BMAD 62 Report post 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. Share this post Link to post Share on other sites

0 BMAD 62 Report post 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. Share this post Link to post Share on other sites

0 TimeSpaceLightForce 11 Report post Posted February 23, 2014 Share this post Link to post Share on other sites

0 bonanova 77 Report post 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. Share this post Link to post Share on other sites

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

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