Representation of vectors in $\mathbb{R}^2$ via differences of small vectors. Is the following fact true?  

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 $w_0=w_k$)

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.  
 A: 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 $\sqrt{5}-\varepsilon$.
A: You can do it with $\|w_i\| \leq \sqrt{5}/2$ and $\sqrt{5}/2 \simeq 1.12$. Indeed, a refinement of the polygonal confinement theorem of Steinitz (see [1]) says that you can reorder your vectors in a way that the partial sums $v_1 + \cdots + v_j$ all satisfy $\| v_1 + \cdots + v_j \| \leq \sqrt{5}/2$. You can now take $w_j = v_1 + \cdots + v_j$. 
[1] Banaszczyk, Wojciech.The Steinitz constant of the plane. J. Reine Angew. Math. 373 (1987), 218--220. 
Here's the MathSciNet review of that paper:
The Steinitz constant of a finite-dimensional real normed linear space $E$ is the minimum constant $S(E)$ satisfying: For any set of vectors $u_1,\cdots,u_n$ such that $\|u_i\|\leq1$ and $\sum^n_{i=1}u_i=0$, there exists a permutation of $\{1,\cdots,n\}$ such that $\|\sum^k_{i=1}u_{p(i)}\|\leq S(E)$ for $k=1,\cdots,n$. In this article the author proves that $S(E)=\sqrt{5}/2$ if $E$ is the Euclidean plane and that $S(E)\leq\frac32$ for any 2-dimensional space.
A: In $\mathbb R^3$, I have a counterexample:
Pick $n$ big.
The vectors $v_1,\ldots,v_{4n}$ are defined as follows:
$v_1=\ldots=v_n=(1-\frac{1}{n},\frac{1}{2n},0)$
$v_{n+1}=\ldots=v_{2n}=(-1+\frac{1}{n},\frac{1}{2n},0)$
$v_{2n+1}=(0,-1,0)$
$v_{2n+2}=\ldots=v_{3n}=(0,0,-1)$
$v_{3n+1}=\ldots=v_{4n}=(0,0,1-\frac{1}{n})$
These vectors can't be put back-to-back inside a unit ball.
