I found a general solution that proves this for all n. I'll post it after I make some pictures (they're pretty necessary)

I found a general solution that proves this for all n. I'll post it after I make some pictures (they're pretty necessary)

Posted November 7th, 2017
by Xhin

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.

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, 2017
by Xhin

Glad you find it interesting Xhin. The shape you are describing is called the convex hull of the points. It is the smallest convex set containing all the points. The boundary of that set will consist, as you alluded, of a number of line segments. Each of these could however contain 3 points or more. How does one get a single line that contains two and only two points from the initial set?

Glad you find it interesting Xhin. The shape you are describing is called the convex hull of the points. It is the smallest convex set containing all the points. The boundary of that set will consist, as you alluded, of a number of line segments. Each of these could however contain 3 points or more. How does one get a single line that contains two and only two points from the initial set?

Posted November 7th, 2017
by The Fly

What about the Fano plane?
[img]https://upload.wikimedia.org/wikipedia/commons/5/5b/Fano_plane.jpg[/img]

What about the Fano plane?

Edited November 8th, 2017
by EN

The line joining 1 and 2 contains none of the other points. (Also 2 and 4, or 1 and 4.)

The line joining 1 and 2 contains none of the other points. (Also 2 and 4, or 1 and 4.)

Posted November 9th, 2017
by The Fly

The part that looks like a circle is actually a line that contains 1, 2, and 4. From the wiki page:
=In finite geometry, the Fano plane (after Gino Fano) is the finite projective plane of order 2, having the smallest possible number of points and lines, 7 each, with 3 points on every line and 3 lines through every point.

The part that looks like a circle is actually a line that contains 1, 2, and 4. From the wiki page:

**In finite geometry, the Fano plane (after Gino Fano) is the finite projective plane of order 2, having the smallest possible number of points and lines, 7 each, with 3 points on every line and 3 lines through every point.**

Posted November 9th, 2017
by EN

Yes, but the Sylvester-Gallai Theorem is for the usual Euclidean plane with the usual straight lines. (Not for all projective spaces).

Yes, but the Sylvester-Gallai Theorem is for the usual Euclidean plane with the usual straight lines. (Not for all projective spaces).

Posted November 10th, 2017
by The Fly