# Use Szemerédi-Trotter to show that $n$ points in the plane determine at most $O(n^{7/3})$ triangles that contain a fixed acute angle $\alpha$.

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?

• 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 $\ge \alpha$ ? – Jean Marie Mar 25 at 16:15
• I mean triangles with at least one angle that is exactly $\alpha$. – Milo Mar 25 at 23:09

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.