Density of connecting lines inside a circle

Question

A circle can be seen as an infinite number of points which are all placed at the same distance from the center. Let's start with a polygon of $n$ points. The polygon approaches a circle for $n \rightarrow \infty$. Now lets connect every corner point of the polygon with a line. We should end up with $\frac{n^2}{2}-n$ lines. Is there an analytical solution for the density distribution these lines create inside the circle for $n \rightarrow \infty$? By density I mean the number of lines crossing some area within the circle.

Answer 1

If I got your question right, no there isn't. You might convince yourself looking at the Bertrand Paradox.

If you just want to know if your sequence converges in law, and identify the limit, I think it does : for oriented lines, the sequences of beginnings and ends does converge toward the uniform law. That's just a lead, I have no formal proof.

Answer 2

Let us work in a unit circle. You can compute the closest approach radii as a function of $n$. For example, if $n$ is even there are $\frac n2$ diameters among the diagonals. The next $n$ diagonals come from diagonals that are offset by one vertex from a diameter, like $AG$ in the diagram. The next $n$ come from diagonals that are offset by two vertices from a diameter, like $AJ$.
The angle at $A$ subtended by the arc is $\pi-k\frac {2\pi} n$ where $k$ is the number of vertices the diagonal is offset from the diameter, so $1$ for $AG$ and $2$ for $AJ$. If we let $O$ be the center of the circle and $P$ be the midpoint of the diagonal, we can find $OP=\sin OAP=\sin \frac {k\pi}n$. When $k \ll n$ we can see the miss distance will be about $\frac {k\pi}n$ so you have one set of diagonals that passes $\frac {k\pi}n$ from the center for each $k$. As $k$ gets larger they will get more tightly packed.

Answer 3

I assume that you mean taking unique pairs of points to form a chord. If so then you have the number of chords wrong. Here $n$ must be even, and There are $n$ choices for the first point and $n−1$ for the second. But for any chord it doesn't matter which is chosen first so the total number of chords is $(n(n−1))/2$.

Now to answer your question. As $n \rightarrow \infty$, then this is the random angle method from Bertrand's paradox. However to get the density of chords you have to convert the formulas which are typically given as chord length into a density for the midpoints via a change of variable.

If you want a complete graph over the $n$ points it is still the random angle method for Bertrand's paradox, but (1) $n$ could be even or odd and (2) it just changes the number of chords per $n$ points on the circle.