In a bichromatically colored plane, is it always possible to construct any regular polygon such that all vertices are the same color?

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Jerry Kidd · Accepted Answer

Step 1Start with this:Colour   (      R    ∖      Q    )  ×  (      R    ∖      Q    ) and       Q    ×      Q   red,and the other points blue.Then any line segment (a line segment being more than just a point) which is monochromatic is a segment of a rational line, i.e. defined by   a  x  +  b  y  +  c  =  0 , where a, b and c are rational and either   a  ≠  0 or   b  ≠  0 .In order to end up with a colouring where none of these is monochromatic, we are going to colour some of the points on some of them (which are now red) back to blue, with the following procedure:There are countably many rational lines. Enumerate them as       T    n   , that is to say that   n  ↦      T    n   is a bijection from       N   .Let       L    n    =      T    n    ∩  (      Q    ×      Q    ) . We just want to look at rational points for this procedure.Choose a dense subset       S    1   of       L    1   , such that       L    1    −      S    1   is also dense in       L    1   . Colour       S    1   blue.Choose a dense subset       S    2   of       L    2   , such that       S    2   does not intersect       L    1   anywhere and such that       L    2    −      S    2   is also dense in       L    2   . Colour       S    2   blue.Choose a dense subset       S    3   of       L    3  , such that       S    2   does not intersect       L    1    ∪      L    2   anywhere and such that       L    3    −      S    3   is also dense in       L    3   . Colour       S    3   blue.Continue this recursively. Every rational line segment will have both colours.

In a bichromatically colored plane, is it always possible to

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