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A set $V$ has 2010-vectors: $V=\{v_{1}, \ldots,v_{2010}\}$ and these vectors create another set with the lengths of these vectors: $B=\{1,2,\ldots,2010\}$. Each vector is parallel to one of $2$ given concurrent lines (see picture). Prove that the sum of these vectors is${}\neq 0$ regardless of their directions.

For me this is strange. Why are those two concurrent lines given ?

Thanks for help :) I will try to do solve this problem even if it is strange.

two concurrent lines

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1 Answer

My interpretation: The two given lines give us two unit vectors $\vec{u}$ and $\vec{v}$ that are not parallel (i.e. linearly independent). Then the constraints are that for all $n$ the vector $\vec{w}_n$ of length $n$, $n=1,2,\ldots,2010,$ has to be one of the four possibilities: $n\vec{u}$, $-n\vec{u}$, $n\vec{v}$ or $-n\vec{v}$.

Hint (assuming that I correctly guessed what the question is about): The sum of the lengths is an odd number.

Spoiler solution:

Let the starting point of the first vector be $P$, and place the vectors after each other as in the usual vector addition. The end points of all the vectors will lie at some point of the form $P_{k,\ell}=P+ k\vec{u}+\ell\vec{v}$. Color the point $P_{k,\ell}$ black, if $k+\ell$ is even and color it white, if $k+\ell$ is odd - i.e. chessboard style. Because the sum $1+2+3+\cdots+2010=1005\cdot2011$ is an odd number, after all the vectors have been placed, the last one ends at a white point. Therefore it cannot end at $P$.

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Yes, seeing this hint it seems that the problem is a very cumbersome wrapper around the fact that every partitioning of an odd set must have an odd-sized constituent. In other words: The problem also works in three dimensions, i.e. with each vector parallel to one of three linearly independant directions (and also in any higher dimension). –  Hagen von Eitzen Feb 19 '13 at 9:06
@Hagen, my educated guess would be a contest problem from the year 2010 (rolls eyes). –  Jyrki Lahtonen Feb 19 '13 at 11:42
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