# Can we split a polygon in half (vertex-wise) by a diagonal, but with a constant maximum difference?

This is a followup to my last question - Can we prove that simple polygons can always be split in half (vertex-wise) by diagonals?

Is there a constant natural number K for which the following is true:

For any simple polygon with more than 3 vertices there always exists a diagonal which:

• is inside the polygon
• doesn't intersect with any edges
• splits the polygon in two polygons in such a way that the difference between their vertex counts is smaller than K
• No, by a slight modification of the example answering the previous question: just place most of the vertices evenly between A and A' and the other two pairs. Gerhard "Imagine Using A Circular Saw Blade" Paseman, 2019.10.02. – Gerhard Paseman Oct 2 at 17:01

We start with an equilateral triangle $$ABC$$ and a larger equilateral triangle $$A'B'C'$$ with the same center and parallel edges. We consider the hexagon $$AA'BB'CC'$$.
Instead of this hexagon, we replace $$A'$$ with a sequence of $$2K$$ points all clustered near the original $$A'$$. Then we do the same for $$B'$$ and $$C'$$, so that the new polygon has $$6K+3$$ vertices.
Any division of this polygon into nearly equal halves will have to divide up one the clusters of $$2K$$ vertices. But starting from the cluster of vertices near $$A'$$, the only diagonals that avoid edges either stay within the cluster or go to $$B$$, so no diagonals can avoid the edges and provide the desired split.
• A diagonal from $A$ to $B$ has $2K$ vertices on one side, $4K+1$ on the other, so the minimum difference (which I think is what you meant to say) is $2K+1$. – Matt F. Oct 2 at 17:15