6. On the table, there are three spheres and a cone (with its base $k$ on the table), touching each other externally. The radii of the spheres are 20, 40, and 40, and the radius of the base of the cone is 21. Find the height of the cone.
Answer: 28.
![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-06.jpg?height=490&width=1298&top_left_y=1970&top_left_x=390)

Solution. Let $O$ be the center of the base of the cone, $h$ be its height, and $2 \alpha$ be the angle of inclination of the cone's generators to the table. Consider the section passing through the axis of symmetry of the cone and the center of one of the spheres (see the left figure). Let $K$ be the point of contact of the sphere with the table, and $R$ be the radius of the sphere. Then

$$
O K=R \cdot \operatorname{tg} \alpha+21
$$

Let $A, B, C$ be the points of contact of the spheres with the table (see the right figure). From the condition of the spheres touching, $B C=80$ and

$$
A B=A C=\sqrt{(40+20)^{2}-(40-20)^{2}}=40 \sqrt{2}
$$

Thus, the height $A D$ of the triangle $A B C$ is the perpendicular bisector of $B C$. From (*) it follows that the point $O$ is equidistant from $B$ and $C$, i.e., it lies on $A D$, and $O A=20 \operatorname{tg} \alpha+21$. Note that

$$
B D=40, \quad A D=\sqrt{A B^{2}-B D^{2}}=40, \quad O D=A D-O A=19-20 \operatorname{tg} \alpha
$$

The equality (*) gives $O B=40 \operatorname{tg} \alpha+21$, and by the Pythagorean theorem

$O B^{2}=O D^{2}+B D^{2} \Longleftrightarrow(40 \operatorname{tg} \alpha+21)^{2}=(19-20 \operatorname{tg} \alpha)^{2}+40^{2} \Longleftrightarrow 30 \operatorname{tg}^{2} \alpha+61 \operatorname{tg} \alpha-38=0 \Longleftrightarrow \operatorname{tg} \alpha=\frac{1}{2}$.

From this,

$$
\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{4}{3} \quad \text { and } \quad h=21 \cdot \operatorname{tg} 2 \alpha=28
$$

## Variant 3