* Tom Karzes <karzes@sonic.net> [Apr 03. 2013 08:34]:
Spoilier spoiler:
Place two equilateral triangles edge-to-edge, and consider the properties of the resulting rhombus.
That was the first thing I tried, but the configuration B R R G suggests "no fish" to me: the distance R--R is unequal to the distances R--g and R--B, a case of "either d or sqrt(3)/2*d" (if I got the constant right). However, Charles' pointer to http://en.wikipedia.org/wiki/Chromatic_number_of_the_plane (especially the image "Solomon W. Golomb's ten-vertex four-chromatic unit distance graph") convinces me (3-color wise). Given the problem has a name (and having been considered by heavy duty mathematicians) I consider myself only semi stupid now. Thanks everybody! Best, jj
Tom
Joerg Arndt writes:
* 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.
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