Let $\Omega$ be a polygonal domain in $\mathbb{R}^3$. Assume $\Omega$ is partitioned into tetrahedra using the most common admissible triangulation, that is, roughly speaking, two adjacent tetrahedra meet vertex-to-vertex, or edge-to-edge, or face-to-face.
The question is:
Suppose $z$ is an interior vertex in $\Omega$ and there are $m$ tetrahedra having $z$ as a vertex. Can we determine the number of faces adjacent to $z$ ? If so, what is it (expressed using $m$) ?
There are $3m/2$ adjacent faces, and $m$ must be an even number. Else you would have a face that is adjacent to only one tetrahedra and then $z$ wouldn't be an interior point.