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I recently watched a movie (A Brilliant Young Mind) in which this problem appeared:

Let the vertices of a regular $72$-gon be colored red, blue, and green in equal parts. Show that there are $4$ vertices of each color such that the resulting monochromatic quadrilaterals formed are congruent to each other.

I don't know the solution to this problem, nor do I even know if the problem is actually true (it is from a movie after all). But I would love to see a proof, if one exists, or otherwise a counter-example.

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    $\begingroup$ I feel like the proof would involve the pigeonhole principle somehow. $\endgroup$ – Akiva Weinberger Aug 27 '15 at 7:00
  • $\begingroup$ It seems that I can prove a counterpart of the claim for the congruent monochromatic triangles and for the quadrilaterals for $n$-gon, provided $n\ge 81$. $\endgroup$ – Alex Ravsky Sep 1 '15 at 8:59
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    $\begingroup$ A quick proof of Alex's statement: View the points as integers mod $72$. To each triple $(r,g,b)$ of one point of each color we associate the pair $(g-r, b-r)$ (taken modulo $72$). There's $24^3$ triples, and $71^2$ pairs, so by pigeonhole some pair is associated to $3$ distinct triples, which must be disjoint. This corresponds to a red, green, and blue triangle that are rotations of each other. Replace $72$ by $81$ and $24$ by $27$, and now you have four triples, which gives a quadrilateral. An obvious place for improvement would be to account for reflections somehow... $\endgroup$ – Kevin P. Costello Sep 1 '15 at 17:26
  • $\begingroup$ @KevinCostello That's a wonderful argument. Please add it as an answer. If there are no other answers by the time the bounty is about to expire, I would happily award the bounty to you. $\endgroup$ – EuYu Sep 1 '15 at 21:03
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This is essentially Problem #7 from Tournament of the Towns, Fall 2011, Junior A-Level and Problem #2 from USAMO 2012.

Tournament of the Towns, Fall 2011, Junior A-Level, Problem #7. Each vertex of a regular 45-gon is red, yellow or green, and there are 15 vertices of each colour. Prove that we can choose three vertices of each color so that the three triangles formed by the chosen vertices of the same color are congruent to one another.

The official solution is here.

USAMO 2012, Problem #2. A circle is divided into 432 congruent arcs by 432 points. The points are colored in four colors such that some 108 points are colored Red, some 108 points are colored Green, some 108 points are colored Blue, and the remaining 108 points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.

The official solution is here. There is discussion at AoPS here.

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  • $\begingroup$ Great! Thank you for the references. $\endgroup$ – EuYu Sep 2 '15 at 20:47

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