6. Three cones with vertex $A$ touch each other externally, and the first two are identical, while the third has an angle at the vertex of $\frac{\pi}{4}$. Each of the cones touches internally a fourth cone with vertex at point $A$ and an angle at the vertex of $\frac{3 \pi}{4}$. 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: $2 \operatorname{arctg} \frac{2}{3}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-38.jpg?height=492&width=1472&top_left_y=1291&top_left_x=360)

Solution. Let $2 \alpha$ be the desired angle, $\beta=\frac{\pi}{8}, \gamma=\frac{3 \pi}{8}$. 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 have the same length, 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 $B$ in the plane containing the generatrix $A B$ and tangent to the fourth cone). Therefore, the points $O_{1}, O_{2}, O_{3}$ lie on the segments $O B, O C, O D$ respectively, and $O O_{1}=O O_{2}$. Let $H$ be the point of tangency of the spheres with centers $O_{1}$ and $O_{2}, \varphi=\angle H A D$. The segments $A H, O H$ and $O_{3} H$ are perpendicular to $O_{1} H$ as medians of isosceles triangles $O_{1} O_{2} A, O_{1} O_{2} O$ and $O_{1} O_{2} O_{3}$. Therefore, the points $A, O, O_{3}, D$ lie in a plane passing through point $H$ perpendicular to $O_{1} H$. In particular, $A O \perp O_{1} H$ and $A O_{3} \perp O_{1} H$. Thus, points $H$ and $O_{1}$ have the same projections on the line $A O$ and on the line $A O_{3}$ (denote these projections by $E$ and $F$ respectively). Hence

$$
\begin{aligned}
& A H \cdot \cos (\varphi-\gamma)=A H \cdot \cos \angle H A E=A E=A O_{1} \cdot \cos \angle O_{1} A E=A O_{1} \cdot \cos (\gamma-\alpha)=\frac{A B}{\cos \alpha} \cdot \cos (\gamma-\alpha) \\
& A H \cdot \cos (\varphi-\beta)=A H \cdot \cos \angle H A F=A F=A O_{1} \cdot \cos \angle O_{1} A F=A O_{1} \cdot \cos (\alpha+\beta)=\frac{A B}{\cos \alpha} \cdot \cos (\alpha+\beta)
\end{aligned}
$$

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

$$
\left\{\begin{array} { l } 
{ \operatorname { c o s } \varphi \operatorname { c o s } \gamma + \operatorname { s i n } \varphi \operatorname { s i n } \gamma = \operatorname { c o s } \gamma + \operatorname { t g } \alpha \operatorname { s i n } \gamma , } \\
{ \operatorname { c o s } \varphi \operatorname { c o s } \beta + \operatorname { s i n } \varphi \operatorname { s i n } \beta = \operatorname { c o s } \beta - \operatorname { t g } \alpha \operatorname { s i n } \beta }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
\operatorname{tg} \alpha=\operatorname{ctg} \gamma(\cos \varphi-1)+\sin \varphi \\
\operatorname{tg} \alpha=\operatorname{ctg} \beta(1-\cos \varphi)-\sin \varphi
\end{array}\right.\right.
$$

Excluding $\operatorname{tg} \alpha$ from the system, we get $2 \sin \varphi=(1-\cos \varphi)(\operatorname{ctg} \beta+\operatorname{ctg} \varphi)$, from which

$$
\operatorname{ctg} \frac{\varphi}{2}=\frac{\sin \varphi}{1-\cos \varphi}=\frac{1}{2}(\operatorname{ctg} \beta+\operatorname{ctg} \gamma)=\sqrt{2}
$$

since

$$
\operatorname{ctg} \beta=\frac{1+\cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}=\sqrt{2}+1, \quad \operatorname{ctg} \gamma=\frac{1+\cos \frac{3 \pi}{4}}{\sin \frac{3 \pi}{4}}=\sqrt{2}-1
$$

Now from the second equation of the system

$$
\operatorname{tg} \alpha=(1-\cos \varphi)\left(\operatorname{ctg} \beta-\operatorname{ctg} \frac{\varphi}{2}\right)=\frac{2\left(\operatorname{ctg} \beta-\operatorname{ctg} \frac{\varphi}{2}\right)}{1+\operatorname{ctg}^{2} \frac{\varphi}{2}}=\frac{2(\sqrt{2}+1-\sqrt{2})}{3}=\frac{2}{3} \Longrightarrow 2 \alpha=2 \operatorname{arctg} \frac{2}{3}
$$