Triangle in Randomly Colored PlaneDate: 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/ |
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