* Gareth McCaughan <gareth.mccaughan@pobox.com> [Apr 01. 2013 07:42]:
On 31/03/2013 20:33, Tom Karzes wrote:
You can strengthen that to three colors, and there's still a nice solution:
Show that for every possible coloring of the plane with three colors there exit, for every distance d, two points of distance d with the same color.
Me. Very. Stupid. I cannot see how to do this. (Only: "for each d, two points of distance either d or sqrt(2)*d"). I am not asking for the solution (yet!), but just a slightly spoilier spoiler than that below.
And [spoilerish comment follows] ...
[...]
... the proof also involves simple manipulations of equilateral triangles.
-- g
[...]
Can anyone *disproof*: "For every coloring of the plane with finitely many colors and every d there are always two points of the same color with distance d" ?