Is the following fact true? 



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> Let $v\_1,\ldots, v\_k \in \mathbb{R}^2$, $\|v\_i\|\leq 1$, be vectors that add up to zero. Does there exist a permutation $\sigma\in S\_k$ and vectors $w\_1,\ldots, w\_k \in \mathbb{R}^2$, $\|w\_i\|\leq 1$, such that $v\_{\sigma(i)}=w\_i-w\_{i-1}$? (here, I assume $w\_0=w\_k$)
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**Edit. Related (known) facts:**  

1. The same fact in $\mathbb{R}^1$ is true. (can be easily proven by choosing $w\_0=0$ and $\sigma(i)$ such that $\|w\_i\|\leq 1$ for $w\_i:=w\_{i-1}+v\_{\sigma(i)}$)  

2. For each $\epsilon>0$ there exists a family of vectors $v\_i$, such that $\max\_i\|w\_i\|>1-\epsilon$. See my comment below for the proof. 



Representation of vectors in $\mathbb{R}^2$ via differences of small vectors.
I have a counterexample in $\mathbb R^2$.
Here's how it goes.  



Pick two numbers $n$ and $N$, with $N>>n>>1$.  

The collection {$v\_i$} consists of:


* $N(n-1)$ times the vector $(\frac{n-2}{n},\frac{1}{N(2n-3)})$
* $N(n-2)$ times the vector $(-\frac{n-1}{n},\frac{1}{N(2n-3)})$
* The vector $(0,-1)$ once.


The smallest ball into which those vectors can be fit back-to-back has diameter $\sqrt{5}-\varepsilon$.


