2.25. Two opposite edges of a tetrahedron are perpendicular, and their lengths are $a$ and $b$; the distance between them is $c$. A cube is inscribed in the tetrahedron, four of its edges are perpendicular to these two edges of the tetrahedron, and on each face of the tetrahedron, exactly two vertices of the cube lie. Find the edge of the cube.
2.25. The common perpendicular to the given edges is divided by planes parallel to them and the faces of the cube into segments of length $y, x$ and $z$ ( $x$ - the length of the edge of the cube; the segment of length $y$ is adjacent to edge a). The planes of the cube's faces, parallel to the given edges, intersect the tetrahedron in two rectangles. The smaller sides of these rectangles are equal to the edge of the cube $x$. Since the sides of these rectangles are easily calculated, we get $x=b y / c \quad$ and $\quad x=a z / c$. Therefore, $c=x+y+z=x+$ $+c x / b+c x / a$, i.e., $x=a b c /(a b+b c+c a)$.