Fred said:
Despite appearances, there are in practice rather a lot of 5-vertex 3-dim cases which satisfy MK's "isosceles" constraint --- say AD = AE, BD = BE, CD = CE: 21 of the first 63 may be cast in this form; and 12 of those have sufficient symmetry to do so in two distinct ways.
Upon further reflection, I suppose these are the simplest example of the construction "glue two tetrahedra together along a shared triangle." Given any tetrahedron with an altitude that is an integer or half-integer, you can glue it to its mirror image -- I didn't intend "upon further reflection" above to be a pun, honest! -- and get an isosceles 5-vx-3-d. Is the altitude of an integer-edged tetrahedron always degree <=2 over the rationals, as for triangles? That would make clear why these are so common.
PS --- after laboriously transcribing between MK ordering and WFL ordering, I belatedly realised that all that was required was to reverse the vector ...
Oof, sorry for the sign error. (Of course you chose the convention where adding a sixth point just involves tacking five more distances onto the end; duh.) Having run overnight, I've searched through length 64 for the longest edge touching one of the separation-1 points. Only found five so far. The list (in WFL-compatible order, I think!): [30, 32, 46, 16, 23, 24, 16, 23, 24, 1] [24, 24, 42, 2, 25, 25, 2, 25, 25, 1] [40, 42, 12, 26, 31, 40, 26, 31, 40, 1] [70, 64, 54, 31, 41, 41, 31, 41, 41, 1] [86, 54, 54, 46, 46, 49, 46, 46, 49, 1] Thus I can state for sure that the 3rd one on that list, with longest edges of length 40, has the smallest possible maximal edge length. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.