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Suppose that we are given 40 points equally spaced around the perimeter of a square, so that four of them are located at the vertices and the remaining points divide each side into ten congruent segments. If P, Q, and R are chosen to be any three of these points which are not collinear, then how many different possible positions are there for the centroid of triangle PQR?

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    $\begingroup$ Did you forget the password to your old account which asked the same question? $\endgroup$ – Rahul Nov 2 '12 at 10:39
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    $\begingroup$ Wow, someone really didn't like that comment. I just got hit by a bunch of downvotes. $\endgroup$ – Rahul Nov 2 '12 at 10:45
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    $\begingroup$ I think that if you take coordinates so your square is $[0,10]\times[0,10]$ and use that the centroid of $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is $$\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right),$$then everything reduces to a combinatorics problem. My conjecture is that all points $(x,y)\in(0,10)\times(0,10)$ such that $3x,3y\in\mathbb{Z}$ are centroids of some triangle. $\endgroup$ – Manzano Nov 2 '12 at 10:58

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