6. A cone, whose base diameter equals its slant height, contains three identical spheres that touch each other externally. Two of the spheres touch the lateral surface and the base of the cone. The third sphere touches the lateral surface of the cone at a point lying in the same plane as the centers of the spheres. Find the ratio of the radii of the base of the cone and the spheres.
Answer: $\frac{5}{4}+\sqrt{3}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-12.jpg?height=480&width=1106&top_left_y=2026&top_left_x=536)

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, $B$ be the projection of $O_{1}$ on $PO$, $C$ be the point of tangency of the third sphere with the cone, $R$ and $r$ be the radii of the base of the cone and the sphere, respectively. 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 $2r$. The point $C$ lies in the plane $O_{1}O_{2}O_{3}$ and, by symmetry, in the plane $AO_{3}P$. Therefore, it lies on the line $AO_{3}$. But the radius $CO_{3}$ is perpendicular to the generatrix $PN$ (see the right diagram), so $AC \perp PN$. Then $AC = r(1+\sqrt{3})$. By the condition $\angle ADC = \angle ONP = \frac{\pi}{3}$, we have

$$
AB = AD - BD = \frac{AC}{\sin \frac{\pi}{3}} - \left(R - r \operatorname{ctg} \frac{\pi}{3}\right) = 2r\left(1 + \frac{1}{\sqrt{3}}\right) + \frac{r}{\sqrt{3}} - R = r(2 + \sqrt{3}) - R
$$

On the other hand, from the left diagram,

$$
BO_{1} = EO = R - EF = R - r \operatorname{ctg} \frac{\pi}{6} = R - r \sqrt{3} \quad \text{and} \quad AB = \sqrt{BO_{1}^{2} - r^{2}} = \sqrt{(R - r \sqrt{3})^{2} - r^{2}}
$$

since $AB$ is the height of the isosceles triangle $BO_{1}O_{2}$ with base $2r$. Thus, we arrive at the equality $\sqrt{(R - r \sqrt{3})^{2} - r^{2}} = r(2 + \sqrt{3}) - R$. Squaring it under the condition $R \leqslant r(2 + \sqrt{3})$, we get

$$
R^{2} - 2Rr\sqrt{3} + 2r^{2} = r^{2}(7 + 4\sqrt{3}) - 2Rr(2 + \sqrt{3}) + R^{2} \Longleftrightarrow r^{2}(5 + 4\sqrt{3}) = 4Rr \Longleftrightarrow \frac{R}{r} = \frac{5}{4} + \sqrt{3}
$$

The condition on the radii is clearly satisfied.

## Variant 4