In a regular Cartesian coordinate system find the coordinates of six lattice points that are the vertices of a regular hexagon (being a lattice point means that the point has integer coordinates).
Solution: One has to be careful in how one interprets this question (and I had to be careful in how I worded it). One might give the following "proof" that the desired configuration in impossible. If it were possible we could then take the average of the six points (or of just two opposite points) to find the center of the hexagon. Then, together with any two adjacent vertices, we would have an equilateral triangle with all three points having rational coordinates. But this is impossible - we may assume that one vertex of the equilateral triangle is at (0,0) and one is at (a,b). Then to get to the third vertex you could start at (a/2, b/2) and go perpendicularly to this side a distance equal to root(3)/2 of the side length. Alternatively we could get to the third vertex by rotating the point (a,b) 60 degrees in either direction. Either way, we end up with the third point being, which is obviously not a point with rational coordinates.
To paraphrase Mr. Spock in the Wrath of Khan this solution shows some innate talent, however it betrays a tendency for 2 dimensional thinking. Once one gets over this tendency, many solutions become possible. My favorite is to pick your six points to be the six permutations of (1,2,3). I.e. One solution is the hexagon with vertices located at (1,2,3), (1,3,2), (2,3,1), (3,2,1), (3,1,2), and (3,2,1). If one is just specifying the vertices then the order chosen is irrelevant. However, the order I picked should help one verify the validity of the claim since I listed the points in adjacent order. One can check that each successive pair are root (2) apart from each other and that the angle determined by any three of these in order is 120 degrees.
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