# Deformations that flatten small curvature

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.

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