Ed Pegg
unread,Mar 19, 2026, 12:48:09 AM (13 days ago) Mar 19Sign in to reply to author
Sign in to forward
You do not have permission to delete messages in this group
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to SeqFan
Here's a nice tetrahedron
{{0, 0, 0}, {2, 0, 0}, {0, 4, 0}, {0, 3, 4}}
A lot of usually nasty centers are rational
{{"Incenter", {8/13, 16/13, 8/13}}, {"Centroid", {1/2, 7/4, 1}},
{"Circumcenter", {1, 2, 13/8}},
{"Monge", {0, 3/2, 3/8}}, {"12Point", {1/3, 5/3, 19/24}},
{"Symmedian", {64/93, 74/93, 32/93}}, {"Spieker", {6/13, 25/13, 44/39}}}
The incenter is based on face areas, so unless all the faces have integer area, the incenter will be highly complicated.
For the vertices to be integers, all the faces need to be heronian, but edges can be square roots. Doing that with different scalene triangles is non-trivial... and then the Gram determinant has to be a perfect square The only set of edges that worked for max edge<sqrt(30) was {4,16,25,20,29,17}
and {{0, 0, 0}, {2, 0, 0}, {0, 4, 0}, {0, 3, 4}} is a representation.
I'm not sure yet if there's a non-messy way to find more of these. There's
probably a sequence driven by these.
{{0, 0, 0}, {2, 0, 0}, {0, 4, 0}, {0, 3, 4}}
looks very simple but was difficult for me to find.