Two Colors - Two Points

Points in the plane are each colored with one of two colors: red or blue. Prove that, for a given distance d, there always exist two points of the same color at the distance d from each other.

Solution

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Copyright © 1996-2018 Alexander Bogomolny

Points in the plane are each colored with one of two colors: red or blue. Prove that, for a given distance d, there always exist two points of the same color at the distance d from each other.

Solution 1

Select a point O. With O as the center, draw a circle of radius d. There are just two possibilities:

  1. The circle thus drawn contains a point of the same color as O. If that's the case we are finished.
  2. All points of the circle have a color different from the O's. Then any chord of length d connects two points of the same color.

A question may be asked by a student or suggested by a teacher, Is it always the case that in a circle of radius d there exists a chord of length d? The answer may at first seem obvious but, depending on a student's level, may eventually prove insurmountable. A discussion here may be extremely useful.

For the lower grades, it's possible to simply stipulate existence of such a chord. For more advanced students, a better axiom would be the one that asserts that two circles, each passing through the center of another, intersect at two points.

Solution 2

Consider any equilateral triangle ABC of a given side d. Of the three points A, B, C, two must be of the same color. These two solve the problem.

Remark

Observe that there are colorings of a line with two colors with no monochromatic pait of points at a given distances. For example, the line is the union of half closed, half open segments [n, n+1). Color these segments so that any adjacent pair is colored with different colors. Then no pair of monochromatic points is at distance 1.

Elsewhere, I discuss a similar problem of finding two points of different colors. Do have alook.

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|Contact| |Front page| |Contents| |Coloring Plane|

Copyright © 1996-2018 Alexander Bogomolny

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