31 Mar
2013
31 Mar
'13
1:53 p.m.
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.
And [spoilerish comment follows] ... ... ... ... ... ... ... ... ... ... ... ... ... ... the proof also involves simple manipulations of equilateral triangles. -- g