Triangle in Randomly Colored Plane

Date: 10/28/2002 at 15:07:05
From: Scott
Subject: Triangle in randomly colored plane

I have a problem I need to solve for my teacher, but I am having a bit 
of trouble answering it. Can you give me any help?

Prove:

If I assume that all points in the real plane are colored white or 
black at random, no matter how the plane is colored (even all white or 
all black) there is always at least one triangle whose vertices and 
center of gravity (all 4 points) are of the SAME color.

Date: 10/29/2002 at 06:43:09
From: Doctor Floor
Subject: Re: Triangle in randomly colored plane

Hi, Scott,

Thanks for your question.

Assume that in such a plane the vertices and center of gravity of 
each triangle need two colours.

It must be possible to find three noncollinear points A, B, and C of 
one and the same colour. If that were not the case, then the plane 
could only consist of a black and a white line. Without loss of 
generality we assume A, B, and C are black, and thus its center of 
gravity G must be white by the above assumption.

Now there exists a point A' such that A is the center of gravity of 
A'BC. This is the point on AG such that GA:GA' = 1:4 because then 
A'A:AM = 2:3 where M is the midpoint of BC (and AG:GM = 2:1).

Among the vertices of triangle A'BC and its center of gravity A there 
are three black points. This means that A' must be white.

In the same way we find white points B' and C' on GB and GC 
respectively, with ratios GB:GB'= GC:GC'= 1:4.

So the vertices of triangle A'B'C' are all white. Since A'B'C' is ABC 
"multiplied" from G with factor 4, it is clear that A'B'C' has G as 
its center of gravity. But G is white as well. This contradicts our 
assumption, and thus the assumption cannot be true, and your theorem 
is proved.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/