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 5, 4, and 4, and the height of the cone is to the radius of its base as 4:3. Find the radius of the base of the cone.
# Answer: $\frac{169}{60}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-03.jpg?height=448&width=554&top_left_y=1078&top_left_x=388)

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-03.jpg?height=474&width=468&top_left_y=1051&top_left_x=1205)

Solution. Let $O$ be the center of the base of the cone, $r$ be its radius, and $2 \alpha$ be the angle between the generatrix of the cone and the table. According to the problem, $\operatorname{tg} 2 \alpha = \frac{4}{3}$, from which we have

$$
\sin 2 \alpha = \frac{4}{5}, \quad \cos 2 \alpha = \frac{3}{5}, \quad \operatorname{tg} \alpha = \frac{1 - \cos 2 \alpha}{\sin 2 \alpha} = \frac{1}{2}
$$

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 tangency of the sphere with the table, and $R$ be the radius of the sphere. Then

$$
O K = R \cdot \operatorname{tg} \alpha + r = \frac{1}{2} R + r
$$

Let $A, B, C$ be the points of tangency of the spheres with the table (see the right figure). From the condition of tangency, $B C = 8$ and

$$
A B = A C = \sqrt{(5+4)^{2} - (5-4)^{2}} = \sqrt{80}
$$

Thus, the height $A D$ of triangle $A B C$ is the perpendicular bisector of $B C$. From (*), it follows that point $O$ is equidistant from $B$ and $C$, meaning it lies on $A D$, and $O A = \frac{5}{2} + r$. Note that

$$
B D = 4, \quad A D = \sqrt{A B^{2} - B D^{2}} = 8, \quad O D = A D - O A = \frac{11}{2} - r
$$

Equation (*) gives $O B = 2 + r$, and by the Pythagorean theorem

$$
O B^{2} = O D^{2} + B D^{2} \Longleftrightarrow (2 + r)^{2} = \left(\frac{11}{2} - r\right)^{2} + 16 \Longleftrightarrow r = \frac{169}{60}
$$

## Variant 2