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 3 colors (Posted on 2003-05-01)
Imagine that a painter went down to a mathematical plane and colored all of the points on that plane one of three colors.

Prove that there exist two points on this plane, exactly one meter apart, that have the same color.

 See The Solution Submitted by Jonathan Waltz Rating: 4.1000 (10 votes)

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 Proof by contradiction | Comment 16 of 27 |
This is a proof by contradiction.

Lets assume the statement "there exist two points on this plane, exactly one meter apart, that have the same color" is false. Then every equilateral triangle with sides of 1 meter have three different colors at the verticies. Consider two adjacent triangles. The two triangles can be ABC and BCD and share side BC.

Points A, B, and C are all different and points B, C, and D are all different. The only way this happens is when A and D are the same color.

Consider a second pair of triangles DEF and EFG with common side EF. Using the same argument as with points A, B, C and D, D is the same color as G. This means that A is the same color as G.

But points E, F, and G can be chosen such that the distance between A and G is 1 meter. A contradiction is formed, therefore the statement "there exist two points on this plane, exactly one meter apart, that have the same color" must be true.
 Posted by Brian Smith on 2003-05-01 08:37:44

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