Let $C_1, ..., C_{n+1} \in R^{n}$ be the vertices of a simplex and $D_1, ..., D_{n+1} \in \mathbb{R}^n$ form another simplex. Then what is the upper bound on the number of the vertices of $$\text{conv}(C1, ..., C_{n+1}) \cap \text{conv}(D1, ..., D_{n+1})$$
My try
Let $\sigma_1$ and $\sigma_2$ denote the simplices. I think that since each simplex has at most $n+1$ vertices and facets then an upper bound would be $(n+1)\cdot n$ since each vertex $\sigma_1$ would correspond to possibly $n$ vertices that are formed on a facet of $\sigma_2$ which "cuts out" the vertex.