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Consider $A_1A_2A_3\dots A_n$ regular polygon is inscribed in circle with $r$ radius. I want to find $S$ versus $r$ and angles if needed.

\begin{align} S = & (A_1A_2 + A_1A_3 +A_1A_4 + \dots+ A_1A_n) \\ & {} + (A_2A_3 + A_2A_4 +A_2A_5 + \dots + A_2A_n)+\dots +A_{n-1}A_n \end{align}

Note : $A_1A_2, A_2A_3 ,\dots ,A_{n-1}A_n$ are sides of polygon and $A_1A_3, A_1A_4,\dots$ are diagonals of it.

I tried many ways but didn't get result. Please Help!

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    Is it a regular polygon? – 2017-01-17
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    Yes it is regular polygon – 2017-01-18
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    @Jens Can you help me now ? – 2017-01-18
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    So essentially you want to find the sum of all sides and all diagonals? – 2017-01-18
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    @IvanNeretin Yes , you're right. – 2017-01-18
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    Then it should be quite simple. Wait a sec. – 2017-01-18

1 Answers 1

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OK, so we want to find the sum of all sides and all diagonals of a regular $n$-gon with circumradius $r$. How do we go about that?

First, what is the length of the side (say, $A_1A_2$)? It subtends an angle of $2\pi\over n$, so by the law of cosines $A_1A_2=\sqrt{r^2+r^2-2r^2\cos{2\pi\over n}}=2r\sin{\pi\over n}$. By similar reasoning, the length of the smallest diagonal $A_1A_3=2r\sin{2\pi\over n}$, then comes $A_1A_4=2r\sin{3\pi\over n}$, etc.

Now how many of each kind of diagonals are there? Again, that's simple: we have a set of $A_iA_j$ with both indexes running from 1 to $n$ (inclusive). For every fixed $i$, by varying $j$ we get each angle exactly once. Note that we've counted each chord twice (first as $A_iA_j$, then as $A_jA_i$), so we'll have to divide by 2. All in all, it is $n\cdot r\cdot\sum\limits_{i=1}^n\sin{\pi i\over n}$.

So how do we find a closed form of this? There are different approaches; one is based on representation of all these sines via complex exponents, and another is given below:

$$\sin1+\sin2+\dots+\sin n={1\over\sin{1\over2}}\left(\sin{1\over2}\sin1+\sin{1\over2}\sin2+\dots+\sin{1\over2}\sin n\right)=\\ ={1\over2\sin{1\over2}}\left(\Big(\cos{1\over2}-\cos{3\over2}\Big)+\Big(\cos{3\over2}-\cos{5\over2}\Big)+\dots+\Big(\cos{2n-1\over2}-\cos{2n+1\over2}\Big)\right)=\\ ={1\over2\sin{1\over2}}\left(\cos{1\over2}-\cos{2n+1\over2}\right)= {\sin{n\over2}\sin{n+1\over2}\over\sin{1\over2}}$$

Putting all together, we have $$S =n\cdot r\cdot{\sin{\pi n\over2n}\sin{\pi(n+1)\over2n}\over\sin{\pi\over2n}} =n\cdot r\cdot{\sin{\pi(n+1)\over2n}\over\sin{\pi\over2n}} =n\cdot r\cdot{\cos{\pi\over2n}\over\sin{\pi\over2n}} =n\cdot r\cdot\cot{\pi\over2n}$$

So it goes.