BMAD Posted February 18, 2014 Report Share 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. Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted February 24, 2014 Author Report Share 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. Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted February 23, 2014 Report Share Posted February 23, 2014 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted February 23, 2014 Report Share 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. Quote Link to comment Share on other sites More sharing options...
Question
BMAD
Is it possible to find six points on a square lattice that form the vertices of a regular hexagon?
Link to comment
Share on other sites
3 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.