Wednesday, December 3, 2014

Monochromatic Equilateral Triangle, IMO problem

This problem appeared in the International Math Olympiad (IMO) in the 1970/80s.

Each point of the 2D plane, is coloured either black or white. Show that there are three points $A,B,C$ which are of the same colour, and form the vertices of an equilateral triangle.

In other words, there is a monochromatic equilateral triangle, no matter how you colour each point of the plane, with one of two colours.

A variant:

Given three positive angles $\alpha + \beta + \gamma = 180^{\circ}$, show that there is a monochromatic triangle with those angles.

[Solution]

4 comments:

  1. Assume colors are B & R. Take two similar colored points x, y both colored B (without loss of generality). They can form an equilateral triangle with two other points (z, z') on either side of line segment xy. Assuming triangles xyz and xyz' are not monochromatic (else condition is satisfied), then both z, z' are colored R. Now the mid-point of xy (say m) is either B or R. If m is colored B, then we have 3 points (x,m,y) comprising a line segment such that its end points and mid-point are the same color. Similary if m is colored R then the 3 points z,m,z' satisfy the condition that mid-point and end-points of a line segment are of the same color.

    So, in such a plane, we have have at least one line segment with end-points and mid-point of the same color (say B). Now on one side of xmy, form two equilateral triangles xma, xmb. Then with ab form another equilateral triangle abc (such that xcy is also an equilateral triangle).

    c

    a b

    x m y
    B B B

    If a or b is B then we have monochromatic triangle below with the other Bs.
    If a and b are R, then
    if c is R, then abc is monochromatic
    if c is B, then xcy is monochromatic

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  2. Begin with negation that there is no monochromatic ..

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    Replies
    1. Don't worry about the other comments :-) I can delete them.

      Delete