 Circumcenter of a tetrahedron

For independent edge vectors $u=B-A$, $v=C-A$, $w=D-A$, the displacement $x=O-A$ solves $x\cdot u=|u{|}^2/2$, $x\cdot v=|v{|}^2/2$, $x\cdot w=|w{|}^2/2$. The reciprocal cross-product basis gives the displayed formula. Its squared distances to all four vertices are equal. It need not lie inside the tetrahedron.