Triangulation of Polytope

Question

It seems to me that, in two dimensions, a triangulation of a polytope with odd amount of vertices gives us an odd amount of simplices while a triangulation of a polytope with even amount of vertices gives us an even amount of simplices. (We can add vertices to the boundary of $P$ during the triangulation, but then they must be counted as vertices of $P$.) See the following images for example. Just count the vertices on the boundary of the polytope and the number of simplices that it is composed of.

Does anyone know if this claim is true? If yes, where can I find a proof for it? Otherwise, a counterexample would be nice. Thanks.

Answer 1

This is a simple proof that, in two dimensions, a triangulation of a polytope with odd amount of vertices gives us an odd amount of simplices while a triangulation of a polytope with even amount of vertices gives us an even amount of simplices.

You have a plane graph, bounded by a cycle $C$, in which every internal region has three edges. Let $C$ have length $k$, and let there be $n-k$ vertices in the interior, so n vertices altogether. Let this graph be $G$.

If we add another vertex and make it adjacent to every vertex of $C$ we obtain a maximal planar graph $H$. It is known (an easy consequence of Euler's formula) that every maximal planar graph with $p>3$ vertices has exactly $3p-6$ edges and $2p-4$ regions. $H$ has $n+1$ vertices, so it has $3n-3$ edges and $2n-2$ regions. But $H$ has $k$ regions more than $G$, so $G$ has $2n-2-k$ regions (not counting the infinite region). This is even if and only if $k$ is even.