XII OBM

Opposite edges of the tetrahedron have the same length. Let $AB$, $CD$ have length $a$. Take parallel planes through $AB$ and $CD$. Project $A$, $B$ onto the other plane, and $C$, $D$ onto the other plane. Then we get $8$ points at the vertices of a rectangular block. If the sides have lengths $x$, $y$, $z$, then we have $a^2=y^2+z^2$, $b^2=x^2+z^2$, $c^2=x^2+y^2$. A long diagonal of the block has length $\sqrt{x^2+y^2+z^2}$ and is a diameter of the circumsphere. So $x^2+y^2+z^2=4$. Hence $a^2+b^2+c^2=2(x^2+y^2+z^2)=8$.