For a convex polygon $P$, draw all the diagonals of $P$ and consider the intersection points made by those diagonals. Let $f(n)$ be the minimal number of such intersections where $P$ ranges over all convex $n$-gons. (If three or more diagonals intersect at one point, we count it as a single point)

For example, $f(3) = 0$, $f(4) = 1$, $f(5) = 5$, but $f(6) = 13$ since there are only thirteen intersections in a regular hexagon.

Question:Is $f(n) \sim \frac{1}{24} n^4$? If not, then what is the asymptotic behavior of $f(n)$?

**Comment 1.** Obviously, $f(n) \le \binom{n}{4}$. On the other hand, it is possible to show that $f(n) \gt c n^4$ for an absolute constant $c \gt 0$ using the Szemerédi–Trotter theorem.

**Comment 2.** Poonen and Rubinstein proved that for a regular $n$-gon, the number of intersections made by its diagonals is $\frac{1}{24} n^4 + O(n^3)$.

("Number of Intersection Points Made by the Diagonals of a Regular Polygon." SIAM J. Disc. Math. 11, 135-156, 1998)

**Edit 1.** (August 17, 2015) In the comments, Boris Bukh suggested to consider the $n$-gon determined by the points $(i, i^2)$ for $-n/2 \le i \lt n/2$. Because I was unable to make a nice estimate in this case, I calculated them up to $n=200$. Here is the result.

The blue, orange, and green lines each represents the general case, the regular $n$-gon, and the parabola polygon; the values plotted are $(\# \text{ of intersection})/\binom{n}{4}$. This clearly shows that the parabola polygon has far less intersections compared to the regular $n$-gon. Still, the quotient increases after about $n=50$, and it might be heading towards $1$.

Subquestion:Is $(\# \text{ of intersections}) \sim \frac{1}{24} n^4$ for the parabola polygon?