I'm trying to show that any 3-dimensional polyhedron with many vertices can be mildly deformed so that its vertices are no longer convexly independent. I suspect it suffices to look at a vertex with low angular defect (guaranteed to exist by Descartes' theorem of total angular defect).

Specifically, I conjecture the following: for all $\epsilon > 0$, there exists $\delta > 0$ such that if $v$ is a vertex with angular defect at most $\delta$, one can construct a map $\rho : \mathbb{R}^3 \to \mathbb{R}^3$ such that:

- $\rho$ preserves all Euclidean distances in $\mathbb{R}^3$ up to a multiplicative $(1 + \epsilon)$.
- If the neighboring vertices of $v$ are $u_1, \ldots, u_d$, then $\rho(v)$ is a convex combination of $\rho(u_1), \ldots, \rho(u_d)$.

This seems like a statement that should have a simple proof if true, but I'm not sure how to prove it.