The Canonical Triangle of a Tetrahedron 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.
 A: I generated thousands of triangles and tetrahedra, then normalized them to have the orthocenter $O=(0,0,0)$ the centroid $G=(1,0,0)$, and an incenter of the form $I=(x,y,0)$.  Where is the incenter?  
In thousands of random triangles, here's a plot of incenter locations:
 
For $O=(0,0)$, $G=(1,0)$, we have $I=|(x,y)-(2,0)|\le1$
In thousands of random tetrahedra, here's a plot of incenter locations: 
 
Various points are in harmonic range on the Euler line.  Here are some of them, I give the full list below. 

The crucial triangle centers are $X_{381}$, centroid $G$ $X_2$, and incenter $I$ $X_1$. In a tetrahedron, if $I$ is further from $X_{381}$ than $G$, a simple mapping of the major triangle centers to a triangle is impossible.
Harmonic triangle centers on the Euler Line.
{{2, 20}, {3, 0}, {4, 60}, {5, 30}, {20, -60}, {140, 
  15}, {376, -20}, {381, 40}, {382, 120}, {546, 45}, {547, 
  25}, {548, -15}, {549, 10}, {550, -30}, {631, 12}, {632, 18}, {1656,
   24}, {1657, -120}, {2041, 120 + 60 Sqrt3}, {2042, 
  120 - 60 Sqrt3}, {2043, -20 Sqrt3}, {2044, 20 Sqrt3}, {2045, 
  1/13 (240 - 60 Sqrt3)}, {2046, 1/13 (240 + 60 Sqrt3)}, {2675, 
  1/59 (-900 + 1440 Sqrt[5])}, {2676, 1/109 (4500 - 720 Sqrt[5])}}
