6. Three cones with vertex $A$ touch each other externally, and the first two are identical, while the angle at the vertex of the third cone is $2 \arcsin \frac{1}{4}$. Each of the cones touches the fourth cone internally, with the vertex at point $A$. Find the angle at the vertex of the first two cones, if it is half the angle at the vertex of the fourth cone. (The angle at the vertex of a cone is defined as the angle between its generators in the axial section.)
Answer: $\frac{\pi}{6}+\arcsin \frac{1}{4}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-23.jpg?height=480&width=1034&top_left_y=891&top_left_x=561)

Solution. Let $2 \alpha$ be the desired angle, $\beta=\arcsin \frac{1}{4}$. 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 the same, 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 a certain 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 $B$ on 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 $A O_{3}$. Thus, $H$ 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

$$
\begin{aligned}
\cos \beta\left(1-\cos ^{2} 2 \alpha\right)=\sin \beta \sin 2 \alpha & (2+\cos 2 \alpha) \Longleftrightarrow \cos \beta \sin 2 \alpha=\sin \beta(2+\cos 2 \alpha) \Longleftrightarrow \\
& \Longleftrightarrow \sin (2 \alpha-\beta)=2 \sin \beta \Longleftrightarrow 2 \alpha=\arcsin (2 \sin \beta)+\beta=\frac{\pi}{6}+\arcsin \frac{1}{4}
\end{aligned}
$$

## Variant 7