6. Two spheres are lying on a table, touching each other externally. A cone touches the table with its lateral surface and both spheres (externally). The vertex of the cone is located on the segment connecting the points of contact of the spheres with the table. It is known that the rays connecting the vertex of the cone with the centers of the spheres form equal angles with the table. Find the maximum possible 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: $2 \operatorname{arcctg} 2$.
![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-43.jpg?height=488&width=1244&top_left_y=1669&top_left_x=424)

Solution. Let $O_{1}$ and $O_{2}$ be the centers of the spheres, $R$ and $r$ be the radii of the spheres, $A_{1}$ and $A_{2}$ be the points of tangency of the spheres with the table, $C$ be the vertex of the cone, and $2 \alpha$ be the angle at the vertex. By the problem's condition, $\angle O_{1} C A_{1} = \angle O_{2} C A_{2}$, denote the common value of these angles 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

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

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,

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

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

From 1) with account of (*), 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,

$$
\sin \alpha = \cos \angle K C H = \operatorname{ctg} \varphi \cos \alpha - \sin \alpha = \frac{2 \sqrt{R r}}{R+r} \cos \alpha - \sin \alpha \Longleftrightarrow \operatorname{ctg} \alpha = \frac{R+r}{\sqrt{R r}}
$$

By the Cauchy inequality, $\operatorname{ctg} \alpha \geqslant 2$, that is, $\alpha \leqslant \operatorname{arcctg} 2$. Equality is achieved in the case $R = r$.