155. $n$ equal spheres of radius $R$ touch the lateral surface from the inside and the plane of the base of a cone, and each sphere touches two adjacent ones; $n$ spheres of radius $2R$ are arranged similarly, touching the lateral surface from the outside. Find the volume of the cone.
155. Without limiting the generality, we will assume that all the generators of the cone, touching the spheres, touch two spheres simultaneously - the inner and the outer one. Let's make a section through the vertex of the cone $S$ and the centers of two spheres touching one generator

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-094.jpg?height=279&width=559&top_left_y=823&top_left_x=344)

Fig. 34.

(Fig. 34, the notation is clear from the figure). From the condition that $n$ spheres of radius $R$ touch each other, it follows that $|0 A|=\frac{R}{\sin \frac{\pi}{n}}$, similarly $|O B|=\frac{2 R}{\sin \frac{\pi}{n}}$. Therefore, $|A B|=a=\frac{R}{\sin \frac{\pi}{n}}$.

Let $|A C|=x$. Then $\operatorname{tg} \alpha=\frac{R}{x}, \operatorname{ctg} \alpha=\frac{2 R}{a-x}$. Multiplying these equations, we get the equation for $x$:

$$
x^{2}-a x+2 R^{2}=0
$$

from which $x_{1}=\frac{a-\sqrt{a^{2}-8 R^{2}}}{2}, x_{2}=\frac{a+\sqrt{a^{2}-8} R^{\bar{i}}}{2}$, where $a=$ $=\frac{R}{\sin \frac{\pi}{n}}$.

The condition $a^{2}-8 R^{2} \geqslant 0$ leads to the inequality $\sin \frac{\pi}{n} \leqslant \frac{1}{2} \frac{1}{\sqrt{2}}$. In addition, the inequality $\operatorname{tg} \alpha=\frac{R}{x}<1$ must be satisfied. Now it is not difficult to obtain that the root $x_{1}$ is suitable if $\frac{1}{3}<\sin \frac{\pi}{n} \leqslant$ $\leqslant \frac{1}{2 \sqrt{2}}$. For the root $x_{2}$, there remains one restriction: $\sin \frac{\pi}{n} \leqslant \frac{1}{2 \sqrt{2}}$.

It can be proven that $\frac{1}{3}<\sin \frac{\pi}{n} \leqslant \frac{1}{2 \sqrt{2}}$ only for $n=0$.

The volume of the cone will be equal to $\frac{1}{3} \pi(a+x)^{3} \operatorname{tg} 2 \alpha$. Expressing $a, x$ and $\operatorname{tg} 2 \alpha$ by the corresponding formulas through $R$ and $n$, we get

Answer:

$$
V=\frac{\pi R^{3}\left(3+\sqrt{1-8 \sin ^{2} \frac{\pi}{n}}\right)^{3}\left(1+\sqrt{1-8 \sin ^{2} \frac{\pi}{n}}\right)}{12 \sin ^{2} \frac{\pi}{n}\left(1-6 \sin ^{2} \frac{\pi}{n}+\sqrt{1-8 \sin ^{2} \frac{\pi}{n}}\right)},
$$

In addition, for $n=9$ there is another value:

$$
\frac{\pi R^{3}\left(3-\sqrt{1-8 \sin ^{2} \frac{\pi}{9}}\right)^{3}\left(1-\sqrt{1-8 \sin ^{2} \frac{\pi}{9}}\right)}{12 \sin ^{2} \frac{\pi}{9}\left(1-6 \sin ^{2} \frac{\pi}{9}-\sqrt{1-8 \sin ^{2} \frac{\pi}{9}}\right)}
$$