From: kanad@quip.eecs.umich.edu (Kanad Chakraborty)
Subject: Re: An(other) *easy* puzzle.
Organization: University of Michigan EECS Dept., Ann Arbor
Date: Sun, 10 Jan 1993 20:18:37 GMT

In article <1993Jan10.043342.24730@smds.com> rh@ishmael.UUCP (Richard Harter) writes:
>In article <1993Jan9.163703.24700@zip.eecs.umich.edu> kanad@quip.eecs.umich.edu (Kanad Chakraborty) 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.
>
>>Proof. WLOG, let the colors be red and blue.
>>Choose a point A in the colored plane. Suppose, WLOG, that 
>>A is colored blue. Find the (or a) point colored red _nearest_ to A 
>>(Euclidean distance-wise). (This is always possible because both
>>colors have been used.)
>
>Stop there.  The assertion is false.  For a counter example, color all
>points with rational coordinates red and those with irrational coordinates
>blue.
>
>For a simple proof, consider this figure (equilateral triangular net);:
>
>                 0
>                1 2
>               3 4 5
>              6 7 8 9
>
>This figure contains a same-colored equilateral triangle.  Proof:
>Suppose it does not.   Then 1-5-7 contains both colors; one is the same
>as the color of 4.  Without loss of generality we can suppose that 1 and
>4 are red and that at least one of 5 and 7 are blue, which gives us:
>
>
>                 0
>                R 2
>               3 R 5
>              6 7 8 9
>
>Then it follows that 2 and 3 must be blue and, to keep 2-3-8 multi-colored,
>8 must be red.  In turn this implies that 5 and 7 are both blue, giving us
>
>                 0
>                R B
>               B R B
>              6 B R 9
>
>Now from triangle 3-6-7 6 must be R, from triangle 2-7-9 9 must be R,
>and from 0-3-5 0 must be R, so that 0-6-9 is all R, contrary to the
>assumption.  Q.E.D.
>
>-- 
>Richard Harter: SMDS Inc.  Net address: rh@smds.com Phone: 508-369-7398 
>US Mail: SMDS Inc., PO Box 555, Concord MA 01742.    Fax: 508-369-8272
>In the fields of Hell where the grass grows high
>Are the graves of dreams allowed to die.

My apologies to all you rec.puzzlers for the erroneous approach. OK, then,
here's a solution obtained by my friend L. Chandar of Syracuse University.
There was a mistake in his original reply to me and I am posting his solution
here with the correction. It is a bit too verbose but I could understand
it clearly, hence all you folks can, for sure. Here it goes :

========================================================================
Given any two points of the same colour (say 1 unit apart),
on either side along perpendicular bisector at sqrt(3)/2 units away,
we should have another colour if given statement is not true.
Thus, along the line joining first two points at 1 unit further 
away from either point on the outside of the segment (note :
sqrt(3) * cos 30 - 0.5 = 1), we should
have the same two original colours because either of these points
along with the second two points forms an equilateral triangle.
Continuing between the two new such pairs of points one unit 
apart which have same colour, we get now three parallel lines
such that at one unit intervals along the middle line we have
a particular colour, while in the outer two lines (sqrt(3)/2 away on
either side), at one unit intervals with a given point lying
on the perpendicular bisector of the segment joining the
nearest two points of the middle line, we should have the other
colour.  But this is a contradiction for we can construct an
equilateral triangle based only on points of the outer two lines
of side 2 units (2 of them lying on one of the line and the third
on the other) and all of them have the same colour by construction.
Rather messy, huh?

- chandar
========================================================================

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