Let $S$ be a set of $n$ points in the plane in general position. Each 3 points of S span a triangle. Total number of triangles spanned by S:

$$\binom{n}{3}=\frac{n(n-1)(n-2)}{6}=\frac{1}{6} n^3-O(n^2 )$$

We know that Every set of $n$ points in general position in the plane can be partitioned into $6$ parts of almost equal size by $3$ concurrent lines (at least $\frac{n}{6}-c$(where $c$ is constant) points in each part).

Let $x$ be the point of intersection of the $3$ lines.

I need to prove that *x is contained in at least $\frac{n^3}{27}-O(n^2 )$ triangles.*

My approach:

•Every triangle of the form ACE contains $x$, the number of the triangles of this form $\frac{n^3}{216}.$

•Every triangle of the form BDF contains x, the number of the triangles of this form $\frac{n^3}{216}.$

•For every edge of the form AD, either adding any point of E makes a triangle that contains x, or adding any point of C, the number of the triangles of this form $\frac{n^2}{36}(\frac{n}{6}+\frac{n}{6})=\frac{n^3}{108}.$

How to proceed after then? I have stucked at this stage.