12.240. The ratio of the volume of a sphere inscribed in a cone to the volume of a circumscribed sphere is $k$. Find the angle between the slant height of the cone and the plane of its base and the permissible values of $k$.
## Solution.

Let $\triangle A B C$ be the axial section of the cone, $D$ be the center of its base, $B D$ be the height, $O$ be the center of the inscribed sphere (Fig. 12.107), $\angle B A D$ be the angle we are looking for between the generatrix $B A$ of the cone and the plane of its base.

$\angle B A O=\alpha, r$ and $R$ are the radii of the inscribed and circumscribed spheres.

In $\triangle A D O\left(\angle A D O=90^{\circ}\right):$

$$
D A=D O \operatorname{ctg} \angle O A D=r \operatorname{ctg} \frac{\alpha}{2} \cdot A C=2 r \operatorname{ctg} \frac{\alpha}{2}
$$

The radius of the circle circumscribed around triangle $A B C$,

$$
R=\frac{A C}{2 \sin \angle A B C}=\frac{2 r \operatorname{ctg} \frac{\alpha}{2}}{2 \sin \left(180^{\circ}-2 \alpha\right)}=\frac{r \operatorname{ctg} \frac{\alpha}{2}}{\sin 2 \alpha}
$$

The volumes of the two spheres are in the ratio of the cubes of their radii.

Thus,

$$
\begin{aligned}
& \frac{r^{3}}{R^{3}}=k \Leftrightarrow \frac{r}{R}=\sqrt[3]{k} \Leftrightarrow \frac{r \sin 2 \alpha}{r \operatorname{ctg} \frac{\alpha}{2}}=\sqrt[3]{k} \Leftrightarrow \frac{2 \sin \alpha \cos \alpha \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}=\sqrt[3]{k} \Leftrightarrow \\
& \Leftrightarrow 4 \sin ^{2} \frac{\alpha}{2} \cos \alpha=\sqrt[3]{k} \Leftrightarrow 2(1-\cos \alpha) \cos \alpha=\sqrt[3]{k} \Leftrightarrow 2 \cos ^{2} \alpha- \\
& -2 \cos \alpha+\sqrt[3]{k}=0
\end{aligned}
$$

$\cos \alpha=\frac{1 \pm \sqrt{1-2 \sqrt[3]{k}}}{2} ; \alpha=\arccos \frac{1 \pm \sqrt{1-2 \sqrt[3]{k}}}{2}$

Since $k>0$ and $1-2 \sqrt[3]{k} \geq 0$, then $0<k \leq \frac{1}{8}$.

Answer: $\arccos \frac{1 \pm \sqrt{1-2 \sqrt[3]{k}}}{2}, 0<k \leq \frac{1}{8}$.