## Task A-3.4.

Determine the maximum possible value of the ratio of the volume of a sphere to the volume of a right circular cone circumscribed about it.
## First Solution.

In the figure, a cross-section of a sphere and a right circular cone circumscribed around it is shown.

Let's mark the points as in the figure, and let $|O D|=r, |P C|=R, |A P|=h$.

Since

$$
\begin{aligned}
& \varangle O A D=\varangle C A P \text{ and} \\
& \varangle A D O=\varangle A P C=90^{\circ},
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_2903a2297dc3986740b2g-6.jpg?height=717&width=716&top_left_y=264&top_left_x=1087)

triangles $A D O$ and $A P C$ are similar.

From this, it follows that $\frac{|A O|}{|D O|}=\frac{|A C|}{|P C|}$,

or $\frac{h-r}{r}=\frac{\sqrt{h^{2}+R^{2}}}{R}$.

After squaring, we get $\frac{h^{2}-2 h r+r^{2}}{r^{2}}=\frac{h^{2}+R^{2}}{R^{2}}$

from which it follows that $h^{2} R^{2}-2 h r R^{2}=h^{2} r^{2}$, and finally $h=\frac{2 r R^{2}}{R^{2}-r^{2}}$.

The observed ratio is

$$
\begin{aligned}
\frac{V_{K}}{V_{S}} & =\frac{\frac{4}{3} r^{3} \pi}{\frac{1}{3} R^{2} \pi \cdot h}=\frac{4 r^{3}}{R^{2} h} \\
& =\frac{4 r^{3}}{R^{2}} \cdot \frac{R^{2}-r^{2}}{2 r R^{2}}=\frac{2 r^{2}\left(R^{2}-r^{2}\right)}{R^{4}} \\
& =-2\left(\frac{r}{R}\right)^{4}+2\left(\frac{r}{R}\right)^{2} \\
& =\frac{1}{2}-2\left(\left(\frac{r}{R}\right)^{2}-\frac{1}{2}\right)^{2}
\end{aligned}
$$

so the desired maximum is $\frac{1}{2}$ (and it is achieved for $R=r \sqrt{2}$).