6. On the table lie two spheres with radii 4 and 1, with centers $O_{1}$ and $O_{2}$, touching each other externally. A cone touches the table's surface and both spheres (externally). The vertex $C$ of the cone is on the segment connecting the points of contact of the spheres with the table. It is known that the rays CO $_{1}$ and CO $_{2}$ form equal angles with the table. 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: $2 \operatorname{arctg} \frac{2}{5}$.

Solution. Let $A_{1}, A_{2}$ be the points of contact of the spheres with the table, and $2 \alpha$ be the angle at the vertex of the cone. By the problem's condition, the angles $O_{1} C A_{1}$ and $O_{2} C A_{2}$ are equal; denote their common value by $\varphi$. Note that the points $O_{1}, O_{2}, A_{1}, A_{2}, C$ lie in the same plane, since $O_{1} A_{1} \| O_{2} A_{2}$, and $C$ lies on the segment $A_{1} A_{2}$. Then

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

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

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

Drop perpendiculars $K M$ and $K N$ from point $K$ to $C O_{1}$ and $C O_{2}$, respectively. The right triangles $K C M$ and $K C N$ are equal, so $C M=C N$. Note several simple facts.

1) $\mathrm{MH} \perp \mathrm{CO}_{1}$ and $\mathrm{NH} \perp \mathrm{CO}_{2}$. This follows from the converse of the theorem of three perpendiculars.
2) $\angle M C H=\angle N C H$. Indeed, by 1) the right triangles $M C H$ and $N C H$ are equal by hypotenuse and leg, so their corresponding angles are equal. Denote the common value of angles $M C H$ and $N C H$ by $\psi$.
3) The line $C H$ is perpendicular to the table. Indeed,

$$
180^{\circ}=2 \varphi+2 \psi, \quad \text { from which } \quad \angle A_{1} C H=\varphi+\psi=90^{\circ} \text {. }
$$

Therefore, the line $C H$ is parallel to $A_{1} O_{1}$ and, hence, perpendicular to the table.

From 1) with account for $(*)$ we get

$$
\cos \angle K C H=\frac{C H}{C K}=\frac{C M}{\cos \angle O_{1} C H}: \frac{C M}{\cos \angle K C O_{1}}=\frac{\cos (\varphi+\alpha)}{\cos \psi}=\frac{\cos (\varphi+\alpha)}{\sin \varphi}=\operatorname{ctg} \varphi \cos \alpha-\sin \alpha
$$

The ray $C K$ forms an angle $\alpha$ with the table, and from 3) it follows that $\angle K C H=90^{\circ}-\alpha$. Therefore

$$
\sin \alpha=\cos \angle K C H=\operatorname{ctg} \varphi \cos \alpha-\sin \alpha=\frac{4}{5} \cos \alpha-\sin \alpha \Longleftrightarrow \operatorname{tg} \alpha=\frac{2}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-40.jpg?height=486&width=1244&top_left_y=1502&top_left_x=424)