This problem was first put to me by Luke Pebody (who did not know the answer at the time) and after some work I am yet to find a proof or counterexample. I would be grateful of any insights.


Call a vector $v$ in $\mathbb{R}^2$ 'short' if it has modulus less than 1. Let $v\_1,\dots,v\_6$ be short vectors such that $\sum\_{i=1}^6 v\_i = 0$. Prove that some three of the $v\_i$ have a short sum.



An elementary problem in Euclidean geometry
Lemma 1:


If the angle, $\theta$, between two 'short' vectors, $v\_1$ and $v\_2$ (placed head to tail) satisfies $\theta \leq \pi/3$, then their sum is a short vector.


Proof:


Place $v\_1$ at the origin, then the terminal point of $v\_1$ lies within the unit ball. Placing the $v\_2$ at the tail of $v\_1$ creates an angle $\theta \leq \pi/3$. Any arc of radius $r < 1$ traced between $\pi/3$ and $-\pi/3$ from the terminal point of $v\_1$ lies within the unit ball as well. Thus $v\_1+v\_2$ is a 'short' vector. $\blacksquare$


Since $\Sigma v\_i = 0$ we can order the vectors in such a way as to create a convex polygon. If any of the interior angles satisfy the conditions for lemma 1, then we can reduce the problem to a polygon with fewer sides.


Lemma 2:


In a convex n-gon (n=4,5 or 6), with interior angles $\theta\_i$, either $\theta\_i \leq \pi/3$ for some $i$ or at least one pair of adjacent sides (as vectors) may be interchanged so that an angle less than $\pi/3$ is created.


Proof:


If $v\_1$ and $v\_2$ create interior angle $\alpha$ and $v\_2$ and $v\_3$ create angle $\beta$, then interchanging $v\_2$ and $v\_3$ creates an angle $\alpha+\beta-\pi$ between $v\_1$ and $v\_3$ (which are now adjacent by exchanging $v\_2$ and $v\_3$). Aiming for a contradiction, assume that this new angle, $\alpha+\beta-\pi > \pi/3$. Thus, $\alpha+\beta > 4\pi/3$. If this were true for every pair of adjacent angles in a quadrilateral, then $16\pi/3 < 2(\Sigma^4\_{i=1} \theta\_i)$. $\sharp$ For a 5-gon, $20\pi/3 < 2(\Sigma^5\_{i=1} \theta\_i)$. $\sharp$ And for a 6-gon $8\pi < 2(\Sigma^6\_{i=1} \theta\_i)$. $\sharp$ Thus, there exists at least one pair of adjacent vectors which may be exchanged to get an interior angle less than or equal to $\pi/3$. $\blacksquare$


Thus, in a 6-gon, we can always exchange two vectors to get the sum of two adjacent vectors to be 'short'. Replacing $v\_1, ... ,v\_5, v\_6$ with $v\_1, ... , v\_4, v\_5+v\_6$ to get 5 'short' vectors whose sum is zero. If $v\_5+v\_6$ is pair-able with another vector in such a way that their sum is 'short', then we are done. Otherwise, we get two other vectors whose sum is a 'short' vector, and we reduce to the four vector case. Because we can again reduce the quadrilateral case, we are done unless we must pair the two non-"sum" vectors to get a 'short' vector. If this is the case, I claim we still have the sum of three of the original six vectors is 'short'.


Proof of claim:


Looks like fedja just beat me to it, but this was just too much to write to delete it. :P


