Let, $ P_1P_2P_3\ldots P_n$ be a regular polygon whose circumcentre is $O$. Prove that,
\[\sum_{i=1}^n \overrightarrow{OP_i}=0\]
Vectors around Regular Polygon
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.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
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, \[(2-k)\sum_{i=1}^n \overrightarrow{OP_i} = 0 =\sum_{i=1}^n \overrightarrow{OP_i}\]
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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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: Vectors around Regular Polygon
Two words:
"Everything should be made as simple as possible, but not simpler." - Albert Einstein
Re: Vectors around Regular Polygon
Hmm, Nayel Vai... Complex Numbers definitely give a straightforward solution. But, Mahi's approach was very neat.
Solution using complex numbers:
Solution using complex numbers: