most recent 30 from http://mathoverflow.net 2013-05-22T22:56:17Z http://mathoverflow.net/feeds/question/63056 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63056/an-elementary-problem-in-euclidean-geometry Chris Taylor 2011-04-26T16:07:32Z 2011-04-27T05:20:09Z

<p>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.</p> <p>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.</p>

http://mathoverflow.net/questions/63056/an-elementary-problem-in-euclidean-geometry/63063#63063 hungrygrad 2011-04-26T17:29:00Z 2011-04-26T17:49:08Z

<p>I think this was a problem in my graph theory class. We used Ramsey theory... I don't remember the details, but you have to force a condition on the angles that ensures modulus less than 1.</p>

http://mathoverflow.net/questions/63056/an-elementary-problem-in-euclidean-geometry/63112#63112 fedja 2011-04-27T04:22:54Z 2011-04-27T04:22:54Z

<p>One word: <a href="http://www.artofproblemsolving.com/Forum/viewtopic.php?f=73&amp;t=393696" rel="nofollow">AoPS</a> </p>

http://mathoverflow.net/questions/63056/an-elementary-problem-in-euclidean-geometry/63113#63113 Andrew Parker 2011-04-27T04:26:40Z 2011-04-27T05:20:09Z

<p>Lemma 1:</p> <p>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.</p> <p>Proof:</p> <p>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 $r &lt; 1$ 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. $\blacksquare$</p> <p>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.</p> <p>Lemma 2:</p> <p>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.</p> <p>Proof:</p> <p>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 $16\pi/3 &lt; 2(\Sigma^4_{i=1} \theta_i)$. $\sharp$ For a 5-gon, $20\pi/3 &lt; 2(\Sigma^5_{i=1} \theta_i)$. $\sharp$ And for a 6-gon $8\pi &lt; 2(\Sigma^6_{i=1} \theta_i)$. $\sharp$ Thus, there exists at least one pair of adjacent vectors which may be exchanged to get an interior angle less than or equal to $\pi/3$. $\blacksquare$</p> <p>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'.</p> <p>Proof of claim:</p> <p>Looks like fedja just beat me to it, but this was just too much to write to delete it. :P</p>