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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- 6$\begingroup$ 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. $\endgroup$Noam D. Elkies– Noam D. Elkies2012-12-16 02:30:39 +00:00Commented Dec 16, 2012 at 2:30
- $\begingroup$ 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. $\endgroup$Aaron Meyerowitz– Aaron Meyerowitz2012-12-16 09:51:01 +00:00Commented Dec 16, 2012 at 9:51
- $\begingroup$ 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). $\endgroup$Pietro Poggi-Corradini– Pietro Poggi-Corradini2012-12-16 09:55:15 +00:00Commented Dec 16, 2012 at 9:55
- $\begingroup$ @Pietro, then you should have mentioned that the lines are vertical, horizontal, or with slopes $p/q$ (where $p,q\in\{-6,-5,-4,-3,-2,-1,1,2,3,4,5,6\}$). :) $\endgroup$JRN– JRN2012-12-16 10:18:43 +00:00Commented Dec 16, 2012 at 10:18
- $\begingroup$ Hmm, so it seems that Noam's example above works. $\endgroup$JRN– JRN2012-12-16 10:20:12 +00:00Commented Dec 16, 2012 at 10:20
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1 Answer
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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?
- The centroid is the arithmetic average of the coordinates, and hence rational.
- 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.
- 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.