# Integer triangle

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?

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Sure. Start with a 3-4-5 triangle, with vertices $(0,0)$, $(4,0)$, and $(3,0)$. The classical centers are all rational: centroid clearly (it's at $(4/3,1)$), orthocenter at the the origin, circumcenter at $(2,3/2)$, and incenter at $(1,1)$. Now scale by a factor of 6, Q.E.F. – Noam D. Elkies Dec 16 '12 at 2:30
The question which remains is, can this be done with a triangle whose sides have no common divisor? I don't think so, but I am not sure. – Aaron Meyerowitz Dec 16 '12 at 9:51
Noam, thanks! To be perfectly honest the question arose from trying to draw a picture using the Latex picture environment where only some rational slopes are allowed. In particular, I wanted a triangle in general position, where all the points mentioned are distinct (and inside the triangle preferably). – Pietro Poggi-Corradini Dec 16 '12 at 9:55
@Pietro, then you should have mentioned that the lines are vertical, horizontal, or with slopes $p/q$ (where $p,q\in\{-6,-5,-4,-3,-2,-1,1,2,3,4,5,6\}$). :) – Joel Reyes Noche Dec 16 '12 at 10:18
Hmm, so it seems that Noam's example above works. – Joel Reyes Noche Dec 16 '12 at 10:20

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.