On Sat, Nov 11, 2017 at 11:06 AM, James Propp <jamespropp@gmail.com> wrote:
Veit's proof collapses the 24 edge-meets-edge terms into a single term whose value is the constant 2! What's more, his argument can be used to show that for any centrally symmetric 2n-gon, the expected number of vertices in the intersection of the 2n-gon with a translate of itself, conditioned on the event that the intersection is nonempty, is exactly n+2.
I think this needs the additional assumption that the polygon is convex. Take a polygon with a sawtooth top edge, and then a thin horizontal bar above the sawtooth. translate this down a bit, and the intersection isn't connected, and has lots more sides; each tooth of the saw, which has 2 sides, contributes an additional 4-sided quadrilateral to the intersection.
The part of his argument that I still don't understand is this: how do wet know that (generically) the boundaries of the two 2n-gons intersect exactly twice?
This isn't true in general; it's not true for the intersections of of a short fat rectangle and the translates of a long skinny rectangle. I believe it is true for the intersection of a convex polygon and its translates, but don't have a proof yet. But the fact that the proof needs to use both the fact that the polygons are translates and that they are convex should help guide the search for a proof. Let's say that P and Q are two points on the boundary of A + v, neither of which are in A. We want to show there is a path from P to Q along the boundary of A + v that does not intersect A. This will show that the boundary of A+v has only one component outside of A. I don't have the proof yet, but I can "see" that this is true, with the relationship between v and P-Q telling us which of the two paths from P to Q stays outside of A. Andy