A *triangulation* of a convex polytope $P\subset\Bbb R^n$ is a partition of $P$ into $n$-simplices $\{\Delta_1,...,\Delta_m\}$ each of which has all its vertices among the vertices of $P$. A polytope may have many different triangulations.

> **Question I:** do all combinatorially equivalent polytopes have the same triangulations?

More precisely, let $P,Q\subset\Bbb R^n$ be combinatorially equivalent, $\phi:\mathcal F(P)\to\mathcal F(Q)$ be a face-lattice isomorphism, and $\{\Delta_1,...,\Delta_m\}$ a triangulation of $P$.
If $\Delta_i$ has vertices $p_1,...,p_{n+1}\in\mathcal F_0(P)$, then let $\phi(\Delta_i)\subset Q$ be the simplex with vertices $\phi(p_1),...,\phi(p_{n+1})$. Do the simplices $\phi(\Delta_1),...,\phi(\Delta_m)$ form a triangulation of $Q$? That is, do they have disjoint interiors and cover all of $Q$?

> **Question II:** if not, is there always a **universal triangulation**? That is, a special triangulation for $P$ that carries over to every combinatorially equivalent polytope in the way described above?