Conference rusure::math

> It is a well-known theorem that given any coloring of the plane by two
>    colors, there exists an equilateral triangle with monochromatic
>    vertices. 



This seems so uninterestingly trivial, that I fear I don't interpret
what you mean.

Given a 2-coloring of the plane into, say, black and white, I can
merely pick a tiny enough triangle (heck, make it equilateral if you like)
so that all 3 corners are on any chosen black region of the plane.

Please clarify...

        Let R denote the set of real numbers, let the plane be R^2,
        then a coloring of the plane by some set C of colors is just a
        function f : R^2 -> C.  You are assigning a color to each
        point of the plane.  The function f can be completely
        arbitrary, for example, if both coordinates are rational the
        point is red, otherwise the point is blue.  There is no
        requirement that the set of points of a particular color be
        open or closed or measurable or connected or have smooth
        boundaries or anything.  In particular, any line segment
        "bigger than a point" (i.e., non-zero length) may contain
        points of both colors.
        
        With the coloring given above, there is an equilateral
        triangle with one edge parallel to the x-axis and with three
        blue corners, one at (pi, pi), one at (pi + 1, pi), and the
        third at (pi + 1/2, pi + (sqrt(3))/2).
        
        Dan
        

o.k. I see now.  You're talking about colorings that we might not be able
to see at all, namely "states" in a country that might be only a geometric
point in size.

I'd hate to discuss things like the 4-color map theorem for such a map.
I mean, how many colors are required to "color" 3 "adjacent" geometric
points ?  

re .3
    
    I'm unclear on how one could define "adjacent points".

re .0
    
Not answering the question but I thought it might be useful to attempt to prove
the "well-known" theorem mentioned in .0 (as it isn't well known to me).

I use the term "monochrome equilateral triangle" below as an abbreviation for
"equilateral triangle with monochrome vertices".

Choose an arbitrary point and wlog assume that it is red. If all other points
of the plane are blue then certainly there is a monochrome equilateral
triangle. Otherwise there is a second red point. Now assume that the theorem is
false and colour the points as we are forced to do as shown in the diagram
below. The numbering gives the order in which the points are labelled.


                   B3



              R1       R2       R5



                   B4       B6



                       ?7

There is no way to label point 7. Hence the assumption that the theorem is
false fails.

Can the theorem be proved using less points?

How many colours would ensure that no monochrome equilateral triangle exists?
The above proof would seem to require radical surgery for more than two
colours.

Now I must go and consider the actual question posed in .0...

>    It is a well-known theorem that given any coloring of the plane by two
>    colors, there exists an equilateral triangle with monochromatic
>    vertices.  As a generalization, show that given any two-coloring of the
>    plane and any triangle T, there exists a triangle similar to T with
>    monochromatic vertices.

	Step (1) Find three monochromatic points evenly spaced on a straight
line. Call these 3 points A,B,C.

	Method:
	Pick two points of the same colour. WLOG, say both (-1,0) & (1,0) are
red. Now what about (3,0)? If it's red, then we're home. So suppose it's
blue. Similarly suppose (-3,0) is blue. Similarly, (0,0) is blue. But then
(-3,0), (0,0) & (3,0) are all blue evenly spaced.

	Step (2) (This bit is like Eric's in -.1) Draw the following triangle,
similar to T.

		A	B	C

		  D       E         

		    F      

	where B,D,E are the midpoints of AC,AF,CF respectively. Then A,B,C
are monochromatic (red, say) so D & E are blue. F cannot be coloured without
admitting a monochromatic triangle similar to T. 

Andrew.