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This is a problem that I have spent a good 2 hours on but can seem to come up with any rigorous solution. If someone could provide one that would be great!

If we color 21 vertices of a 50-gon, how do we go about proving that will always exist 3 vertices such that if you connect them you get an isosceles triangle?

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I've re-titled the question to be more accurate and hopefully generate more traffic; the original title was misleading, as the problem may not be of any particular interest to some readers. –  Arkamis Aug 28 '12 at 20:35
    
Presumably the 50-gon is regular? –  Jyrki Lahtonen Aug 28 '12 at 20:38
    
Depends on how many colors you are coloring with. If I have 50 different colors to choose from, then it obviously isn't true. –  Rijul Saini Aug 28 '12 at 20:38
    
Sorry I should have been more clear. We have to connect 3 vertices of the same color, and all 21 are colored the same color. Also the 50-gon is regular –  Nitesh Aug 28 '12 at 20:44
    
Do you mean that after you "color" 21 vertices, you can always find three colored vertices that form an isosceles triangle? Or do you mean that if you color 21 vertices one color and 29 a second color, you can always find three vertices of the same color that form an isosceles triangle? –  MJD Aug 28 '12 at 21:27
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1 Answer 1

up vote 6 down vote accepted

Join alternate vertices of the $50$-gons to produce two $25$-gons. One of the two $25$-gons must have at-least $11$ colored vertices. Further partition this polygon into $5$ - regular pentagons by joining every fifth vertex. One of the pentagons must contain at least three colored vertices. But then those three vertices necessarily form an isosceles triangle.

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Beautiful! There are probably clunkier pigeonhole principle arguments, but this is an amazingly elegant one. +1 for sure. –  Steven Stadnicki Aug 28 '12 at 21:41
    
Thank you very much! –  Nitesh Aug 28 '12 at 21:45
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