TITLE: $n$ lines in a general position and the number of empty triangles
QUESTION [7 upvotes]: Question. Consider $n \geq 5$ lines in a general position (i.e. no two lines are parallel and no triple intersections are allowed) in $\mathbb{R}^2$. Let $T(n)$ denote the maximal number of empty triangles (here empty triangle means that it does not contain other triangle). What would be best upper and lower bounds for $T(n)$? I know $(n-2) \leq T(n)$ holds, but I am hoping for a better lower bound. Is it true that $n \leq T(n)$? Also, is it possible to compute $T(n)$ it for small $n$ (where small means $6 \leq n \leq 10$)? I think $T(6) = 6$, but I am not able to show $6$ is an upper bound as well.

REPLY [2 votes]: You can easily show that $3k$ lines can be arranged to have at least $\frac34 k^2$ triangles, which proves that asymptotically, $T(n) \ge n^2/12$. 
The following picture shows the example for $k=4$ -- there are three groups of $k$ lines each, group 1 and group 2 forming a "squarish" $k\times k$ chessboard, which is being then intersected by the lines from group 3. They will produce $(k+1)+(k+1)+k+k+(k-1)+(k-1)+...$ triangles which comes to approximately $\frac34k^2$ triangles. And I am not even counting the ones that these groups will produce far away from the chessboard, inside their "own domain" so to speak.
picture with 3 $k$ lines: