## Vectors around Regular Polygon

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sowmitra
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### Vectors around Regular Polygon

Let, $P_1P_2P_3\ldots P_n$ be a regular polygon whose circumcentre is $O$. Prove that,
$\sum_{i=1}^n \overrightarrow{OP_i}=0$
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Nirjhor
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### Re: Vectors around Regular Polygon

We choose the coordinate system so that $\overrightarrow{OP_1}$ is parallel to $x$-axis and pointed to the positive direction. Breaking each vector into components we get $\overrightarrow{OP_i}=r\cos\dfrac{2\pi(i-1)}{n}\hat{\mathbf{i}}+r\sin\dfrac{2\pi(i-1)}{n}\hat{\mathbf{j}}$ so $\sum_{i=1}^n \overrightarrow{OP_i}=r\sum_{i=0}^{n-1}\cos\dfrac{2\pi i}{n}\hat{\mathbf{i}}+r\sum_{i=0}^{n-1}\sin\dfrac{2\pi i}{n}\hat{\mathbf{j}}.$ The proof that the first sum (and the second analogously) is equal to $0$ is proven in example $2$ here.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

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*Mahi*
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### Re: Vectors around Regular Polygon

\begin{align*} 2\sum_{i=1}^n \overrightarrow{OP_i} &= \sum_{i=1}^n (\overrightarrow{OP_i} + \overrightarrow{OP_{i+2}} )\\ &= k \sum_{i=1}^n \overrightarrow{OP_{i+1}} \end{align*}
Where $k < 2$ as $\overrightarrow{OP_i}$ and $\overrightarrow{OP_{i+2}}$s are not in the same line.
So, $(2-k)\sum_{i=1}^n \overrightarrow{OP_i} = 0 =\sum_{i=1}^n \overrightarrow{OP_i}$

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nayel
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### Re: Vectors around Regular Polygon

Two words:
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

sowmitra
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