Does there exist an isometry from a triangle to a circle?

Question

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Setup:

Let $C_n$ be a closed $n$-simplex in $\mathbb{R}^n$ and let $r \in (0,R)$ where $R$ is the distance any one of the vertices $\{v_1,\cdots , v_{n+1}\}$ of $C_n$ to the centroid $\frac{v_1+ \cdots v_{n+1}}{n+1}\in C_n$.

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Question:

Is there a way or removing a connected open set $A$ from the interior of $C_n$ such that for every $c \in \partial C_n$ $$ r= d(\partial A, c) \left(\triangleq \inf_{a \in \partial A}d(a,c)\right)? $$

Answer 1

This is impossible. If such $A$ existed, then for every $c\in \partial C_n$ there exists $a\in \partial A$ such that $|c-a|=r$. For all other $c'\in \partial C_n$ we have $|c'-a|\ge r$. Thus, the sphere of radius $r$ centered at $a$ is contained in the simplex and passes through its boundary point $a$.

But there is no such sphere when $c$ is a vertex of the simplex, or lies on its edge (or on any face of dimension $<n-1$).