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I'm doing exercises in Lectures on Discrete Geometry by Jiri Matousek. There's an application of the Szemerédi-Trotter Theorem to a problem of acute-angled triangles:

Fix an acute angle $\alpha$. Use Szemerédi-Trotter to show that $n$ points in the plane determine at most $O(n^{7/3})$ triangles with each triangle having at least one angle $\alpha$.

How do I begin this?

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  • $\begingroup$ Please explain what you mean by saying "triangles that contain the angle $\alpha$". Do you mean "triangles having at least one of their angles' measure $\ge \alpha$ ? $\endgroup$ – Jean Marie Mar 25 at 16:15
  • $\begingroup$ I mean triangles with at least one angle that is exactly $\alpha$. $\endgroup$ – Milo Mar 25 at 23:09
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Here is an outline. Consider a fixed point $p$ and consider any other point $q$. For any other point $r$ to satisfy $\angle rpq = \alpha$, we have that $r$ must lie on one of two lines (why?). This means that we can count all the occurrences of these triangles that include the point $p$ by the incidents between the $O(n)$ lines ($2$ for every other point $q$) and our set of points. This gives $O(n^{4/3})$ per point $p$ and so $O(n^{7/3})$ altogether.

This bound can actually be improved to $O(n^2 \log n)$ many triangles with angle $\alpha$. See "Repeated angles in the plane and related problems" by Pach and Sharir.

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  • $\begingroup$ "Repeated angles in the plane and related problems" by Pach and Sharir can be found here $\endgroup$ – Jean Marie Mar 26 at 2:10

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