Posted February 18, 2014 Is it possible to find six points on a square lattice that form the vertices of a regular hexagon? A square lattice is invariant under rotation by π/2 around any lattice point. 0 Share this post Link to post Share on other sites

0 Posted February 24, 2014 hexmesh.JPG For any hexagon, I see a 30-60-90-degree triangle necessarily coming into play. So somewhere there will be an x - spacing of 1 and a y - spacing of sqrt(3), or vice versa. These two distances are not rationally commensurate. By definition all x-y distances on a rectangular grid are rationally commensurate. So my guess is that it's not possible. 0 Share this post Link to post Share on other sites

0 Posted February 23, 2014 For any hexagon, I see a 30-60-90-degree triangle necessarily coming into play. So somewhere there will be an x - spacing of 1 and a y - spacing of sqrt(3), or vice versa. These two distances are not rationally commensurate. By definition all x-y distances on a rectangular grid are rationally commensurate. So my guess is that it's not possible. 0 Share this post Link to post Share on other sites

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Is it possible to find six points on a square lattice that form the vertices of a regular hexagon?

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