6. On the table lie spheres with radii 2, 2, 1, touching each other externally. The vertex of the cone is located midway between the points of contact of the identical spheres with the table, and the cone itself touches all the spheres externally. Find the angle at the vertex of the cone. (The angle at the vertex of the cone is the angle between its generators in the axial section.)
Answer: $2 \operatorname{arcctg} 8$.
![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-11.jpg?height=498&width=1302&top_left_y=776&top_left_x=390)

Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A_{1}, A_{2}, A_{3}$ be the points of contact of the spheres with the table, $C$ be the vertex of the cone, $2 \alpha$ be the angle at its vertex, $\varphi=\angle O_{1} C A_{1}, \psi=\angle O_{3} C A_{3}$. From the condition of the spheres touching,

$$
C A_{1}=2 \quad \text { and } \quad A_{1} A_{3}=A_{2} A_{3}=\sqrt{(2+1)^{2}-(2-1)^{2}}=\sqrt{8} .
$$

Then

$$
C A_{3}=\sqrt{A_{1} A_{3}^{2}-C A_{1}^{2}}=2 \quad \text { and } \quad \operatorname{tg} \psi=\frac{O_{3} A_{3}}{C A_{3}}=\frac{1}{2}
$$

Moreover, $O_{1} A_{1}=2=C A_{1}$, from which $\varphi=45^{\circ}$.

Choose a point $K$ on the axis of symmetry of the cone such that its projection $H$ onto the plane $C O_{1} O_{2}$ falls on the segment $O_{1} O_{2}$ (see the left figure). The generators along which the cone touches the identical spheres lie in the planes $K C O_{1}$ and $K C O_{2}$, hence

$$
\angle K C O_{1}=\angle K C O_{2}=\varphi+\alpha=45^{\circ}+\alpha
$$

Moreover, $C O_{1}=2 \sqrt{2}=C O_{2}$. Let's note a few simple facts.

1) $H$ is the point of contact of the equal spheres. Indeed, triangles $K C O_{1}$ and $K C O_{2}$ are equal, from which $K O_{1}=K O_{2}$. Since $K H \perp O_{1} O_{2}$, we get $H O_{1}=H O_{2}$.
2) The plane $K C H$ is perpendicular to $O_{1} O_{2}$. Indeed, triangle $O_{1} C O_{2}$ is isosceles, and $H$ is the midpoint of $O_{1} O_{2}$, hence $C H \perp O_{1} O_{2}$. It is also obvious that $K H \perp O_{1} O_{2}$.
3) If $K M$ is the perpendicular dropped from point $K$ to $C O_{1}$, then $M H \perp C O_{1}$. This follows from the converse of the theorem of three perpendiculars.

Let $K M$ be the perpendicular dropped from point $K$ to $C O_{1}$. From 1) and 2) it follows that the line $C H$ touches the equal spheres, so $\angle O_{1} C H=\angle O_{1} C A_{1}=45^{\circ}$. Then, by 3),

$$
\cos \angle H C K=\frac{C H}{C K}=\frac{C M}{\cos \angle O_{1} C H}: \frac{C M}{\cos \angle K C O_{1}}=\frac{\cos \angle K C O_{1}}{\cos \angle O_{1} C H}=\frac{\cos \left(45^{\circ}+\alpha\right)}{\cos 45^{\circ}}=\cos \alpha-\sin \alpha
$$

On the other hand, by 1) and 2) the plane $H C K$ consists of points equidistant from $O_{1}$ and $O_{2}$. Therefore, $H C K$ contains the point $O_{3}$. Then (see the right figure)

$$
\cos \alpha-\sin \alpha=\cos \angle H C K=\cos \left(90^{\circ}-2 \psi-\alpha\right)=\sin (2 \psi+\alpha)=\sin 2 \psi \cos \alpha+\cos 2 \psi \sin \alpha
$$

from which

$$
\operatorname{tg} \alpha=\frac{1-\sin 2 \psi}{1+\cos 2 \psi}=\frac{(\sin \psi-\cos \psi)^{2}}{2 \cos ^{2} \psi}=\frac{(\operatorname{tg} \psi-1)^{2}}{2}=\frac{1}{8}
$$

## Variant 4