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Let $P$ be a set of $n$ points in the plane. Show that there are at most $O(n^{7/3})$ triangles of unit area whose vertices are from $P$ by using Szemeredi-Trotter theorem.

kimchi lover
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trung chum
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    Fix a point $p \in P$. Then for any other point $q$, if $pqr$ make a unit triangle then $r$ must lie on one of two lines (that are parallel to the segment $pq$). That means the number of such triangles that include the point $p$ are counted by incidence between $P$ and $O(n)$ lines so Szemeredi-Trotter gives $O(n^{4/3})$. Then multiply by $n$ for all possible $p$ to get $O(n^{7/3})$. – Sandeep Silwal Mar 29 '20 at 23:47

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Sandeep has sketched the solution in the comments. Let me fill in the details (and also get this question out of the "unanswered questions" queue).

Let us recall the Szemerédi-Trotter theorem, first published in their paper Extremal problems in discrete geometry (Combinatorica 1983):

Theorem (Szemerédi and Trotter 1983) Suppose there are $n$ points and $\ell$ lines on the plane $\mathbb{R}^2$. The number of incidences between points and lines is at most $c( (n \ell)^{2/3} + n + \ell$) for some absolute constant $c>0$.

Here, an incidence just means a pair $(P, L)$ where $P$ is a point and $L$ is a line from our collection and $P\in L$. When $\ell=n$, we see that the number of incidences is $O( n^{4/3})$. Of course, when $\ell = 2n$ or more generally $\ell = \lambda n$ (for some fixed $\lambda$), then again the number of incidences is at most $O(n^{4/3})$. A particularly elegant proof of Szemerédi-Trotter theorem was given by Székely in his paper Crossing Numbers and Hard Erdös Problems in Discrete Geometry (Combinatorics, Probability and Computing 1993). Székely's proof used the crossing lemma (see here for a statement).

We can now prove the following result.

Proposition. Given $n$ points on the plane, they determine at most $O(n^{7/3})$ triangles of unit area.

By "determine", we mean that the triangle has vertices coming from the collection of points under consideration.

Proof. Let us fix a point $P$ from our collection, and try to understand how many unit area triangles are possible where $P$ is one of the vertices. Well, we first need to connect $P$ to another point $Q$ in the collection to get a side of the triangle. Consider the line $\overline{PQ}$ joining them. For a third point $R$ to be the remaining vertex of the triangle, it needs to be positioned anywhere on a particular line paralell to $\overline{PQ}$ with a prescribed distance away (and there are two possible parallel lines, one for each side of the line $\overline{PQ}$). In other words, the number of unit triangles where one of the vertices is $P$ can be counted by analyzing the number of incidences between the points in the collection (other than $P$, so there are $n-1$ points) and the 2 parallel lines joining $P$ to other points $Q$ (and again, there are at most $2(n-1)$ such lines). By corollary to Szemerédi-Trotter theorem, there are at most $O(n^{4/3})$ such incidences, which gives an upper bound for the number of unit area triangles where $P$ is one of the vertices. But now we must vary over $P$ among all $n$ different points in the collection, giving us a total of at most $n\cdot O(n^{4/3}) = O(n^{7/3})$ many triangles of unit area all of whose vertices come from our collection.

Prism
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