6. On the table lie spheres with radii 2, 2, 1, touching each other externally. The vertex of the cone \( C \) is on the table, and the cone itself touches all the spheres externally. Point \( C \) is equidistant from the centers of the two equal spheres, and the cone touches the third sphere along a generatrix perpendicular to the table. Find the angle at the vertex of the cone. (The angle at the vertex of the cone is the angle between its generatrices in the axial section.)
Answer: $2 \operatorname{arctg} \frac{\sqrt{5}-2}{3}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-21.jpg?height=486&width=1520&top_left_y=899&top_left_x=290)

Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A_{1}, A_{2}, A_{3}$ be the points of tangency of the spheres with the table, $2 \alpha$ be the angle at the vertex of the cone, $\varphi=\angle O_{1} C A_{1}$. The line passing through $C$ perpendicular to the table touches the sphere with center $O_{3}$, hence $\angle A_{3} C O_{3}=45^{\circ}$ and $A_{3} C=1$. Let $A_{3} D$ be the height of the triangle $A_{1} A_{2} A_{3}$ (see the middle figure). From the condition of tangency of the spheres,

$$
A_{1} A_{3}=A_{2} A_{3}=\sqrt{(2+1)^{2}-(2-1)^{2}}=\sqrt{8}, \quad A_{1} D=2
$$

from which

$$
A_{3} D=\sqrt{A_{1} A_{3}^{2}-A_{1} D^{2}}=2, \quad \text { and } \quad C D=A_{3} D-A_{3} C=1
$$

Since $A_{1} C=2 \operatorname{ctg} \varphi$, by the Pythagorean theorem we get

$$
A_{1} C^{2}=A_{1} D^{2}+C D^{2} \Longleftrightarrow 4 \operatorname{ctg}^{2} \varphi=5 \Longleftrightarrow \operatorname{ctg} \varphi=\frac{\sqrt{5}}{2}
$$

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

$$
\angle K C O_{1}=\angle K C O_{2}=\varphi+\alpha .
$$

Moreover, by the condition $C O_{1}=C O_{2}$. Note several simple facts.

1) $H$ is the point of tangency of the equal spheres. Indeed, triangles $K C O_{1}$ and $K C O_{2}$ are equal, hence $K O_{1}=K O_{2}$. Since $K H \perp O_{1} O_{2}$, we get $H O_{1}=H O_{2}$.
2) The plane $H C K$ 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. Then $\angle O_{1} C H=\angle O_{1} C A_{1}=\varphi$, and by 3)

$$
\cos \gamma=\frac{C H}{C K}=\frac{C M}{\cos \angle O_{1} C H}: \frac{C M}{\cos \angle K C O_{1}}=\frac{\cos (\varphi+\alpha)}{\cos \varphi}=\cos \alpha-\operatorname{tg} \varphi \sin \alpha=\cos \alpha-\frac{2}{\sqrt{5}} \sin \alpha
$$

Since $C H$ and $C A_{1}$ are segments of tangents to the sphere, we get

$$
C H=C A_{1}=2 \operatorname{ctg} \varphi=\sqrt{5}, \quad \text { hence } \quad \sin \beta=\frac{2}{C H}=\frac{2}{\sqrt{5}} \text { and } \cos \beta=\frac{1}{\sqrt{5}}
$$

By 1) and 2) the plane $H C K$ consists of points equidistant from $O_{1}$ and $O_{2}$, so it contains the point $O_{3}$. Therefore, $\gamma=90^{\circ}-\beta-\alpha$ (see the right figure). Then

$$
\cos \alpha-\frac{2}{\sqrt{5}} \sin \alpha=\cos \gamma=\sin (\beta+\alpha)=\sin \beta \cos \alpha+\cos \beta \sin \alpha=\frac{\sin \alpha+2 \cos \alpha}{\sqrt{5}}
$$

from which $\operatorname{tg} \alpha=\frac{\sqrt{5}-2}{3}$.

## Variant 6