6. Three cones with vertex $A$ and slant height $\sqrt{8}$ touch each other externally. For two of the cones, the angle between the slant height and the axis of symmetry is $\frac{\pi}{6}$, while for the third cone, it is $\frac{\pi}{4}$. Find the volume of the pyramid $O_{1} O_{2} O_{3} A$, where $O_{1}, O_{2}, O_{3}$ are the centers of the bases of the cones.
Answer: $\sqrt{\sqrt{3}+1}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-04.jpg?height=488&width=1146&top_left_y=2020&top_left_x=520)

Solution. Let $\alpha=\frac{\pi}{6}, \beta=\frac{\pi}{4}, l=\sqrt{8}, B-$ be the midpoint of segment $O_{1} O_{2}, A D-$ be the common generatrix of the second and third cones. Then

$$
A O_{1}=A O_{2}=l \cos \alpha, \quad A O_{3}=l \cos \beta=l \frac{\sqrt{2}}{2}
$$

Quadrilateral $A O_{2} D O_{3}$ is inscribed in a circle with diameter $A D$, from which, by the Law of Sines,

$$
O_{1} O_{3}=O_{2} O_{3}=A D \cdot \sin \angle O_{2} A O_{3}=l \sin (\alpha+\beta)
$$

Similarly,

$$
O_{1} O_{2}=l \sin 2 \alpha \quad \text { and } \quad B O_{2}=l \sin \alpha \cos \alpha=l \frac{\sqrt{3}}{4} .
$$

Drop a perpendicular $B C$ to line $A O_{3}$. Note that segments $A B$ and $O_{3} B$ are perpendicular to $O_{1} O_{2}$ as medians of isosceles triangles $O_{1} O_{2} A$ and $O_{1} O_{2} O_{3}$. Therefore, segment $O_{1} O_{2}$ is perpendicular to the plane $A B O_{3}$ and, hence, to line $A O_{3}$. Therefore, edge $A O_{3}$ is perpendicular to the plane $O_{1} O_{2} C$ and, in particular, to segment $\mathrm{CO}_{2}$. Therefore,

$$
C O_{2}=A O_{2} \cdot \sin \angle O_{2} A C=l \cos \alpha \sin (\alpha+\beta)
$$

By the Pythagorean theorem, we get

$$
B C=\sqrt{C O_{2}^{2}-B O_{2}^{2}}=l \cos \alpha \sqrt{\sin ^{2}(\alpha+\beta)-\sin ^{2} \alpha}=l \cos \alpha \sqrt{\frac{\cos 2 \alpha-\cos 2(\alpha+\beta)}{2}}=\frac{l \sqrt{3} \sqrt{\sqrt{3}+1}}{4} .
$$

Let $V-$ be the volume of the pyramid $O_{1} O_{2} O_{3} A, A H-$ be its height. Then

$$
\begin{aligned}
V=\frac{1}{6} \cdot O_{1} O_{2} \cdot B O_{3} \cdot A H=\frac{1}{3} \cdot O_{1} O_{2} \cdot & S_{A B O_{3}}=\frac{1}{3} \cdot B O_{2} \cdot A O_{3} \cdot B C= \\
= & \frac{l^{3}}{3} \cdot \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{4} \cdot \sqrt{\sqrt{3}+1}=\frac{l^{3}}{16 \sqrt{2}} \cdot \sqrt{\sqrt{3}+1}=\sqrt{\sqrt{3}+1}
\end{aligned}
$$

## Variant 2