[ Sphere circumscribed around a tetrahedron ]

Two opposite edges of a triangular pyramid are equal to $a$, two other opposite edges are equal to $b$, and the two remaining edges are equal to $c$. Find the radius of the circumscribed sphere.
We will complete the given tetrahedron to a parallelepiped by drawing three pairs of parallel planes through the opposite edges. Since the opposite edges of the tetrahedron are pairwise equal, the diagonals of each face of the resulting parallelepiped are also equal. Therefore, all faces of the parallelepiped are rectangles. Hence, the parallelepiped is a rectangular one. A sphere can be circumscribed around it. This sphere passes through all the vertices of the tetrahedron.

Thus, the problem reduces to finding the radius of the sphere circumscribed around the rectangular parallelepiped. Let \(A B C D A 1 B 1 C 1 D 1\) be the resulting rectangular parallelepiped, and \(A C B 1 D 1\) be the original tetrahedron, in which

\[
A C = B 1 D 1 = a, \quad A D 1 = B 1 C = b, \quad A B 1 = D 1 C = c.
\]

Let \(A B = x\), \(A D = y\), \(A A 1 = z\). Then

\[
\left\{\begin{array}{l}
x^{2} + y^{2} = a^{2} \\
y^{2} + z^{2} = b^{2} \\
z^{2} + x^{2} = c^{2}
\end{array}\right.
\]

from which we find that

\[
x^2 + y^2 + z^2 = \frac{1}{2}(a^2 + b^2 + c^2)
\]

Let \(R\) be the desired radius. Then

\[
R = \frac{1}{2} A C 1 = \frac{1}{2} \sqrt{A B^2 + A D^2 + A A_1^2} = \frac{1}{2} \sqrt{x^2 + y^2 + z^2} =
\]

\[
= \frac{1}{2} \sqrt{\frac{1}{2}(a^2 + b^2 + c^2)} = \frac{1}{4} \sqrt{2(a^2 + b^2 + c^2)}
\]

## Answer \(\frac{1}{4} \sqrt{2(a^2 + b^2 + c^2)}\)