the plane of this cross-section, the cross-section of each sphere will be a "great" circle (that is, the largest possible circular cross-section

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-128.jpg?height=605&width=440&top_left_y=825&top_left_x=1493)
of the sphere).

Because the top of the larger sphere is just level with the top of the cone, then the sphere "touches" the circular top of the cone at its centre $C$.

Finally, because the spheres touch the cone all the way around, then the circles will be tangent to the triangle in the cross-section.

We label the triangular cross section as $A B P$.

Note that $C P$ is perpendicular to $A B$ at $C$.

Draw radii from $O$ and $Q$ to the points $T$ and $U$, respectively, on $A P$ where the circles with centre $O$ and $Q$ are tangent to $A P$.

Note that $O T$ and $Q U$ are perpendicular to $A P$, with $O T=2 r$ and $Q U=r$.

Also, since the two circles are just touching, then the line segment joining their centres, $O Q$, passes through this point of tangency, and so $O Q=2 r+r=3 r$.

Now $\triangle O T P$ is similar to $\triangle Q U P$, since each is right-angled and they share a common angle at $P$.

Since $\frac{O T}{Q U}=\frac{2 r}{r}=2$, then $\frac{O P}{Q P}=2$ or $O P=2 Q P$.

Since $O P=O Q+Q P=3 r+Q P$, then $3 r+Q P=2 Q P$ or $Q P=3 r$.

Thus, the height of the cone is $C P=C O+O Q+Q P=2 r+3 r+3 r=8 r$.

(Note that $C O$ is a radius of the larger circle, so $C O=2 r$.)

We also see that $\triangle A C P$ is similar to $\triangle Q U P$, since $\triangle A C P$ is also right-angled (at $C$ ) and shares the angle at $P$.

Thus, $\frac{A C}{C P}=\frac{Q U}{U P}$.

We know that $C P=8 r$ and $Q U=r$.

To calculate $U P$, we use the Pythagorean Theorem to get

$$
U P=\sqrt{Q P^{2}-Q U^{2}}=\sqrt{(3 r)^{2}-r^{2}}=\sqrt{8 r^{2}}=\sqrt{8} r
$$

since $r>0$.

Therefore, $A C=\frac{C P \cdot Q U}{U P}=\frac{8 r \cdot r}{\sqrt{8} r}=\sqrt{8} r$.

Finally, we can use the given information. We are told that the volume of water remaining after the full cone has the two spheres added is $2016 \pi$. This is equivalent to saying that the difference between the volume of the cone and the combined volumes of the spheres is $2016 \pi$. Using the given volume formulae, we obtain

$$
\begin{aligned}
\frac{1}{3} \pi(A C)^{2}(C P)-\frac{4}{3} \pi(Q U)^{3}-\frac{4}{3} \pi(O T)^{3} & =2016 \pi \\
\frac{1}{3} \pi(\sqrt{8} r)^{2}(8 r)-\frac{4}{3} \pi r^{3}-\frac{4}{3} \pi(2 r)^{3} & =2016 \pi \\
64 \pi r^{3}-4 \pi r^{3}-32 \pi r^{3} & =6048 \pi \\
28 r^{3} & =6048 \\
r^{3} & =216 \\
r & =6
\end{aligned}
$$

Therefore, the radius of the smaller sphere is 6 .

ANSWER: (B)