Actually I found an even more elegant solution while trying to organize my notes. This one might not require pictures.

For any scattering of n points where they're not all on the same line, you will have an overall "shape" of the outer boundary. This can be a triangle, a quadrilateral, etc. Let's call the number of sides of that shape d.

If you make lines to make whatever that shape is, then you have found solutions. There may be some points inside those shape lines that make that particular solution fail. Let's call the number of those points s.

So then, for any d-sided shape whatsoever, no matter what the points look like inside there will always be at least d-s solutions.

What if d = s? Well in that case you can form a new shape of the same number of sides where all the s points form the outer boundary. Then you apply the same exact rules to this shape.

Whatever your boundary shape is will infinitely recurse as you try to break all the old solutions, inscribing the same n-gon in itself over and over.

Posted November 7th
by Xhin