6. A cone with a base radius of 2 and a slant height of c contains three spheres of radius r. They touch each other (externally), the lateral surface of the cone, and the first two spheres touch the base of the cone. Find the maximum value of $r$.
Answer: $\sqrt{3}-1$.
![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-04.jpg?height=480&width=1302&top_left_y=2022&top_left_x=430)

Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A$ be the point of tangency of the first two, $PO$ be the height of the cone, and $B$ be the projection of $O_{1}$ onto $PO$. On the left diagram, part of the section of the cone by the plane $POO_{1}$ is shown, and on the right - the section by the plane $POA$. Note that $AO_{3}=r\sqrt{3}$ as the height of the equilateral triangle $O_{1}O_{2}O_{3}$ with side $2r$. Additionally, $\cos \angle PFO = \frac{OF}{PF} = \frac{1}{2}$, hence $\angle PFO = \frac{\pi}{3}$ and

$$
BO_{1} = EO = 2 - EF = 2 - r \operatorname{ctg} \frac{\pi}{6} = 2 - r \sqrt{3}
$$

Let $\alpha$ be the angle between $AO_{3}$ and the perpendicular $AD$ to the segment $PN$, and $AC \parallel MN$ (see the right diagram). Then $AD = r + AO_{3} \cos \alpha = r(1 + \sqrt{3} \cos \alpha)$. The height $BA$ of the isosceles triangle $BO_{1}O_{2}$ is $\sqrt{BO_{1}^2 - r^2} = \sqrt{(2 - r \sqrt{3})^2 - r^2}$, which decreases as $r$ increases. Since $BO = r$, the length $BC$ also decreases as $r$ increases. Therefore, the expressions $AD = (AB + BC) \frac{\sqrt{3}}{2}$ and $\cos \alpha = \frac{AD - r}{r \sqrt{3}}$ are also decreasing with respect to $r$. Thus, the maximum value of $r$ is achieved when $\alpha$ is the largest, i.e., when the sphere with center at $O_{3}$ is either inscribed in the cone or touches its base. Let's consider these cases.

1) If the sphere with center at $O_{3}$ is inscribed in the cone, then $O_{1}O_{3} \parallel PF$, so $BO_{1} = 2r \cos \frac{\pi}{3} = r$ (see the left diagram). Then $2 - r \sqrt{3} = r$, so $r = \frac{2}{\sqrt{3} + 1} = \sqrt{3} - 1$.
2) If the sphere with center at $O_{3}$ touches the base of the cone, then point $B$ is the center of the equilateral triangle $O_{1}O_{2}O_{3}$, so $BO_{1} = \frac{2r}{\sqrt{3}}$. Therefore, $2 - r \sqrt{3} = \frac{2r}{\sqrt{3}}$, so $r = \frac{2\sqrt{3}}{5} < \sqrt{3} - 1$.

Thus, the maximum $r$ is achieved in the first case and is equal to $\sqrt{3} - 1$.

## Variant 2