6. Two spheres with a radius of 12 are lying on a table, touching each other externally. A cone touches the table's surface and both spheres (externally). The distances from the vertex of the cone to the points where the spheres touch the table are 13. 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: $2 \operatorname{arcctg} \frac{4}{3}$ or $2 \operatorname{arcctg} 3$.
![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-35.jpg?height=490&width=1478&top_left_y=1966&top_left_x=310)

Solution. Let $O_{1}, O_{2}$ be the centers of the spheres, $A_{1}, A_{2}$ be the points of tangency of the spheres with the table, $C$ be the vertex of the cone, $2 \alpha$ be the angle at its vertex, $\varphi=\angle O_{1} C A_{1}$. By the condition $\operatorname{tg} \varphi=\frac{12}{13}$. 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 $\mathrm{O}_{1} O_{2}$ (see the left figure). The generators along which the cone touches the identical spheres lie in the planes $K C O_{1}$ and $K_{C} O_{2}$, from which

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

Moreover, $\mathrm{CO}_{1}=\mathrm{CO}_{2}$. Let's note several simple facts.

1) $H$ is the point of tangency of the 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 $NCK$ is perpendicular to $O_{1} O_{2}$. Indeed, triangle $O_{1} C O_{2}$ is isosceles, and $H$ is the midpoint of $O_{1} O_{2}$, from which $C H \perp O_{1} O_{2}$. It is also obvious that $K H \perp O_{1} O_{2}$.
3) If $M$ 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 both spheres. Then $\angle O_{1} C H=\angle O_{1} C A_{1}=\varphi$, and by 3)

$$
\cos \angle K C H=\frac{C H}{C K}=\frac{C M}{\cos \angle O_{1} C H}: \frac{C M}{\cos \angle K C O_{1}}=\frac{\cos (\varphi+\alpha)}{\cos \varphi}=\cos \alpha-\operatorname{tg} \varphi \sin \alpha=\cos \alpha-\frac{12}{13} \sin \alpha
$$

Let $\beta$ be the angle between $C H$ and the table. Since the segment $O_{1} O_{2}$ is parallel to the table, from 2) it follows that $\beta$ is the angle between $C H$ and the line of intersection of the plane $H C K$ with the table. Since $C H$ and $C A_{1}$ are segments of tangents to the sphere, we get

$$
\sin \beta=\frac{12}{C H}=\frac{12}{C A_{1}}=\frac{12}{13} \text { and } \cos \beta=\frac{5}{13} \text {. }
$$

There are two cases.

1) $\angle K C H=180^{\circ}-\alpha-\beta$ (see the middle figure). Then

$$
\cos \alpha-\frac{12}{13} \sin \alpha=\cos \angle K C H=-\cos (\alpha+\beta)=\sin \alpha \sin \beta-\cos \alpha \cos \beta=\frac{12}{13} \sin \alpha-\frac{5}{13} \cos \alpha
$$

from which $\operatorname{ctg} \alpha=\frac{4}{3}$.
2) $\angle K C H=\beta-\alpha$ (see the right figure). Then

$$
\cos \alpha-\frac{12}{13} \sin \alpha=\cos \angle K C H=\cos (\beta-\alpha)=\cos \alpha \cos \beta+\sin \alpha \sin \beta=\frac{5}{13} \cos \alpha+\frac{12}{13} \sin \alpha
$$

from which $\operatorname{ctg} \alpha=3$.

## Variant 9