A polygon with n vertices (not necessarily regular) is inscribed in a circle of radius R. The center of the circle is the centroid of the n-gon. Find the sum of the squares of all sides and diagonals of the n-gon.

- anonymous

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- anonymous

Two things I'm unsure on. I am assumign that the center of the circle being the centroid means the inscribed polygon is regular, right? Or can it possibly still not be regular despite that condition? And when it says all diagonals, does that assume all the diagonals that can be drawn from a single point, or all the unique diagonals connecting all points?
If I can be certain about those two things, then I have an idea where to start but not how to finish. If the polygon might not be regular despite the centroid condition, then I'm not really sure at all, lol. Either way, this is what I have so far:
\[\sum_{i=i}^{n} A_{i} = \sum_{i = 1}^{n}2R\sin(\frac{ \alpha_{i} }{ 2 })\]
The \(A_{i}'s\) are the measure of each chord (not necessarily a diagonal), assuming they may not be of the same length, and the \(\alpha_{i}'s\) are the central angles opposite each \(A_{i}\). But that's as far as I got. Of course if the n-gon is regular then this is easier but even then, I wouldnt know how to manage that infinite sum. Any ideas?

- ganeshie8

are you suggesting that if circumcenter and centroid are equal for a polygon, then it must be regular ?

- anonymous

I dont know the answer to that. I'm good with centroids and such for triangles, but I never really extended that to an n-gon. I wanted to approach this as if the n-gon was not regular, but that part about the circle's center being the centroid made me think it might be a regular n-gon

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## More answers

- ganeshie8

yeah,
"not ncessarily regular" part makes me think we should consider it as a convex polygon whose centroid happens to be same as the circumcenter

- anonymous

Yeah, thats why I tried to notate the summation from i to n, trying to say they could be different measures. But even so, assuming I didnt mess up the summation I came up with, Im not sure where that gets me. Seemed more promising than anything else I had come up with.

- anonymous

Is there a formula, or at least a way to get a formula, for a sum of sines beyond just two? Like some sort of \(\sin(\alpha_{1}) + \sin(\alpha_{2}) + ... + \sin(\alpha_{n}) = ?\)

- ganeshie8

can we use complex numbers?

- anonymous

Thats allowed, yes.

- ganeshie8

if it is regular, the sum is easy to evaluate :
\[\begin{align}\sum_{k=i}^{n} A_{i} &= \sum_{k = 1}^{n}2R \sin (k2\pi/n)\\~\\
&=2R*\Im \sum_{k = 1}^{n} e^{i(k2\pi/n) }\\~\\
\end{align}\]
thats just a geometric series which can be evaluated easily

- ganeshie8

but wait a sec, aren't we supposed to find the sum of "squares" ?

- anonymous

Yeah. But I was just setting it up as a regular summation. But I figure we would then just have:
\[\sum_{k = 1}^{n}A^{2}_{k} = 4R^{2}\sum_{k=1}^{n}\sin^{2}(\frac{ 2k\pi }{ n })\]
Is it really just 2kpi/n, though? I was thinking, using what I had before, that if it were regular it'd be
\[4R^{2}\sum_{k=1}^{n}\sin^{2}(\frac{ \alpha_{k} }{ 2 }+\frac{ n\pi }{ 2 })\]

- ganeshie8

|dw:1441535193941:dw|

- ganeshie8

the sum of squares of lengths of all the chords that have \(V1\) as one endpoint is given by
\[\sum\limits_{k=1}^{n} (2R\sin(k\pi/n))^2\]
yes ?

- anonymous

I think I was just looking at it from a different perspective. Starting with a side length and adding to it as opposed to a vertex and going from there. But okay, I gotcha.

- ganeshie8

I see.. i was thinking it would be easy if we can find the sum for one vertex,
multiply that by \(n\),
then fix the duplicates if needed ..

- anonymous

That would probably be much easier to do. If it still works, then sure :P Okay, so we have
\[4R^{2}\sum_{k=1}^{n}\sin^{2}(\frac{ k\pi }{ n })\]
So is this still geometric? I know going to n would give us an nth power in exponential form, but where would the cosine term go in the normal euler's formula? Don't we need a \(\cos(\frac{2k\pi}{n})\) to go to \(e^{i2k\pi/n}\)?

