6. Three cones with vertex $A$ touch each other externally, with the first two being identical, and the third having an angle at the vertex equal to $\frac{\pi}{2}$. All cones touch the same plane passing through point $A$ and lie on the same side of it. Find the angle at the vertex of the first two cones. (The angle at the vertex of a cone is defined as the angle between its generators in the axial section.)
Answer: $2 \operatorname{arctg} \frac{4}{5}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-26.jpg?height=336&width=1124&top_left_y=2066&top_left_x=524)

Solution. Let $2 \alpha$ be the desired angle, $\beta=\frac{\pi}{4}, \Pi$ be the plane tangent to the cones. 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 $\sin \psi=\frac{r}{A H}=\operatorname{tg} \alpha$. Consider two cases.

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

$$
\begin{aligned}
& \cos (\pi-\beta-\psi)=\frac{\cos (\alpha+\beta)}{\cos \alpha} \Longleftrightarrow \sin \beta \sin \psi-\cos \beta \cos \psi=\cos \beta-\sin \beta \operatorname{tg} \alpha \Longleftrightarrow \\
& \Longleftrightarrow \cos \beta(1+\cos \psi)=2 \sin \beta \sin \psi \Longleftrightarrow \operatorname{tg} \frac{\psi}{2}=\frac{1}{2} \operatorname{ctg} \beta=\frac{1}{2}
\end{aligned}
$$

From this

$$
\operatorname{tg} \alpha=\sin \psi=\frac{2 \operatorname{tg} \frac{\psi}{2}}{1+\operatorname{tg}^{2} \frac{\psi}{2}}=\frac{4}{5}
$$

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

$$
\begin{aligned}
& \cos (\psi-\beta)=\frac{\cos (\alpha+\beta)}{\cos \alpha} \Longleftrightarrow \cos \beta \cos \psi+\sin \beta \sin \psi=\cos \beta-\sin \beta \operatorname{tg} \alpha \Longleftrightarrow \\
& \Longleftrightarrow \cos \beta(1-\cos \psi)=2 \sin \beta \sin \psi \Longleftrightarrow \operatorname{ctg} \frac{\psi}{2}=\frac{1}{2} \operatorname{ctg} \beta=\frac{1}{2}
\end{aligned}
$$

From this

$$
\operatorname{tg} \alpha=\sin \psi=\frac{2 \operatorname{ctg} \frac{\psi}{2}}{1+\operatorname{ctg}^{2} \frac{\psi}{2}}=\frac{4}{5}
$$

Thus, in both cases we get $2 \alpha=2 \operatorname{arctg} \frac{4}{5}$.

## Variant 8