6. The region is bounded by two cones with a common base, the height of which is 4, and the radius of the base is 3. Three spheres are placed in the region, each touching the others externally. Two of the spheres are identical and touch both cones, while the third touches the boundary of the region. What is the minimum radius of the third sphere?
# Answer: $\frac{27}{35}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-28.jpg?height=494&width=1140&top_left_y=1181&top_left_x=525)

Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $BO$ be the height of one of the cones, $r$ and $x$ be the radii of the first two and the third sphere, respectively, and $\alpha$ be the angle between the generatrix and the height of the cones. From the left diagram, it is clear that

$$
3 = AO = AO_{1} + O_{1}O = r\left(\frac{1}{\cos \alpha} + 1\right) = \frac{9}{4} r, \quad \text{so} \quad r = \frac{4}{3}.
$$

Let $OD$ in the right diagram be the perpendicular to the generatrix $BC$ of the cone, and $\beta$ be the angle between $OD$ and $OO_{3}$. Note that $OO_{3}$ is the height of the isosceles triangle $O_{1}O_{2}O_{3}$ with legs $r + x$ and base $2r$, so $OO_{3} = \sqrt{(x + r)^2 - r^2} = \sqrt{x^2 + 2rx}$. Then

$$
CO \cdot \cos \alpha = OD = OO_{3} \cdot \cos \beta + O_{3}E \Longleftrightarrow \frac{12}{5} = \cos \beta \sqrt{x^2 + 2rx} + x \Longleftrightarrow \cos \beta = \frac{\frac{12}{5} - x}{\sqrt{x^2 + \frac{8}{3}x}}
$$

The right-hand side of the last equation decreases as $x$ increases, so the minimum value of $x$ is achieved when $\cos \beta$ is maximized, i.e., when $\beta = 0$. In this case, $x \leq \frac{12}{5}$ and

$$
\sqrt{x^2 + \frac{8}{3}x} = \frac{12}{5} - x \Longleftrightarrow \frac{8}{3}x = \frac{144}{25} - \frac{24}{5}x \Longleftrightarrow x = \frac{27}{35}
$$

## Variant 8