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Given the set of $37$ points in $\mathbf{Z^3} $ in which no 4 points are on the same plane. Show that there exist 3 point A,B,C in this set such that $ (\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3},\frac{z_A+z_B+z_C}{3}) \in \mathbf{Z^3} $.

This is my approach:
Consider the coordinate mod 3 of those points, which can only be 0, 1 or 2.

In 37 points, there should be at least 13 points with the same x-coordinate mod 3. In these 13, there should be at least 5 points with the same y-coordinate mod 3. And in these 5 there are at least 2 points with same z-coordinate mod 3. let's call the z coordinate mod 3 of these 5 points: $Z_0$ for the first 2, and $Z_1,Z_2,Z_3$ for the last three. If one of $Z_1,Z_2,Z_3$ is equal to $Z_0$ then we have three points with $z \equiv Z_0$ we need. if it's not the case, then at least two in $Z_1,Z_2,Z_3$ are equal (say, $Z_1=Z_2)$ we pick $Z_0, Z_1$ and $Z_3$ and their sum are still divisible by 3.

I haven't used the condition that no 4 points are on the same plane yet (which is quite strong). Is there any mistake in my proof ? if yes then what is the alternative proof ?

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Your proof seems fine to me. – joriki Mar 28 '12 at 15:33
You should mentioning your usage of: – Salech Alhasov Mar 28 '12 at 16:10
oh ok. Thanks everyone! – Geralt of Rivia Mar 28 '12 at 16:18

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