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I’m researching tetratomic molecules and tetrahedra have consequently become a subject of interest.

I’m seeking to prove equivalent congruence theorems for tetrahedra to those for triangles.

Has any research been published on this subject?

I’ve proved that 6 same edge lengths is equivalent to congruence. My theories are that:

  • 6 same planar angles, arranged as 2 each on 3 different faces is equivalent to similarity of tetrahedra
  • 6 same dihedral angles is equivalent to similarity
  • 5 edges and a planar angle; 4 edges and 2 planar angles; or 3 edges and 3 planar angles are equivalent to congruence

I’m uncertain about the case of 2 edges and 4 planar angles. Does the disposition of those edges and planar angles have to be specified to be the same?

Are there any other combinations of elements (e.g. face areas and dihedral angles, or face areas and edges, or dihedral and planar angles) that show congruence or similarity?

It seems that 6 elements are necessary and sufficient to prove congruence or similarity (akin to the 3 needed for triangles).

Thanks

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    $\begingroup$ It depends on what you mean by "pieces". For instance, knowing two faces and their dihedral angle (3 pieces) is enough. $\endgroup$
    – Intelligenti pauca
    Commented Aug 15, 2023 at 22:15
  • $\begingroup$ 4 vertices, 6 edges, 12 planar angles, 6 dihedral angles and various other properties. demonstrations.wolfram.com/… is one dihedral theorem. There is likely room for exploration with this problem. $\endgroup$
    – Ed Pegg
    Commented Aug 15, 2023 at 23:35
  • $\begingroup$ Six appropriately-chosen pieces of info will do. The six edge-lengths; three "concurrent" edge-lengths and the three (planar) angles between pairs of them; three face areas and the three dihedral angles between pairs of them; four face areas alone aren't enough, but including the three of what I call "pseudoface" areas will do. (A pseudoface is the shadow of the tetrahedron in a plane parallel to a pair of opposite edges. The seven values are related by "sum-of-squares-of-face-areas = sum-of-squares-of-pseudoface-areas", so that there are only six independent values.) Etc, etc, etc. $\endgroup$
    – Blue
    Commented Aug 15, 2023 at 23:48
  • $\begingroup$ I haven't formally published anything, but I've archived a bit of on-gong research into "hedronometry" (the dimensionally-enhanced trigonometry of tetrahedra), both Euclidean and non-, here. $\endgroup$
    – Blue
    Commented Aug 16, 2023 at 0:16
  • $\begingroup$ Thanks. I've drafted a proof of the 6 edge-lengths case. I'm struggling slightly with the rest. $\endgroup$
    – TheorVHP
    Commented Aug 24, 2023 at 16:19

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