Triangulation of Hypercube

Question

Assume that we have a triangulation of the hypercube in $\mathbb{R}^n$ (i.e. a partition into simplices that only involves the $2^n$ vertices). Can we always find a vertex $v$ of the hypercube such that $v$ together with the $n$ vertices that are adjacent to $v$ (i.e. share an edge of the hypercube with $v$) form a simplex from the triangulation? This is clearly true for $n\leq2$ but does it hold for higher dimensions as well?

Answer 1

No, this is already not true for $n=3$, as you can see in this image on Wikipedia:

Each vertex has a red line to a non-neighbouring vertex emanating from it.