Combinatorial with point in the plane 
Consider $4n$ points in the plane, with no three points collinear.
  Using these points as vertices, we form $\binom{4n}{3}$ triangles.
  Show that there exists a point $X$ of the plane that belongs to the
  interior of at least $2n^3$ of these triangles.

Let $S$ be the set of $4n$ points.
Let $T$ be the set of all lines through the origin. For each $\ell \in T$, let $f(\ell)$ be a line parallel to $\ell$ that does not pass through any point in $S$ and has $2n$ points of $S$ on each side. Such an $f(\ell)$ exists for each $\ell$ since we can move the line continuously in the plane and have the number of points on each side of the line vary continuously in the discrete sense
Any solution?
 A: Consider a line not parallel to any line between two points in our set $S$. Clearly there is some line $\ell$ parallel to this line which bisects $S$. Then by the discrete version of the Ham Sandwich Theorem, there is a line which bisects each of these two sets, with the added constraint that we may have points on our line, such that there are at most $n$ points in each set on each side not counting these. Note that by translating this line by a small amount or if there are two points on the line rotating it by a small angle around their midpoint, we can make it such that these two lines split our set into 4 quadrants with $n$ points in each. We can also translate these lines by a small amount such that their intersection does not pass through any lines containing two points in $S$. Call the intersection of these two lines $X$. We claim that $X$ is contained in at least $2n^3$ triangles.
Now consider two points $A,B$ in opposite quadrants. Say first that their intersection does not pass through $X$. Note then that in one of the two remaining quadrants, every triangle passing through $A,B$, and one of these points will contain $X$. Thus for every pair of points in opposite quadrants, there are at least $n$ triangles whose vertices these points along with a point from another quadrant such that this triangle contains $X$. Note also that each such triangle will not be counted multiple times, since for each such triangle exactly two points are in opposite quadrants. Then there are $2n^2$ choices for these pairs of points, and $n$ choices from here for a third point, so $X$ is contained in at least $2n^3$ triangles.
