Let $d \ge 2$ be an integer. Prove that there exists a constant $C(d)$ such that the following holds: For any convex polytope $K\subset \mathbb{R}^d$, which is symmetric about the origin, and any $\varepsilon \in (0, 1)$, there exists a convex polytope $L \subset \mathbb{R}^d$ with at most $C(d) \varepsilon^{1-d}$ vertices such that \[(1-\varepsilon)K \subseteq L \subseteq K.\] Official definitions: For a real $\alpha,$ a set $T \in \mathbb{R}^d$ is a [i]convex polytope with at most $\alpha$ vertices[/i], if $T$ is a convex hull of a set $X \in \mathbb{R}^d$ of at most $\alpha$ points, i.e. $T = \{\sum\limits_{x\in X} t_x x | t_x \ge 0, \sum\limits_{x \in X} t_x = 1\}.$ Define $\alpha K = \{\alpha x | x \in K\}.$ A set $T \in \mathbb{R}^d$ is [i]symmetric about the origin[/i] if $(-1)T = T.$

Let K be a bounded, d-dimensional convex polyhedron that is not simplex and P is a point on K. Show that if vertices $P_1 , ..., P_k$ are not all on the same face of K, then one of them can be omitted so that the convex hull of the remaining vertices of K still contains P. [hide=note]caratheodory's theorem might be useful. [/hide]