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This problem is from the book Abstract Algebra with applications by Spindler: Let $P_1,\dots,P_n$ be the vertices of a regular plane $n$-gon centered at $O$ and let $p_i:=\overrightarrow{OP_i}.$ Show that $p_1+p_2+\cdots+p_n=0$ i.e. that the sum of the vectors $p_i$ is the zero vector.

Solution: If $n$ is even, then it is trivial since each $p_i$ has an opposite vector, $-p_i$. There are $n/2$ of both sets of vectors, and thus their sum is zero.

So, suppose $n$ is odd and consider when $n=5.$ Observe that the opposite of each $p_i$ bisects the side of two adjacent vectors e.g. $-\lambda p_1$ bisects side formed by $p_3$ and $p_4$ where $0<\lambda\leq1\enspace[\color{red}1]$

We then have, $$-\lambda p_1=\frac{1}{2}(p_3+p_4)$$ $$-2\lambda p_1=(p_3+p_4)\enspace[\color{red}2]$$

Questions:

$[\color{red}1]$ What is $\lambda$ suppose to be? I know for this case, $n=5$, it has to be less than 1 but when $n=3,\:\lambda=\frac{1}{2}$. But for the triangle its easy to see that $p_1+p_2+p_3=0$ since $p_2+p_3=-p_1$ as can be seen below.

$[\color{red}2]$ I'm not sure what to do after this sense I need to know what $\lambda$ is.

The following is to help visualize the problem: enter image description here enter image description here

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    $\begingroup$ This fact does not require such complicated argument. $p_1 + p_2 + ... + p_n =0$ because when you rotate all the vectors by $\frac{2\pi}{n}$ their sum on the one hand does not change and on the other hand it also rotates by $\frac{2\pi}{n}$, so it must be zero $\endgroup$ Apr 2, 2019 at 20:04
  • $\begingroup$ This is a duplicate of MSE question 2276723 "Property of Vectors of an $n$-gon". $\endgroup$
    – Somos
    Apr 2, 2019 at 20:35
  • $\begingroup$ It is not an exact duplicate, as this question is about a specific proof of this property. $\endgroup$ Apr 2, 2019 at 21:02

1 Answer 1

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The exact value of $\lambda$ is not relevant, it is instead necessary that $\lambda$ be the same for all $n$ vertices. If you add up the $n$ equations you get: $$ -2\lambda(p_1+p_2+\cdots+p_n)=2(p_1+p_2+\cdots+p_n), \quad\text{hence:}\quad p_1+p_2+\cdots+p_n=0. $$

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  • $\begingroup$ How did you make that "jump"? $\endgroup$
    – user480875
    Apr 2, 2019 at 20:44
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    $\begingroup$ You can rewrite the first equation as $$2(1+\lambda)(p_1+p_2+\cdots+p_n)=0$$. $\endgroup$ Apr 2, 2019 at 20:58
  • $\begingroup$ Got it. Thanks. $\endgroup$
    – user480875
    Apr 2, 2019 at 22:28

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