Suppose that we have $N$ equally spaced points. What is the maximal number of chords obtained from connecting the points around the circle so that each chord can not cut across more than one other chord?
I tried with $N=64$, for each adjacent four points, it forms a quadrilateral. When we add another 2 adjacent points from both end sides, we create another attached quadrilateral. Therefore, I obtain total 31 squares. Each square contains 6 connected lines including diagonal with cut across only one point. I have totally 31 quadrilaterals, which give me 186 connected lines. However, there are 30 chords that are shared between quadrilaterals. The total number of chords with one cut across another chord is equal 156.
Now, I am trying to prove that this number is the maximum? Anyone can provide a clue how to solve this for me?
Let $P$ be a convex polygon with $N$ vertices.
Claim. A configuration $\Gamma$ of the described kind can have at most $$\left\lfloor{5N-8\over2}\right\rfloor\tag{1}$$ edges (sides or diagonals of $P$). This number is best possible.
Proof. Let $\Gamma$ be an admissible configuration, and assume that it is maximal in the sense that no further edge can be added in an admissible way. Let $\{a,c\}$ and $\{b,d\}$ be two crossing edges of $\Gamma$. The vertices $a$, $b$, $c$, $d$ then determine a convex quadrilateral $Q\subset P$, and no other edge of $\Gamma$ enters this quadrilateral. On the other hand we can be sure that $\{a,b\}$, $\{b,c\}$, $\{c,d\}$, $\{d,a\}$ are edges of $\Gamma$, since $\Gamma$ is maximal. There will be $r\geq0$ nonoverlapping such quadrilaterals $Q_i\subset P$. If we remove one diagonal from each of these $Q_i$ we obtain a configuration $\Gamma'$ which is a triangulation of $P$, again by maximality of $\Gamma$. Such a triangulation necessarily contains $N-3$ interior edges and $N-2$ triangles.
It follows that any maximal admissible configuration $\Gamma$ can be obtained by starting from a triangulation of $P$ and adding a certain number $r$ of crossing diagonals. Since each such diagonal crosses two adjacent triangles of the given triangulation it follows that $r\leq \left\lfloor{N-2\over 2}\right\rfloor$. The total number of edges then is $$\leq N+(N-3)+\left\lfloor{N-2\over 2}\right\rfloor=\left\lfloor{5N-8\over2}\right\rfloor\ ,$$ as claimed.
On the other hand the number $(1)$ of edges can be realized: Start with a triangulation where all triangles share one vertex $v$ of $P$. The triangles then form a chain of length $N-2$. Beginning at one end you can then form $\left\lfloor{N-2\over 2}\right\rfloor$ pairs of adjacent triangles and add a diagonal to each pair.