Suppose we have $N$ lines in general position (any two lines, but no three lines, meet at a point) ($N\geq 3$). Let the smallest bounded region have area $1$. Determine the minimum (or possibly infimum) of the total bounded area.

For example, $3$ lines create one triangular region while $4$ create one quadrangular region and two triangular regions (the complete region thus bounded is called a complete quadrilateral). I know that $N$ lines in general position will create ${N+1\choose 2}+1$ regions, ${N+1\choose 2}+1-2N=\frac{(N-1)(N-2)}{2}$ of those bounded on all sides by line segments, but I wanted to figure out how to make the regions maximally “equal” and how close it’s possible to get to the ideal minimum of $\frac{(N-1)(N-2)}{2}$ for a given $N$ (where all bounded regions have area $1$). I have little to no experience in the field of maximizing or minimizing geometrical quantities (especially in such a large mathematical space in which to minimize/maximize as this problem gives), so solving this is entirely beyond my experience.

It is easy to show that there is no maximal bounded area for $N\geq 4$ by drawing some three lines arbitrarily close to meeting at a point. In this case, the resulting triangular region becomes arbitrarily close to a point while maintaining an area of $1$ in the problem's scaling, and the area of the remaining regions grows arbitrarily large by comparison. But I really don't know how to do any of this when it comes to minimizing the total bounded area.

For $N=3$, with only one region, the minimum area is trivially $1$. For $N=4$, some fiddling around on Desmos has convinced me that the minimum area is indeed the ideal of $3$. For $N=5$, I believe that the minimum area still remains the ideal of $6$ (although here my fiddling around on Desmos gets much more nonrigorous and guesswork-y). For $N\geq 6$, however, any proofs or even good guesses of any sort elude me entirely.

If finding the exact minimum area (or infimum in case there’s somehow a minimum that can be approached but not actually reached, which I would not expect) is in fact too difficult to do well, I would also be interested in the growth rate of the actual minimal area relative to the ideal as $N\to\infty$. Thank you for your help! :)

  • $\begingroup$ Can the angles in a quadrilateral ( whose four lengths are given) be always changed to make it cyclic... of maximum area? $\endgroup$
    – Narasimham
    Jul 14, 2022 at 22:59
  • $\begingroup$ I thought I understood the problem, but I lost you soon, when you talk about $N=4$. If I build a square, where do the 2 triangles come from? It's a peculiar configurations, but how can you exclude such a configuration from the optimization process? What is the "not enough general configuration" you're using? Or am I missing something? $\endgroup$
    – basics
    Aug 30, 2022 at 21:00
  • $\begingroup$ @basics It's in general position. Any two lines, but no three lines, meet at a point. So no lines parallel to any others, and no three meeting at a point. So a square is impossible. $\endgroup$ Aug 30, 2022 at 22:21
  • $\begingroup$ ok, I didn't understand that any pair of lines MUST meet. I'll start thinking about the problem in the next days $\endgroup$
    – basics
    Aug 30, 2022 at 22:27
  • $\begingroup$ Well, yeah- I did say "any two lines...meet at a point". But thank you. $\endgroup$ Aug 30, 2022 at 22:28


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