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anonymous

  • one year ago

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.

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  1. anonymous
    • one year ago
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    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?

  2. ganeshie8
    • one year ago
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    are you suggesting that if circumcenter and centroid are equal for a polygon, then it must be regular ?

  3. anonymous
    • one year ago
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    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

  4. ganeshie8
    • one year ago
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    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

  5. anonymous
    • one year ago
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    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.

  6. anonymous
    • one year ago
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    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 \(\sin(\alpha_{1}) + \sin(\alpha_{2}) + ... + \sin(\alpha_{n}) = ?\)

  7. ganeshie8
    • one year ago
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    can we use complex numbers?

  8. anonymous
    • one year ago
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    Thats allowed, yes.

  9. ganeshie8
    • one year ago
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    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

  10. ganeshie8
    • one year ago
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    but wait a sec, aren't we supposed to find the sum of "squares" ?

  11. anonymous
    • one year ago
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    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 \[4R^{2}\sum_{k=1}^{n}\sin^{2}(\frac{ \alpha_{k} }{ 2 }+\frac{ n\pi }{ 2 })\]

  12. ganeshie8
    • one year ago
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    |dw:1441535193941:dw|

  13. ganeshie8
    • one year ago
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    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 ?

  14. anonymous
    • one year ago
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    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.

  15. ganeshie8
    • one year ago
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    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 ..

  16. anonymous
    • one year ago
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    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 \(e^{i2k\pi/n}\)?

  17. ganeshie8
    • one year ago
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    We could use this identity : \(2\sin^2x = 1-\cos(2x)\) that gets rid of powers, then complexifying should be easy

  18. anonymous
    • one year ago
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    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?

  19. anonymous
    • one year ago
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    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.

  20. ganeshie8
    • one year ago
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    \[\begin{align}4R^{2}\sum_{k=1}^{n}\sin^{2}(\frac{ k\pi }{ n }) &=2R^{2}\sum_{k=1}^{n}1-\cos(\frac{2 k\pi }{ n }) \\~\\ &=2R^{2}(n-\Re \sum_{k=1}^{n} e^{i2k\pi/n}) \\~\\ \end{align}\]

  21. ganeshie8
    • one year ago
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    do you see anything wrong with that ?

  22. anonymous
    • one year ago
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    Whats the fancy R? Lol, not aware of what that symbol is meant to be, sorry.

  23. ganeshie8
    • one year ago
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    \(\Re(e^{i\theta}) \) = Real part of \(e^{i\theta}\) = \(\cos(\theta)\)

  24. anonymous
    • one year ago
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    Oh, real part, okay. Yeah, thats fine then.

  25. anonymous
    • one year ago
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    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?

  26. ganeshie8
    • one year ago
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    \(\sum_{k=1}^{n} e^{i2k\pi/n} \) it is a geometric series with \(n\) terms and common ratio of \(e^{i2\pi/n}\)

  27. ganeshie8
    • one year ago
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    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

  28. ganeshie8
    • one year ago
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    \(\sum_{k=1}^{n} e^{i2k\pi/n} = 0\)

  29. anonymous
    • one year ago
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    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.

  30. ganeshie8
    • one year ago
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    sum through one vertex = \(2nR^2\) so sum through \(n\) vertices = \(2nR^2*n\)

  31. ganeshie8
    • one year ago
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    next fix the duplicates

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

  33. ganeshie8
    • one year ago
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    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) . . .

  34. anonymous
    • one year ago
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    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?

  35. ganeshie8
    • one year ago
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    im still thinking...

  36. ganeshie8
    • one year ago
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    hmm there are just \(\binom{n}{2}\) diagonals, not \(\frac{n^2}{2}\)

  37. anonymous
    • one year ago
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    Are we counting chords as diagonals?

  38. ganeshie8
    • one year ago
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    yes, lets just pretend that a chord is a diagonal in this thread :)

  39. anonymous
    • one year ago
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    Lol, okay xD

  40. ganeshie8
    • one year ago
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    i think i mean hmm there are just \(\binom{n}{2}\) chords, not \(\frac{n^2}{2}\)

  41. anonymous
    • one year ago
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    So multiply \(2nR^{2}\) by \[\left(\begin{matrix}n \\ 2\end{matrix}\right)\]?

  42. ganeshie8
    • one year ago
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    hey no, looks im making it complicated simply dividing by 2 works fine

  43. anonymous
    • one year ago
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    Oh. Lol, okay. Making sure :)

  44. ganeshie8
    • one year ago
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    each pair of vertices appears in the sums of two different vertices so dividing by 2 takes care of the duplicates

  45. anonymous
    • one year ago
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    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 :)

  46. ganeshie8
    • one year ago
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    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

  47. anonymous
    • one year ago
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    Yeah, not so simple at all. But that helps out a lot. Off to bed finally then, lol. Night :)

  48. ganeshie8
    • one year ago
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    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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