6. Three cones with vertex $A$ touch each other externally, and the angles at the vertex of the first two are $\frac{\pi}{3}$. All cones also touch a plane passing through point $A$, and they lie on the same side of it. Find the angle at the vertex of the third cone. (The angle at the vertex of a cone is defined as the angle between its generators in the axial section.)
Answer: $2 \operatorname{arcctg} 2(\sqrt{3} \pm \sqrt{2})$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-30.jpg?height=332&width=1124&top_left_y=1384&top_left_x=524)

Solution. Let $2 \beta$ be the desired angle, $\alpha=\frac{\pi}{6}, \Pi$ be the plane that the cones touch. Inscribed in the cones are spheres with centers $O_{1}, O_{2}, O_{3}$, touching each other. Denote by $r$ the radius of the first two spheres, and by $H$ the point of their tangency. Since both spheres touch the plane $\Pi$, the distance from $O_{1}$ and $O_{2}$ to $\Pi$ is $r$. Therefore, $O_{1} O_{2} \| \Pi$, so the distance from $H$ to $\Pi$ is also $r$. Let $\varphi=\angle H A O_{3}$, $\psi$ be the angle between the ray $A H$ and the plane $\Pi$. The segments $A H$ and $O_{3} H$ are perpendicular to $O_{1} H$ as medians of isosceles triangles $O_{1} O_{2} A$ and $O_{1} O_{2} O_{3}$. Thus, the segment $O_{1} H$ is perpendicular to the plane $H A O_{3}$ and, in particular, to the line $A O_{3}$. Therefore, the projections of the segments $A H$ and $A O_{1}$ onto the line $A O_{3}$ are equal, from which we have

$$
A H \cdot \cos \varphi=A O_{1} \cdot \cos (\alpha+\beta)=\frac{A H}{\cos \alpha} \cdot \cos (\alpha+\beta)
$$

Note also that

$$
\sin \psi=\frac{r}{A H}=\operatorname{tg} \alpha=\frac{1}{\sqrt{3}}, \quad \cos \psi=\sqrt{\frac{2}{3}} .
$$

Consider two cases.

1) $\varphi=\pi-\beta-\psi$ (see the left figure). Then

$$
\frac{\cos (\alpha+\beta)}{\cos \alpha}=\cos (\pi-\beta-\psi) \Longleftrightarrow \cos \beta-\sin \beta \sin \psi=\cos \beta-\sin \beta \operatorname{tg} \alpha=\sin \beta \sin \psi-\cos \beta \cos \psi
$$

from which

$$
\operatorname{ctg} \beta=\frac{2 \sin \psi}{1+\cos \psi}=\frac{2}{\sqrt{3}+\sqrt{2}}=2(\sqrt{3}-\sqrt{2}) \quad \text { and } \quad 2 \beta=2 \operatorname{arcctg} 2(\sqrt{3}-\sqrt{2})
$$

2) $\varphi=\psi-\beta$ (see the right figure). Then

$$
\frac{\cos (\alpha+\beta)}{\cos \alpha}=\cos (\psi-\beta) \Longleftrightarrow \cos \beta-\sin \beta \sin \psi=\cos \beta-\sin \beta \operatorname{tg} \alpha=\cos \beta \cos \psi+\sin \beta \sin \psi
$$

from which

$$
\operatorname{ctg} \beta=\frac{2 \sin \psi}{1-\cos \psi}=\frac{2}{\sqrt{3}-\sqrt{2}}=2(\sqrt{3}+\sqrt{2}) \quad \text { and } \quad 2 \beta=2 \operatorname{arcctg} 2(\sqrt{3}+\sqrt{2})
$$

## Variant 9