- ganeshie8

We could use this identity : \(2\sin^2x = 1-\cos(2x)\)
that gets rid of powers, then complexifying should be easy

- anonymous

Well, unless \(2k\pi/n = 2m\pi\ \forall\ k, n, m \in\ \mathbb{Z}\) wouldnt we still need a sin term to put it into exponential form?

- anonymous

I would think that if we had something like \(2\pi/3\) as the argument at some point, then the sin term would still be required to be there and we couldnt go straight to exponential form. But idk, lol.

- ganeshie8

\[\begin{align}4R^{2}\sum_{k=1}^{n}\sin^{2}(\frac{ k\pi }{ n }) &=2R^{2}\sum_{k=1}^{n}1-\cos(\frac{2 k\pi }{ n }) \\~\\
&=2R^{2}(n-\Re \sum_{k=1}^{n} e^{i2k\pi/n}) \\~\\
\end{align}\]

- ganeshie8

do you see anything wrong with that ?

- anonymous

Whats the fancy R? Lol, not aware of what that symbol is meant to be, sorry.

- ganeshie8

\(\Re(e^{i\theta}) \) = Real part of \(e^{i\theta}\) = \(\cos(\theta)\)

- anonymous

Oh, real part, okay. Yeah, thats fine then.

- anonymous

Hmm....I guess Im used to seeing geo series differently, lol. So its geometric because the index is k and we can pull out the k, right?

- ganeshie8

\(\sum_{k=1}^{n} e^{i2k\pi/n} \)
it is a geometric series with \(n\) terms and common ratio of \(e^{i2\pi/n}\)

- ganeshie8

but we don't really need to use the partial sum formula to evaluate that
we may use the fact that sum of roots of unity add up to 0

- ganeshie8

\(\sum_{k=1}^{n} e^{i2k\pi/n} = 0\)

- anonymous

Hmm, never realized that, sum of roots of unity add to 0. So the answer is just \(2nR^{2}\) ^_^ So now how do we throw in the diagonals part of the problem? I wasn't sure if it meant all possible diagonals or if it only meant diagonals emminating from 1 vertex.

- ganeshie8

sum through one vertex = \(2nR^2\)
so sum through \(n\) vertices = \(2nR^2*n\)

- ganeshie8

next fix the duplicates

- anonymous

So \(n^{2}R^{2}\). I assume that just dividing by 2 takes care of it, right?

- ganeshie8

not so sure
(1, 2), (1, 3), (1,4), (1, 5) ... (1, n)
(2, 3), (2, 4), (2, 5) ... (2, n)
(3, 4), (3, 5) ... (3, n)
.
.
.

- anonymous

I was thinking that each diagonal would be extended in both directions. V1 goes to V2 but then V2 goes to V1 and we divide that out. Or is this some sort of permutation thing?

- ganeshie8

im still thinking...

- ganeshie8

hmm there are just \(\binom{n}{2}\) diagonals,
not \(\frac{n^2}{2}\)

- anonymous

Are we counting chords as diagonals?

- ganeshie8

yes, lets just pretend that a chord is a diagonal in this thread :)

- anonymous

Lol, okay xD

- ganeshie8

i think i mean
hmm there are just \(\binom{n}{2}\) chords,
not \(\frac{n^2}{2}\)

- anonymous

So multiply \(2nR^{2}\) by \[\left(\begin{matrix}n \\ 2\end{matrix}\right)\]?

- ganeshie8

hey no, looks im making it complicated
simply dividing by 2 works fine

- anonymous

Oh. Lol, okay. Making sure :)

- ganeshie8

each pair of vertices appears in the sums of two different vertices
so dividing by 2 takes care of the duplicates

- anonymous

Okay, awesome. Thanks for spending the time once again ^_^ Now I gotta get back to the sin(z) = z mess, haha. You're always a great help, I appreciate it :)

- ganeshie8

np :) so the final answer is \(n^2R^2\) if we assume the polygon is regular
I think if itsn't regular, there will not be a simple answer

- anonymous

Yeah, not so simple at all. But that helps out a lot. Off to bed finally then, lol. Night :)

- ganeshie8

yesterday i found something about sin(z)=z in mse that uses picard's thm
il dig that up and post the link later
gnite!

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