The faces must therefore all be Heronian triangles (named for Hero of Alexandria).
Every Heronian tetrahedron can be arranged in Euclidean space so that its vertex coordinates are also integers.
[4] Eight examples of Heronian tetrahedra were discovered in 1877 by Reinhold Hoppe.
The integral edge lengths of a Heronian tetrahedron with this volume and surface area are 25, 39, 56, 120, 153 and 160.
[7] However, Starke made an error in reporting its volume which has become widely copied.
[8] Sascha Kurz has used computer search algorithms to find all Heronian tetrahedra with longest edge length at most 600000.
[9] A regular tetrahedron (one with all faces being equilateral) cannot be a Heronian tetrahedron because, for regular tetrahedra whose edge lengths are integers, the face areas and volume are irrational numbers.
[10] For the same reason no Heronian tetrahedron can have an equilateral triangle as one of its faces.
[3] There are infinitely many Heronian tetrahedra, and more strongly infinitely many Heronian disphenoids, tetrahedra in which all faces are congruent and each pair of opposite sides has equal lengths.
[3][11] There are also infinitely many Heronian tetrahedra with a cycle of four equal edge lengths, in which all faces are isosceles triangles.
One method for generating tetrahedra of this type derives the axis-parallel edge lengths
from two equal sums of fourth powers using the formulas For instance, the tetrahedron derived in this way from an identity of Leonhard Euler,
[4] No example of a Heronian trirectangular tetrahedron had been found and no one has proven that none exist.
A complete classification of all Heronian tetrahedra remains unknown.