A right circular cone contains two spheres, as shown. The radius of the larger sphere is 2 times the radius of the smaller sphere. Each sphere is tangent to the other sphere and to the lateral surface of the cone. The larger sphere is tangent to the cone's circular base. Determine the fraction of the cone's volume that is not occupied by the two spheres.

![](https://cdn.mathpix.com/cropped/2024_04_17_56af4dba662dd63eccc5g-3.jpg?height=353&width=249&top_left_y=913&top_left_x=1464)
Let the radius of the small sphere be $r$ and the radius of the large sphere be $2 r$.

Draw a vertical cross-section through the centre of the top face of the cone and its bottom vertex.

By symmetry, this will pass through the centres of the spheres.

In the cross-section, the cone becomes a triangle and the spheres become circles.

![](https://cdn.mathpix.com/cropped/2024_04_30_be08db732f430ab57efag-12.jpg?height=542&width=398&top_left_y=1721&top_left_x=923)

We label the vertices of the triangle $A, B, C$.

We label the centres of the large circle and small circle $L$ and $S$, respectively.

We label the point where the circles touch $U$.

We label the midpoint of $A B$ (which represents the centre of the top face of the cone) as $M$. Join $L$ and $S$ to the points of tangency of the circles to $A C$. We call these points $P$ and $Q$. Since $L P$ and $S Q$ are radii, then they are perpendicular to the tangent line $A C$ at $P$ and $Q$,
respectively.

Draw a perpendicular from $S$ to $T$ on $L P$.

The volume of the cone equals $\frac{1}{3} \pi \cdot A M^{2} \cdot M C$. We determine the lengths of $A M$ and $M C$ in terms of $r$.

Since the radii of the small circle is $r$, then $Q S=U S=r$.

Since TPQS has three right angles (at $T, P$ and $Q$ ), then it has four right angles, and so is a rectangle.

Therefore, $P T=Q S=r$.

Since the radius of the large circle is $2 r$, then $P L=U L=M L=2 r$.

Therefore, $T L=P L-P T=2 r-r=r$.

Since $M C$ passes through $L$ and $S$, it also passes through $U$, the point of tangency of the two circles.

Therefore, $L S=U L+U S=2 r+r=3 r$.

By the Pythagorean Theorem in $\triangle L T S$,

$$
T S=\sqrt{L S^{2}-T L^{2}}=\sqrt{(3 r)^{2}-r^{2}}=\sqrt{8 r^{2}}=2 \sqrt{2} r
$$

since $T S>0$ and $r>0$.

Consider $\triangle L T S$ and $\triangle S Q C$.

Each is right-angled, $\angle T L S=\angle Q S C$ (because $L P$ and $Q S$ are parallel), and $T L=Q S$.

Therefore, $\triangle L T S$ is congruent to $\triangle S Q C$.

Thus, $S C=L S=3 r$ and $Q C=T S=2 \sqrt{2} r$.

This tells us that $M C=M L+L S+S C=2 r+3 r+3 r=8 r$.

Also, $\triangle A M C$ is similar to $\triangle S Q C$, since each is right-angled and they have a common angle at $C$.

Therefore, $\frac{A M}{M C}=\frac{Q S}{Q C}$ and so $A M=\frac{8 r \cdot r}{2 \sqrt{2} r}=2 \sqrt{2} r$.

This means that the volume of the original cone is $\frac{1}{3} \pi \cdot A M^{2} \cdot M C=\frac{1}{3} \pi(2 \sqrt{2} r)^{2}(8 r)=\frac{64}{3} \pi r^{3}$. The volume of the large sphere is $\frac{4}{3} \pi(2 r)^{3}=\frac{32}{3} \pi r^{3}$.

The volume of the small sphere is $\frac{4}{3} \pi r^{3}$.

The volume of the cone not occupied by the spheres is $\frac{64}{3} \pi r^{3}-\frac{32}{3} \pi r^{3}-\frac{4}{3} \pi r^{3}=\frac{28}{3} \pi r^{3}$.

The fraction of the volume of the cone that this represents is $\frac{\frac{28}{3} \pi r^{3}}{\frac{64}{3} \pi r^{3}}=\frac{28}{64}=\frac{7}{16}$.

ANSWER: $\frac{7}{16}$