Consider a $n$ dimensional simplex. Prove (or disprove) that a small enough translation of it in any direction produces a simplex which either has a vertex of the old simplex, either the new simplex has a vertex in the old simplex.
This property seems interesting and is obvious for the two dimensional case, as seen in the picture: 
Here, in any direction one would move the triangle, one of the vertices would be in the interior of the new triangle, or the new triangle would have a vertex in the old one. How can this be properly proved in the general case?
To formalize the claim, let $V_k$ be a vertex of the simplex $\mathcal{S}$ and denote $\Omega_k = \mathcal{B}(V_k,\epsilon) \cap \mathcal{S}$.
Then we basically claim that $$ \left( \bigcup_{k=1}^{n+1} \{x - V_k | x \in \Omega_k\} \right) \cup \left( \bigcup_{k=1}^{n+1} \{-x + V_k | x \in \Omega_k\} \right) = \mathcal{B}(0,\epsilon)$$
Edit
It seems that in general
$$ \left( \bigcup_{k=1}^{n+1} \{x - V_k | x \in \Omega_k\} \right) \cup \left( \bigcup_{k=1}^{n+1} \{-x + V_k | x \in \Omega_k\} \right) \subseteq \mathcal{B}(0,\epsilon)$$ Let
$$ \Lambda = \left( \bigcup_{k=1}^{n+1} \{x - V_k | x \in \Omega_k\} \right) \cup \left( \bigcup_{k=1}^{n+1} \{-x + V_k | x \in \Omega_k\} \right) $$ then can we estimate $$ \frac{\text{vol}(\Lambda)}{\text{vol}(\mathcal{B}(0,\epsilon))}$$
It's not true in three dimensions. Take a regular tetrahedron with unit sides, resting on the $(x,y)-$plane with one vertex on the positive $x$-axis and one vertex on the positive $z$-axis. Now raise it a distance $\epsilon$, and translate it along the $x$-axis by a distance $\epsilon$.