Let $a_1, ...,a_n ,B\in \mathbb{R}^n$ , not all on the same plane.
Prove that for a small enough neighborhood of zero $U$ and $\forall u_1,...,u_n \in U $ there is a point $C \in \mathbb{R}^n$ that
$\forall k : 1\leq k\leq n :|C-a_k|= |B-a_k|+u_k $
moreover, is this point unique?
I tried to use the implicit function theorem for the function $f(X,y)=|X-a_k|-|B-a_k|+y$ for the point $(B,0)$ somehow. How should i continue?
Define $F:\mathbb R^{2n}\to\mathbb R^n$ as $$ F(u_1,\cdots,u_n,C)=\begin{bmatrix} u_1+|B-a_1|-|C-a_1|\\ \vdots\\ u_k+|B-a_k|-|C-a_k| \end{bmatrix}=\begin{bmatrix} F_1(u_1,\cdots,u_n,C)\\ \vdots\\ F_n(u_1,\cdots,u_n,C) \end{bmatrix}\\\\ F(0,\cdots,0,B)=\begin{bmatrix} 0\\ \vdots\\ 0 \end{bmatrix}\\ $$ Use $${\partial |C-a_i|\over\partial C_j}={\partial \sqrt{\sum_{k=1}^n(C_k-a_{ik})^2}\over\partial C_j}={(C_j-a_{ij})\over|C-a_i|}$$ to see that $$J=\begin{bmatrix} {\partial F_1\over\partial C_1}\;\cdots\;{\partial F_1\over\partial C_n}\\ \vdots\\ {\partial F_n\over\partial C_1}\;\cdots\;{\partial F_n\over\partial C_n} \end{bmatrix}\Big{|}_{(0,\cdots,0,B)}\\$$ has $$ {B-a_i\over|B-a_i|} $$ as its $i$-th row. If these vectors are linearly independent $\det J\neq0$. Then according to implicit function theorem $\exists$ open sets $U,V\subset\mathbb R^n$ containing $(0,\cdots,0),B$ respectively and a unique $\mathcal C^1$ function $g:U\to V:$ $$ x\in U\implies F(x,g(x))=[0,\cdots,0]^\top $$ So $g(u_1,\cdots,u_n)$ is the required $C$. Uniqueness of $g$ implies uniqueness of $C$. I think it might be helpful if you take a look at my answer to another implicit function theorem question.