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
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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
2 Answers
19 according to "A lattice point problem and additive number theory" and its references
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1Here is a link: http://www.cs.tau.ac.il/~nogaa/PDFS/centroid.pdf – 2012-09-25
The answer $19$ has already been given, including references, so this is just a numerical check of that result.
The specific shape of the lattice is irrelevant, since a point is a lattice point iff it is an integer linear combination of the lattice vectors. The centre of mass of three lattice points is a lattice point iff all coefficients of the sum are multiples of $3$.
To find the maximal number of points without a centre of mass on the lattice, consider the distribution of the residues of the coefficients of the points $\bmod3$ in $\mathbb Z_3^3$. No three points may add up to $0\in\mathbb Z_3^3$.
We cannot have the same point in $\mathbb Z_3^3$ three times. On the other hand, whatever points in $\mathbb Z_3^3$ we do have, we can have them twice, since the only point that would sum to $0$ with the two copies would be a third copy. Thus, we can reduce the problem to finding the maximal number of different points in $\mathbb Z_3^3$ no three of which add up to $0$; then the desired maximal number of not necessarily different points is twice that.
Here's code that enumerates all subsets of $\mathbb Z_3^3$ and finds the maximal size of a subset that doesn't contain a triple that sums to $0$. That maximal size turns out to be $9$, so the maximal size of a set of not necessarily different points is $2\cdot9=18$. Thus one more than that, $19$, is the number of points required to force a set to contain a triple with centre of mass on the lattice.
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0Thanks to all. My best guess is also 9+9+1, but, is there another way to explain "9"? – 2012-09-26