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$S$ is a set of R(m,m;3) points in the plane in which no 3 points are collinear. I am trying to prove that $S$ contains $m$ points that form a convex $m$-gon.

I have tried using similar logic to the Happy Ending Problem but I can't come up with an argument to apply to this problem.

Edit: I have started trying to apply an argument by defining the coloring of monochromatic triangles of points being based on whether the amount of points inside the triangle is even or odd. For example if a triangle has 2 points in it color it red, and if it has 3 points in it color it blue. I am getting pretty close however I can't seem to connect the dots on why there must be a convex m-gon.

Alex Ravsky
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Mroog
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2 Answers2

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It seems the following.

Enumerate points in an arbitrary order. Color a triangle of the points red, if starting from its vertex with the smallest number and going clockwise we visit its vertices in increasing order, and color it blue, otherwise. Ramsey Theorem implies that there exists a set $S’\subset S$ of $m$ points and a color $c$ such that all the triangles formed by vertices of the set $S’$ are colored by $c$. If $S’$ is not the set of vertices of a convex $m$-gon then, by Carathéodory theorem, :-) there exists a triangle formed by vertices of the set $S’$ containing another vertex of the set $S’$. A simple check shows that it is impossible.

Alex Ravsky
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    Nice. I will just expand a bit on this for a future reader. Note that if a triangle is coloured blue, then we will visit vertices in increasing order when going anti-clockwise, which is analogous to the clockwise case (look at the plane from the other side!) . So for the "simple check" at the end, we can safely assume the clockwise-increasing condition. Also, there are $4$ cases to check, depending on the order of the middle vertex relative to the three others (it's in either position $1$, $2$, $3$ or $4$). They all throw up contradictions with the clockwise condition. – Colm Bhandal Dec 11 '15 at 17:23
  • @ColmBhandal Yes, you are right. But, maybe there are even more simple ways to heck. – Alex Ravsky Dec 11 '15 at 17:46
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    Thank you! I spent so long trying the method I mentioned in the edit but couldn't make it work. This method is much more clear to me. – Mroog Dec 11 '15 at 18:08
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    @AlexRavsky for sure. I just went brute force because my mind was lazy :) Nice solution. – Colm Bhandal Dec 11 '15 at 19:30
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    I heard a story that some clever young Hungarian (don't recall which one) came up with this proof on an exam. The question was something like "prove that a sufficiently large set of points in general position in the plane must contain the vertices of a convex $m$-gon", and the student had missed the lecture where the proof using $(m,5)^4$ had been presented. – bof Apr 13 '18 at 02:48
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I see that this has already been answered and the answer accepted, but for the curious, there is in fact a way to prove it using the original attempt of coloring based on even/odd interior points in the triangle. The proof is due to Scott Johnson, in a paper titled "A New Proof of the Erdos-Szekeres Convex k-gon Result".

As you've said, for each subset of 3 points $a,b,c\in S$, we can color it RED if the triangle $abc$ has an even number of points of $S$ inside it and BLUE otherwise. By Ramsey's Theorem there is a set of $m$ points $X$ such that all 3-subsets of $X$ are the same color.

These $m$ points form a convex $m$-gon. Here's the proof from Johnson: Suppose by way of contradiction that the $m$ points in $X$ are not the vertices of a convex $m$-gon. Then, the convex hull of $X$ contains some other point of $X$ in its interior. By Carathéodory's theorem, this means there are four points $a,b,c,d\in X$ such that $abc$ form a triangle containing $d$. Now, draw line segments from $d$ to each of $a,b,c$ to form three triangles $abd$, $acd$, and $bcd$. Now, if we look at the sum of the numbers of points inside each of these three triangles, it is one less than the total number of points inside $abc$. In symbols (letting, e.g. $|abc|$ denote the number of points of $S$ inside triangle $abc$): $$|abc| = |abd|+|acd|+|bcd|+1$$ Since these three triangles perfectly divide $abc$ (the plus 1 is due to the fact that $d$ is not internal to any of the three triangles). But note that this means that we cannot have all four triangles contain an even number of points, nor all four containing an odd number of points. Since, if all $|abd|,|acd|,|bcd|$ are even, then $|abc|$ must be odd (and vice versa). This contradicts the result of applying Ramsey's Theorem, which gaurantees they are all colored the same.

As a side note, there is also a proof that instead applies to a set of $R(m,5:4)$ points, where we color 4-subsets of $S$ based on whether the 4 points are the vertices of a convex 4-gon. Ramsey's Theorem says there is either a set of $m$ points for which all 4-subsets are the vertices of a convex 4-gon, or a set of 5 for which none of the 4-subsets are the vertices of a convex 4-gon. Klein's original Happy Ending Problem argument implies the latter is impossible, so we must have the former: a set of $m$ points for which all 4-subsets are the vertices of a convex quadrilateral. A fairly straightforward argument suggests that this set of $m$ points are the vertices of a convex $m$-gon (the argument is similar to the above: otherwise, we'd have four points $a,b,c,d$ with $d$ inside the triangle $abc$, but then these 4 would not be the vertices of a convex 4-gon, which is a contradiction). This is the original argument of Erdos and Szekeres.

ngrammel
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