3.103. The ratio of the volume of a cone to the volume of a sphere inscribed in it is $k$. Find the angle between the slant height and the base plane of the cone and the permissible values of $k$.
3.103. Let's depict the axial section of the figure: $S B=S C$ - the generatrix of the cone, $S P$ - its height, $O$ the center of the inscribed sphere (Fig. 3.103), $V_{\text {cone }}: V_{\text {sphere }}=k$; we need to find $\angle S B P$. Let $A$ be the point of tangency of the sphere with the lateral surface of the cone. Let $O A=r, P B=R, S P=H$; then

$$
\frac{\frac{1}{3} \pi R^{2} H}{\frac{4}{3} \pi r^{3}}=3, \text { or } \frac{R^{2} H}{4 r^{3}}=k
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-568.jpg?height=404&width=452&top_left_y=811&top_left_x=701)

Fig. 3.103

Assuming $\angle S B P=\alpha$, from $\triangle S P B$ and $\triangle O P B$ we have $\frac{H}{R}=\operatorname{tg} \alpha, \frac{R}{r}=\operatorname{ctg} \frac{\alpha}{2}$.

But $\frac{R^{2} H}{4 r^{3}}=\frac{1}{4} \cdot \frac{R^{3}}{r^{3}} \cdot \frac{H}{R}=k$ and, therefore,

$$
\frac{1}{4} \operatorname{ctg}^{3} \frac{\alpha}{2} \operatorname{tg} \alpha=k ; \frac{1}{\operatorname{tg}^{3} \frac{\alpha}{2}} \cdot \frac{2 \operatorname{tg} \frac{\alpha}{2}}{4\left(1-\operatorname{tg}^{2} \frac{\alpha}{2}\right)}=k ; \frac{1}{2 \operatorname{tg}^{2} \frac{\alpha}{2}\left(1-\operatorname{tg}^{2} \frac{\alpha}{2}\right)}=\kappa .
$$

Let $\operatorname{tg}^{2} \frac{\alpha}{2}=x$; then $2 x-2 x^{2}-\frac{1}{k}=0 ; 2 k x^{2}-2 k x+1=0$. Hence,

$$
x=\frac{k \pm \sqrt{k^{2}-2 k}}{2 k} ; \operatorname{tg}^{2} \frac{\alpha}{2}=\frac{k \pm \sqrt{k^{2}-2 k}}{2 k} ; \operatorname{tg} \frac{\alpha}{2}=\sqrt{\frac{k \pm \sqrt{k^{2}-2 k}}{2 k}} ;
$$

since $k^{2}-2 k \geq 0$ and $k>0$, then $k \geq 2$. Therefore,

$$
\alpha=2 \operatorname{arctg} \sqrt{\frac{k \pm \sqrt{k^{2}-2 k}}{2 k}}, k \geq 2
$$