6. The sphere touches all the edges of the tetrahedron, two opposite edges of which are equal to $a$ and $b$, and all other edges are equal to each other. Find the radius of this sphere.
Solution 1. Let $BC = a, AD = b, AB = BC = BD = CD = x$. Let $K$ and $M$ be the midpoints of edges $BC$ and $AD$ (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-090.jpg?height=443&width=574&top_left_y=1720&top_left_x=747)

Planes $ADK$ and $BMC$ are perpendicular to edges $BC$ and $AD$, respectively. Therefore, they are planes of symmetry of the tetrahedron, which means that point $O$ - the center of the sphere - is the midpoint of the height

90

$KM$ of the isosceles triangles $ADK$ and $BMC$. Therefore,

$$
\begin{aligned}
R = MO = \frac{1}{2} \sqrt{MB^2 - BK^2} = & \frac{1}{2} \sqrt{BD^2 - DM^2 - BK^2} = \\
& = \frac{1}{2} \sqrt{\frac{(a+b)^2}{4} - \frac{b^2}{4} - \frac{a^2}{4}} = \frac{\sqrt{2ab}}{4}
\end{aligned}
$$