All tetrahedra can be inscribed in a parallelepiped. A tetrahedron is orthocentric if and only if its circumscribed parallelepiped is a rhombohedron. Indeed, in any tetrahedron, a pair of opposite edges is perpendicular if and only if the corresponding faces of the circumscribed parallelepiped are rhombi. If four faces of a parallelepiped are rhombi, then all edges have equal lengths and all six faces are rhombi; it follows that if two pairs of opposite edges in a tetrahedron are perpendicular, then so is the third pair, and the tetrahedron is orthocentric.^[1]

A tetrahedron ABCD is orthocentric if and only if the sum of the squares of opposite edges is the same for the three pairs of opposite edges:^[2][3]

${\displaystyle {\overline {AB}}^{2}+{\overline {CD}}^{2}={\overline {AC}}^{2}+{\overline {BD}}^{2}={\overline {AD}}^{2}+{\overline {BC}}^{2}.}$ 

In fact, it is enough for only two pairs of opposite edges to satisfy this condition for the tetrahedron to be orthocentric.

Another necessary and sufficient condition for a tetrahedron to be orthocentric is that its three bimedians have equal length.^[3]

The characterization regarding the edges implies that if only four of the six edges of an orthocentric tetrahedron are known, the remaining two can be calculated as long as they are not opposite to each other. Therefore the volume of an orthocentric tetrahedron can be expressed in terms of four edges a, b, c, d. The formula is^[4]

${\displaystyle V={\frac {1}{6}}{\sqrt {4(c^{2}+d^{2})s(s-a)(s-b)(s-c)-a^{2}b^{2}c^{2}}}}$ 

where c and d are opposite edges, and ${\displaystyle s={\tfrac {1}{2}}(a+b+c)}$ .

• Disphenoid
• Trirectangular tetrahedron