Vectors around Regular Polygon

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sowmitra
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Vectors around Regular Polygon

Unread post by sowmitra » Sat Feb 21, 2015 11:25 pm

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

Unread post by Nirjhor » Sun Feb 22, 2015 1:59 am

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.
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*Mahi*
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Re: Vectors around Regular Polygon

Unread post by *Mahi* » Sun Feb 22, 2015 6:16 pm

\[\begin{align*}
2\sum_{i=1}^n \overrightarrow{OP_i} &= \sum_{i=1}^n (\overrightarrow{OP_i} + \overrightarrow{OP_{i+2}} )\\
&= k \sum_{i=1}^n \overrightarrow{OP_{i+1}}
\end{align*}\]
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

Unread post by nayel » Mon Feb 23, 2015 3:55 am

Two words:
Complex numbers
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sowmitra
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Re: Vectors around Regular Polygon

Unread post by sowmitra » Tue Feb 24, 2015 10:15 pm

Hmm, Nayel Vai... Complex Numbers definitely give a straightforward solution. But, Mahi's approach was very neat. :)

Solution using complex numbers:
Let, the polygon be inscribed in the unit-circle, and, $OP_1$ be the real axis. Then,
\[\overrightarrow{OP_i}=\cos\frac{2\pi}{n}(i-1)+i\cdot\sin\frac{2\pi}{n}(i-1)=e^{i\cdot\frac{2\pi}{n}(i-1)}=z^{i-1}\]
where, $z=e^{i(\frac{2\pi}{n})}$ and $z^n=e^{2\pi i}=1$.
\[\therefore \sum_{i=1}^n\overrightarrow{OP_i}=\sum_{i=1}^nz^{i-1}=1+z+z^2+\ldots+z^{n-1}=\frac{z^n-1}{z-1}=0\]
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