A regular \(n-gon\) is inscribed in an unit circle. Let \(c_{ij}\) be the length of chord joining vertices \(i\) and \(j\). Show that \[\sum\limits_{i\ne j} {c_{ij}}^2 = n^2\] and \[\sum\limits_{i\ne j}\dfrac{1}{{c_{ij}}^2} = \dfrac{n^2-1}{12}\]

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A regular \(n-gon\) is inscribed in an unit circle. Let \(c_{ij}\) be the length of chord joining vertices \(i\) and \(j\). Show that \[\sum\limits_{i\ne j} {c_{ij}}^2 = n^2\] and \[\sum\limits_{i\ne j}\dfrac{1}{{c_{ij}}^2} = \dfrac{n^2-1}{12}\]

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|dw:1441591152116:dw| how to show: \[c_{ij}=c_{jk}=c_{ki}=\sqrt{3}\]
\[ \operatorname{crd}(x)=2\sin\left(\frac{x}{2}\right)\\ \]
What would be the sum of n=3? \[ \begin{align*} \sum\limits_{i\ne j} {c_{ij}}^2 &= c_{12}+c_{13}+c_{23}?\\ &=c_{12}+c_{13}+c_{23}+c_{21}+c_{31}+c_{32}? \end{align*} \]

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sry i think order doesn't matter freckles has correct interpretation it seems.. there are only \(\binom{n}{2}\) chords, not \(n^2\)
What would be the sum of n=3? \[ \begin{align*} \sum\limits_{i\ne j} {c_{ij}}^2 &= c_{12}+c_{13}+c_{23}?\\ \end{align*} \] i meant this, i see the ambiguity with my notation but i just don't know how to fix it...
|dw:1441592259514:dw| well this my interpretation for a 4-gon...\[c^2_{12}+c^2_{23}+c^2_{34}+c^2_{41}=4^2 \\ \text{ but this a a reg n-gon } \\ \text{ so } c_{12}=c_{23}=c_{34}=c_{41} \\ \text{ so this really becomes } 4 c_{12}^2=4^2 \\ \text{ and so we really just want to show } c_{12}^2=4 \text{ or that } c_{12}=2\]
Oops! Not fully awake. \[ \sum_{i=1}^{n-1}\sum_{j=i+1}^nc_{ij}? \]
that works!
Very ugly I must admit.
|dw:1441592568527:dw|
oh
basically the first identity says : sum of squares of all the possible chords is \(n^2\)
second identity says : sum of squares of reciprocals of all the possible chords is \((n^2-1)/2\)
mainly we need to show how many cycles the chords finish of the unit circle :D like if we have only n=3 then its one cycle and sum c_ij=1
\[ \sum_{k=1}^{n-1}\sum_{j=k+1}^n(c_{ij})^2= \sum_{k=0}^{n-2}\sum_{j=k+1}^{n-1}(c_{ij})^2 \] \[ \operatorname{crd}(x)=2\sin\left(\frac{x}{2}\right)\\ \begin{align*} &\phantom{=}\sum_{k=0}^{n-2}\sum_{j=k+1}^{n-1}\left(2\sin\left(\frac{2\pi (k-j)}{n}\right)\right)^2\\ &=2(n-1)(n)\sum_{k=0}^{n-2}\sum_{j=k+1}^{n-1}\sin^2\left(\frac{2\pi (k-j)}{n}\right)\\ &=2(n-1)(n)\sum_{k=0}^{n-2}\sum_{j=k+1}^{n-1}\left(1-\cos\left(\frac{4\pi (k-j)}{n}\right)\right)\\ &=n^2(n-1)^2-\sum_{k=0}^{n-2}\sum_{j=k+1}^{n-1}\cos\left(\frac{4\pi (k-j)}{n}\right)\\ &=n^2(n-1)^2-\operatorname{Re}\left(\sum_{k=0}^{n-2}\sum_{j=k+1}^{n-1} e^{\frac{4i\pi (k-j)}{n}}\right)\\ &=n^2(n-1)^2-\operatorname{Re}\left(\sum_{k=0}^{n-2}\sum_{j=k+1}^{n-1} e^{\frac{4i\pi k}{n}}e^{\frac{-4i\pi j}{n}}\right)\\ &=n^2(n-1)^2-\operatorname{Re}\left(\sum_{k=0}^{n-2}e^{\frac{4ki\pi }{n}}\left(\frac{1-e^{\frac{-4ni\pi }{n}}}{1-e^{\frac{-4i\pi }{n}}}-\frac{1-e^{\frac{-4(k+2)i\pi }{n}}}{1-e^{\frac{-4i\pi }{n}}}\right)\right)\\ \end{align*} \] I give up!
I think I have an idea for the first identity. Let's start by looking at \[\sum\limits_{j=2}^{n} {c_{1j}}^2 \] That is all the chords starting from the vertex labelled with 1. We'll label vertices counterclockwise, like this: |dw:1441594979932:dw| We note that all other vertices have the same kinds of chords, so \[\frac{n}{2}\sum\limits_{j=2}^{n} {c_{1j}}^2 = \sum\limits_{i\ne j} {c_{ij}}^2 = n^2\] or \[\sum\limits_{j=2}^{n-1} {c_{1j}}^2 = 2n \] Let's define \(\LARGE\alpha=\frac{2\pi}{n}\ \) and use the law of cosines. \[c_{12} = 1^2 + 1^2 - 2cos\alpha \] \[c_{13} = 1^2 + 1^2 - 2cos2\alpha \] \[c_{1n} = 1^2 + 1^2 - 2cos[(n-1)\alpha] \] I think for even n you can prove that the cosine terms pair up to add up to zero, except the middle one, which gives +2 (corresponding to the diagonal).
Here's what I mean by 'pairing up' |dw:1441595804013:dw|
@ganeshie8 @freckles can you check if my reasoning is correct?
Barbarian's approach: \[ \sum_{k=1}^{n-1}\sum_{j=k+1}^n(c_{ij})^2= \frac{1}{2}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}(c_{ij})^2 \quad c_{ij}=c_{ji},\,c_{ii}=0\\ \] \[ \operatorname{crd}(x)=2\sin\left(\frac{x}{2}\right)\\ \begin{align*} &\phantom{=}\frac{1}{2}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\left(2\sin\left(\frac{2\pi (k-j)}{n}\right)\right)^2\\ &=2\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\sin^2\left(\frac{2\pi (k-j)}{n}\right)\\ &=\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\left(1-\cos\left(\frac{4\pi (k-j)}{n}\right)\right)\\ &=n^2+\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\cos\left(\frac{4\pi (k-j)}{n}\right)\\ &=n^2+\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\operatorname{Re}\left( e^{\frac{4i\pi (k-j)}{n}}\right)\\ &=n^2+\operatorname{Re}\left(\sum_{k=0}^{n-1}\sum_{j=0}^{n-1} e^{\frac{4i\pi k}{n}}e^{\frac{-4i\pi j}{n}}\right)\\ &=n^2+\operatorname{Re}\left(\sum_{k=0}^{n-1} e^{\frac{4i\pi k}{n}}\frac{1-e^{\frac{-4i\pi n}{n}}}{1-e^{\frac{-4i\pi }{n}}}\right)\\ &=n^2+\operatorname{Re}\left(\frac{1-e^{-4i\pi}}{1-e^{\frac{-4i\pi }{n}}}\sum_{k=0}^{n-1} e^{\frac{4i\pi k}{n}}\right)\\ &=n^2+\operatorname{Re}\left(\frac{1-e^{-4i\pi}}{1-e^{\frac{4i\pi }{n}}}\frac{1-e^{4i\pi}}{1-e^{\frac{-4i\pi }{n}}}\right)\\ &=n^2+\operatorname{Re}\left(\frac{1-e^{-4i\pi}-e^{4i\pi }+1}{1-e^{\frac{4i\pi }{n}}-e^{\frac{-4i\pi }{n}}+1}\right)\\ &=n^2+\operatorname{Re}\left(\frac{2-\left(e^{-4i\pi}+e^{4i\pi }\right)}{1-e^{\frac{4i\pi }{n}}-e^{\frac{-4i\pi }{n}}+1}\right)\\ &=n^2+\operatorname{Re}\left(\frac{2-2}{1-e^{\frac{4i\pi }{n}}-e^{\frac{-4i\pi }{n}}+1}\right)\\ &=n^2 \end{align*} \]
I feel really old using this method. I have use this method at least 30 times.
