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Forming a regular hexagon from a lattice

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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.

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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.

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Posted · Report post

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.

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