We inscribe a cone around a sphere. What is the maximum fraction of the cone's volume that the sphere can fill?
Let the radius of the base circle of the cone be denoted by $r$, its height by $m$, and the half-aperture angle of the cone by $\alpha$. Let the radius of the sphere be $R$.

$1986-12-440-1 . \mathrm{eps}$

Based on the diagram, we can write the following relationships:

$$
\begin{aligned}
& r=m \cdot \tan \alpha \\
& R=(m-R) \sin \alpha
\end{aligned}
$$

and from this,

$$
R=\frac{m \cdot \sin \alpha}{1+\sin \alpha}
$$

Let's write the volume $V_{g}$ of the sphere and the volume $V_{k}$ of the cone:

$$
V_{g}=\frac{4 \pi}{3} \cdot\left(\frac{m \cdot \sin \alpha}{1+\sin \alpha}\right)^{3} ; \quad V_{k}=\frac{(m \cdot \tan \alpha)^{2} \cdot \pi \cdot m}{3}
$$

The problem asks for the value of the ratio $\frac{V_{g}}{V_{k}}$. According to the above,

$$
\frac{V_{g}}{V_{k}}=\frac{\frac{4 \pi}{3} \cdot\left(\frac{m \cdot \sin \alpha}{1+\sin \alpha}\right)^{3}}{\frac{(m \cdot \tan \alpha)^{2} \cdot \pi \cdot m}{3}}=\frac{4 \cdot \sin \alpha(1-\sin \alpha)}{(1+\sin \alpha)^{2}}
$$

We can write this as follows:

$$
2 \cdot \frac{2 \sin \alpha}{1+\sin \alpha} \cdot \frac{1-\sin \alpha}{1+\sin \alpha}
$$

and notice that

$$
\frac{2 \cdot \sin \alpha}{1+\sin \alpha}+\frac{1-\sin \alpha}{1+\sin \alpha}=1
$$

Since $\alpha$ is an acute angle, the second and third factors in (1) are positive, and according to (2), the sum of these two factors is constant. By the well-known inequality

$$
a \cdot b \leq\left(\frac{a+b}{2}\right)^{2}
$$

we have

$$
\frac{2 \cdot \sin \alpha}{1+\sin \alpha} \cdot \frac{1-\sin \alpha}{1+\sin \alpha} \leq\left(\frac{1}{2}\right)^{2}
$$

and equality holds here, and thus (1) is maximized, when

$$
\frac{2 \cdot \sin \alpha}{1+\sin \alpha}=\frac{1-\sin \alpha}{1+\sin \alpha}
$$

From this, $\sin \alpha=\frac{1}{3}$, and in this case, $\frac{V_{g}}{V_{k}}=\frac{1}{2}$.

Therefore, a sphere can occupy at most half the volume of its circumscribed cone.

Remark. In the found case, the center of the sphere coincides with the centroid of the cone.