Congruent quadrilaterals in a tri-colored $72$-gon

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.

• I feel like the proof would involve the pigeonhole principle somehow. – Akiva Weinberger Aug 27 '15 at 7:00
• It seems that I can prove a counterpart of the claim for the congruent monochromatic triangles and for the quadrilaterals for $n$-gon, provided $n\ge 81$. – Alex Ravsky Sep 1 '15 at 8:59
• 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... – Kevin P. Costello Sep 1 '15 at 17:26
• @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. – EuYu Sep 1 '15 at 21:03