6. Three cones with vertex $A$ touch each other externally, and the angles at the vertex of the first two are $\frac{\pi}{6}$. Each of the cones touches internally a fourth cone with vertex at point $A$ and an angle at the vertex of $\frac{\pi}{3}$. Find the angle at the vertex of the third cone. (The angle at the vertex of a cone is the angle between its generators in the axial section.)
Answer: $2 \operatorname{arcctg}(\sqrt{3}+4)$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-19.jpg?height=482&width=1038&top_left_y=1238&top_left_x=560)

Solution 1. Let $2 \beta$ be the desired angle, $\alpha=\frac{\pi}{12}$. Inscribing spheres with centers $O_{1}, O_{2}, O_{3}$ in the first three cones, which touch each other. Tangents drawn from $A$ to all spheres are equal, since any pair of spheres has a common tangent. 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 some sphere with center $O$, inscribed in the fourth cone. This sphere touches the other spheres (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. Note that $A O$ is the axis of symmetry of the fourth cone, and, therefore, the angle between $A O$ and any generatrix of this cone is equal to the angle at the vertex of two identical cones. Then $A O$ touches these cones, that is, it passes through the point $H$ of tangency of the spheres with centers at $O_{1}$ and $O_{2}$. Segments $A H$ and $O_{3} H$ are perpendicular to $O_{1} H$ as medians of isosceles triangles $O_{1} O_{2} A$ and $O_{1} O_{2} O_{3}$. Therefore, the segment $O_{1} H$ is perpendicular to the plane $A D O_{3}$ and, in particular, to the line $A O_{3}$. This means that $H$ and $O_{1}$, when projected onto the line $A O_{3}$, coincide at the same point $E$. Then

$$
A H \cdot \cos (2 \alpha-\beta)=A H \cdot \cos \angle H A E=A E=A O_{1} \cdot \cos (\alpha+\beta)=\frac{A B}{\cos \alpha} \cdot \cos (\alpha+\beta)
$$

Note that $A H=A B$ as tangents to the sphere with center at $O_{1}$. Therefore,

$$
\cos 2 \alpha \cos \beta+\sin 2 \alpha \sin \beta=\cos \beta-\operatorname{tg} \alpha \sin \beta \Longleftrightarrow \cos \beta(1-\cos 2 \alpha)=\sin \beta(\operatorname{tg} \alpha+\sin 2 \alpha)
$$

Since

$$
\operatorname{tg} \alpha+\sin 2 \alpha=\frac{\sin 2 \alpha}{1+\cos 2 \alpha}+\sin 2 \alpha=\sin 2 \alpha \cdot \frac{2+\cos 2 \alpha}{1+\cos 2 \alpha}
$$

we get

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

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-20.jpg?height=403&width=531&top_left_y=615&top_left_x=820)

Solution 2. Let $2 \beta$ be the desired angle, $\gamma=\frac{\pi}{6}$. Inscribing spheres with centers $O_{1}, O_{2}, O_{3}$ and radii $r, r, \rho$, which touch each other. Tangents drawn from $A$ to all spheres are equal, since any pair of spheres has a common tangent. 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 some sphere with center $O$ and radius $R$, inscribed in the fourth cone. This sphere touches the other spheres (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. Note that $A O$ is the axis of symmetry of the fourth cone, and, therefore, the angle between $A O$ and any generatrix of this cone is equal to the angle at the vertex of two identical cones. Then $A O$ touches these cones, that is, it passes through the point $H$ of tangency of the spheres with centers at $O_{1}$ and $O_{2}$. Segments $A H$ and $O_{3} H$ are perpendicular to $O_{1} H$ as medians of isosceles triangles $O_{1} O_{2} A$ and $O_{1} O_{2} O_{3}$. Therefore, the segment $O_{1} H$ is perpendicular to the plane $A D O_{3}$ and, in particular, to the line $H O_{3}$. Let $N$ be the projection of point $O_{3}$ onto the line $A O, x=R-r, y=R-\rho$. Then

$$
r=O_{1} H=x \cos \gamma, \quad N O_{3}=y \cos \gamma, \quad H N=|x-y| \sin \gamma
$$

By the Pythagorean theorem

$$
\begin{aligned}
& O_{1} O_{3}^{2}=O_{1} H^{2}+H O_{3}^{2}=O_{1} H^{2}+N O_{3}^{2}+H N^{2}=\left(x^{2}+y^{2}\right) \cos ^{2} \gamma+(x-y)^{2} \sin ^{2} \gamma= \\
& =\left(x^{2}+y^{2}-(x-y)^{2}\right) \cos ^{2} \gamma+(x-y)^{2}=2 x y \cos ^{2} \gamma+(x-y)^{2}=2 r y \cos \gamma+(\rho-r)^{2}
\end{aligned}
$$

On the other hand, $O_{1} O_{3}=\rho+r$, from which

$$
2 r y \cos \gamma=(\rho+r)^{2}-(\rho-r)^{2}=4 \rho r \quad \text { and } \quad 2 \rho=y \cos \gamma=(R-\rho) \cos \gamma .
$$

Note that $\rho=A D \cdot \operatorname{tg} \beta, R=A D \cdot \operatorname{tg} \gamma$. Therefore,

$$
2 \operatorname{tg} \beta=\sin \gamma-\operatorname{tg} \beta \cos \gamma \Longrightarrow \operatorname{ctg} \beta=\frac{2+\cos \gamma}{\sin \gamma}=4+\sqrt{3}
$$

## Variant 6