Is there a triangle whose vertices, as well as the four classical points, the centroid, the orthocenter, the incenter, and the circumcenter, all have integer coordinates?



Integer triangle
Clearly it is enough to find a triangle in which all seven points are rational, as then you can make them integral by rescaling. But given *any* triangle with rational coordinates, aren't the centroid, orthocenter, and circumcenter all automatically at rational coordinates?


1. The centroid is the arithmetic average of the coordinates, and hence rational.
2. The slopes of all sides are rational. Hence the altitude of through any vertex has rational slope (-1/the slope of the opposite side) and passes through a rational point, and so has a rational equation. The orthocenter is the intersection of any two altitudes, and the intersection of two lines with rational equations is necessarily rational.
3. By a similar argument, the perpendicular bisectors of each side are rational, and so the circumcenter is rational.


So the only non-automatic point is the incenter. If $A$, $B$, and $C$ are the coordinates of the corners, and $a$, $b$, and $c$ are the side lengths of the respective opposite sides (so that $a = \|B-C\|$, for example) then the incenter is at $(aA + bB + cC) / (a + b + c)$. This is not necessarily rational — the triangle with coordinates $(0,0)$, $(1,0)$, and $(0,1)$ is a counterexample — but is rational as soon as the three sidelengths are.


So Noam D. Elkies's answer in the comments has many generalizations, including all Pythagorean triples, and also including combinations of them.


