🚀 go-pugleaf

In article <1993Jan9.030824.13624@Csli.Stanford.EDU> hiraga@Csli.Stanford.EDU (Yuzuru Hiraga) writes:
>
>A plane is totally colored with two colors.
>* Show that there exist three points of same color that
>  (when connected) form an equilateral triangle.
>
>-YH

My proof is probably easier to show in person but here goes:

Suppose there is no equilateral triangle within the plane such that
its three vertices are all the same color.

Pick two points, A and B, of the same color.  WLOG, suppose that
A and B are both red.  Now refer to the following diagram:

              C     D     E
              *     *     *

                 A     B     F
                 *     *     *

                    G
                    *

Since A and B are both red, and ABD is an equilateral triangle, we
must have that D is blue.

Since A and B are both red, and ABG is an equilateral triangle, we 
must have that G is blue.

Since G is blue, and CGE is an equilateral triangle, we have that
C and E cannot both be blue.  WLOG, suppose that E is red.

Since B and E are both red, and BEF is an equilateral triangle, we
must have that F is blue.

But this now gives us the equilateral triangle DGF whose vertices
are all blue.

This is a contradiction.

Thus, there must be an equilateral triangle in the plane such that
its vertices are all the same color.

                           Bob Murphy (rmurphy@aludra.usc.edu)