We arbitrarily choose n lattice points in a 3-dimensional Euclidean space such that no three points are on the same line. What is the least n in order to guarantee that there must be three points x, y, and z among the n points such that the center of mass of the x-y-z triangle is a lattice point?
center of mass of triangle
4
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combinatorics
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1Still thinking about this; but I've got an upper bound for $n$, namely 55. Consider the congruence classes mod 3 of each of the 3 co-ordinates of a point. For any given point, there are 27 possibilities for the congruence class of each co-ordinate. So if we have 55 points, then the pigeonhole principle implies that there are 3 for which each of the three co-ordinates are congruent mod 3. Those 3 points for a triangle whose centroid is a lattice point. However, I feel that the real answer is way less than 55; I'm still thinking about it. – 2012-09-25
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0Is the center of mass simply the average of the three vertices? – 2012-09-25
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0Is this question already answered at http://math.stackexchange.com/questions/125479/geometry-of-points-in-mathbfz3-and-center-of-mass?rq=1 ? – 2012-09-25
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0There's a useful if inconclusive discussion at http://forums.philosophyforums.com/threads/puzzle-how-many-lattice-points-8780.html – 2012-09-25
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0A lower bound for $n$ is 17. Still considering congruence classes mod 3 of the co-ordinates, choose the 8 combinations of congruence classes where no coordinate $\equiv 2$ mod 3. Choose two points with each of these 8 combinations. Then of those 16 points, no 3 of them make a triangle. So $n$ must be at least 17, and by my earlier comment, at most 55. – 2012-09-25
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0@GerryMyerson the other question that you've linked to indicates that 37 is an upper bound for the answer (although I haven't studied the proof in detail). So it's a partial answer. We don't know $n$ yet, only that it lies between 17 and 37. – 2012-09-25