Every point in 3D-space is colored either red, green or blue. Let R (resp. G and B) be the set of distances between red (resp. green and blue) points. Prove that at least one of R, G, or B, consists of all the non-negative real numbers.

Let's assume both R and G each do not contain one non-negative real number and try to show that B has to be all the non-negative reals.  Let's say R does not contain r and G does not contain g (both r and g are obviously positive, not zero).

Now what does the set of all red points look like?  It has to be composed of one or more components, each of which is contained in the interior of a sphere of radius r/2, and which have a distance of at least r from each other.  Otherwise two red points would have a distance of r, and r would be in R.

The same goes for the set of green points.  So the set of blue points is basically all 3D-space with lots of red and green holes in it.  It takes some spacial imagination to realise that there is no way the green set can "plug the gaps" in the red set or vice versa and that therefore there is a path which runs all the way from any blue point to infinity, all within blue territory.  The distances between the original point and the points on the path are all in B, which therefore contains all non-negative real numbers.

The "spacial imagination" bit is where it gets handwaving: If r=g, the spheres in which the components of the red and green set are contained are of equal size, and one could just fit between two of the other colour.  But that would not block the "blue path", especially since the components cannot extend to the surface of the spheres.  One may go to the length of stacking red and green spheres alternatingly in a cubic grid, but still to no avail.  If r<g or r>g, the smaller components stand even less of a chance to fill out the space between the larger ones.

Posted by vswitchs on 2006-08-25 16:57:44