6. Three identical cones with vertex $A$ touch each other externally. Each of them touches internally a fourth cone with vertex at point $A$ and vertex angle $\frac{2 \pi}{3}$. Find the vertex angle of the identical cones. (The vertex angle of a cone is the angle between its generators in the axial section.)
Answer: $2 \operatorname{arcctg} \frac{4+\sqrt{3}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-12.jpg?height=425&width=275&top_left_y=1729&top_left_x=959)

Solution. Let the desired angle be $2 \alpha$. Inscribed in the equal cones are spheres with centers $O_{1}, O_{2}, O_{3}$ of the same radius $r$. They obviously touch each other. Let the fourth cone touch these spheres at points $B, C, D$. Then $A B=A C=A D$ and, therefore, these points lie on a certain sphere with center $O$ and radius $R$, inscribed in the fourth cone. This sphere touches the spheres inscribed in the equal cones (for example, the spheres with centers at $O$ and $O_{1}$ touch at point $B$ in the plane containing the generatrix $A B$ and touching the fourth cone). Therefore, the points $O_{1}, O_{2}, O_{3}$ lie on the segments $O B, O C, O D$ respectively. Then $O O_{1}=O O_{2}=O O_{3}=R-r$. Moreover, the right triangles $A B O_{1}, A C O_{2}$, and $A D O_{3}$ are equal by two legs, so $A O_{1}=A O_{2}=A O_{3}$. Thus, the line $A O$ passes through the center $H$ of the circumscribed circle of the triangle $O_{1} O_{2} O_{3}$ perpendicular to its plane. Note that this triangle is equilateral and $O_{1} O_{2}=2 r$, hence $O_{1} H=\frac{2 r}{\sqrt{3}}$. On the other hand,

$$
O_{1} H=O O_{1} \cdot \cos \angle O O_{1} H=(R-r) \cos \frac{\pi}{3}=\frac{1}{2}(R-r)
$$

Since $R=A B \cdot \operatorname{tg} \frac{\pi}{3}=r \operatorname{ctg} \alpha \operatorname{tg} \frac{\pi}{3}=r \sqrt{3} \operatorname{ctg} \alpha$, we get

$$
\frac{2 r}{\sqrt{3}}=\frac{r}{2}(\sqrt{3} \operatorname{ctg} \alpha-1) \Longleftrightarrow \sqrt{3} \operatorname{ctg} \alpha=\frac{4}{\sqrt{3}}+1 \Longleftrightarrow 2 \alpha=2 \operatorname{arcctg} \frac{4+\sqrt{3}}{3}
$$

## Variant 4