Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

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 $P$.

share|improve this answer
    
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
1  
@Hagen, my educated guess would be a contest problem from the year 2010 (rolls eyes). –  Jyrki Lahtonen Feb 19 '13 at 11:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.