Consider these 4 points as vertices of a tetrahedron.
$$((0,-15,0),(0,9,-12),(12,9,0),(0,9,12))$$
Consider these 3 points as vertices of a triangle. $$(( \frac{1}{4} \left(49+31 \sqrt{3}\right) , \frac{1}{4} \left(-31+49 \sqrt{3}\right) , 0 ), ( \frac{1}{4} \left(49-31 \sqrt{3}\right) , \frac{1}{4} \left(-31-49 \sqrt{3}\right) , 0 ), ( -\frac{31}{2} , \frac{49}{2} , 0 ))$$
Both have incenter $I =(4, 5, 0)$, circumcenter $O =(0, 0, 0)$, and centroid $G=(3, 3, 0)$. Here's a picture:
For an irregular tetrahedron, most of the major tetrahedral centers are on the same plane. It turns out that a triangle exists with these same centers. Is there an easy to get from the tetrahedron to the triangle, or vice-versa?
Is this an already-known triangle? If not, I would suggest that this would be a good canonical triangle for a given tetrahedron.
What tetrahedral centers can be exactly calculated that are off of the IOG plane?
A larger discussion with code is at Tetrahedron Centers.



