How many $k$-simplices does a non-boundary $(k-1)$-simplex lie in for a triangulation of a $k$-simplex?

Question

A $k$-simplex is defined as the convex hull of $k+1$ affinely independent vectors (vertices). A face of a simplex is a convex hull of a subset of the vertices.

A triangulation of a simplex $X$ is a collection of simplices such that the union of them is $X$ and the intersection of any two of them is a face of each of the two simplices.

For a triangulation of $\sigma_k:=\{(x_1,...,x_{k+1}):x_1+\cdots+x_{k+1}=1, x_i\ge 0\}$ and a non-boundary $(k-1)$-simplex $A$ in the triangulation, i.e., $A$ is not contained in any boundary of $\sigma_k$, I am wondering how many $k$-simplices in the triangulation does $A$ lie in?

And is there any good reference to advance understanding of the simplices?

I think the answer is 2 but I don't know how to prove it rigorous: consider the hyperplane through the vertices $v_1,...,v_k$ of $A$ in $\mathbb{R}^k$. Then some (non-zero many) vertices of the triangulation should be on one side and some (non-zero many) on the other side, as $A$ is not on the boundary of $\sigma_k$. Since the simplices in a triangulation only intersect at faces, there should be only two $k$-simplices sit on $A$, otherwise two of them sit on the same side of the hyperplane and their intersection is larger than $A$ that is in the hyperplane.

Answer 1

Yes, your intuition is right. Work in the $k$-dimensional affine space$~S$ spanned by $\sigma_k$ (so $S=\{\,(x_1,\ldots,x_{k+1})\mid x_1+\cdots+x_{k+1}=0\,\}$). For any $k$-simplex$~\Delta$ (of the triangulation), any face $F$ of $\Delta$ (a $k-1$ simplex of the triangulation), and any interior point $p$ of $F$, it is the case that any sphere in $S$ centred in $p$ is cut in two halves by the affine hyperplane $H$ of $S$ spanned by $F$. Moreover if the sphere is chosen sufficiently small, one of those two halves is contained in$~\Delta$, while the interior of the other does not meet$~\Delta$. If $F$ is not part of a face of $\sigma_k$, then the entire sphere (if sufficiently small) is contained in $\sigma_k$, so there must be another $k$-simplex$~\Delta'$ of the triangulation that has $p$ on its boundary and contains (part of) the interior of the half-sphere that did not meet$~\Delta$. But then, again with a sufficiently small sphere, it contains all of that half-sphere. There cannot be more $k$-simplices of the triangulation that have $p$ on their boundary, as they would have to have interior points in common with either $\Delta$ or $\Delta'$, which is forbidden.