Question

Let $\mathcal{P}$ be a convex lattice polygon with $n$ vertices and let $\mathcal{L}$ be the set of all lattice points inside $\mathcal{P}$. For every $n \geq 5$, does there exist a point in $\mathcal{L}$ such that it also lies in the convex polygon bounded by (all) the diagonals of $\mathcal{P}$? How many such points are there? (//By diagonals I mean of course the lines different from the sidelines of the polygon which are connecting two vertices of $\mathcal{P}$.)

I proved a while ago that for $n=5$ there is such a point in $\mathcal{L}$. I also managed to show this now for $n \geq 6$ using a similar argument, yet it got more involved and I still need to check for potential bugs. Any ideas for the general case?

Answer 1

For $n=5$, this has been shown by Eppstein:
D. Eppstein, Happy endings for flip graphs, Journal of Computational Geometry 1 (2010), no. 1, 3--28.

For odd $n>5$, one could consider the polygon bounded by the longest diagonals.
It may be defined as the intersection of the half-planes containing $(n+1)/2$ vertices of $P$.
For $n=9$, this intersection may be empty (for example, if the nine vertices form three triples and each of the triples is placed very close to a vertex of a regular triangle).
For $n=7$, this intersection is non-empty. However, it may be free of lattice points: take, for example, the polygon $P$ with vertices $[0,1], [1,0], [2,0], [3,2], [3,3], [1,3], [0,2]$.