I messed up the denominator in the last 5 steps but it does not affect the result.
Technically my proof isn't entirely correct since the thing in the real part simplifies to 0/0.
Actually no my proof is correct since \(1-e^{-4\pi i}=0\).
Awesome! @beginnersmind and @thomas5267 you both have the same idea it seems
@beginnersmind is fixing the duplicates in the end @thomas5267 is evaluating the sum straight
Like how many times I have used the identity \(\operatorname{Re}\left(e^{ix}\right)=\cos(x)\) on this site lol...
Can you explain the sums of \(\LARGE e^{\frac{i\pi k}{n}}\) add up the 1?
i think we can just use the result : sum of roots of unity add up to 0
Bring back your awesome analytic proof!
I mean \[\LARGE e^{\frac{i4\pi k}{n}}\]
\(\sum\limits_{k=1}^n e^{i(k2\pi/n)} = 0\) because each term in above sum is a root of the polynomial \(x^n-1\), its easy to see that sum of roots of that polynomial is 0 (vieta formulas)
Yeah, I see how those are equivalent. Not sure I see the easy part. :)
i think you do :) just find the sum of roots of polynomial \(x^n+0x^{n-1}+\cdots +0x-1\)
Ok, I think I see now. Write it as (x-x1)*(x-x2)*...*(x-xn) = 0 Then the sun is the negative of the x^(n-1) term, which is 0.
Exactly..
Okay so one down one more to go
Neat. Plenty of good stuff to remember.
i hope we could approach the other identity same way but there is no easy way to use euler idenity it seems..
Cannot use the same method without modification. x/0 occurs. \[ \operatorname{crd}(x)=2\sin\left(\frac{x}{2}\right)\\ \begin{align*} &\phantom{=}\sum_{i=1}^{n-1}\sum_{j=i+1}^n\frac{1}{\left(c_{ij}\right)^2}\\ &=\frac{1}{2}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\left(2\sin\left(\frac{2\pi (k-j)}{n}\right)\right)^{-2}\text{ with some care}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\left(\sin^2\left(\frac{2\pi (k-j)}{n}\right)\right)^{-1}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\left(1-\cos\left(\frac{4\pi (k-j)}{n}\right)\right)^{-1}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1}{1-\cos\left(\frac{4\pi (k-j)}{n}\right)}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1}{1-\cos\left(\frac{4\pi (k-j)}{n}\right)}\frac{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}{1-\cos^2\left(\frac{4\pi (k-j)}{n}\right)}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}{\sin^2\left(\frac{4\pi (k-j)}{n}\right)}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}{\cos^2\left(\frac{\pi}{2}-\frac{4\pi (k-j)}{n}\right)}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}{\operatorname{Re}\left(e^{\frac{i\pi}{2}-\frac{4i\pi (k-j)}{n}}\right)^2}\\ &=\frac{1}{8}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}{\operatorname{Re}\left(e^{\frac{i\pi}{2}}e^{-\frac{4i\pi (k-j)}{n}}\right)^2}\\ &=\frac{1}{2}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}{\left(e^{\frac{i\pi}{2}}e^{-\frac{4i\pi (k-j)}{n}}+e^{-\frac{i\pi}{2}}e^{\frac{4i\pi (k-j)}{n}}\right)^2}\\ &=\frac{1}{2}\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1+\cos\left(\frac{4\pi (k-j)}{n}\right)}{-e^{-\frac{8i\pi (k-j)}{n}}+2-e^{\frac{8i\pi (k-j)}{n}}}\\ \end{align*} \]
How to count the number of residue classes of 2x+b mod n?
|dw:1441654270423:dw| do this for all n points and you will have found the length of the chords twice, so we need to divide by 2. \[\frac{n}{2} \sum_{k=0}^{n-1} (e^{i\frac{ 2 \pi }{n}k}-1)(e^{i\frac{ 2 \pi }{n}k}-1)^* \] so you can also see in the sum itself I have the term with its complex conjugate since \(zz^* =c^2 \) Do boring algebra: \[\frac{n}{2} \sum_{k=0}^{n-1} 2 - e^{i\frac{ 2 \pi }{n}k}-e^{-i\frac{ 2 \pi }{n}k}\] Separate the sum: \[\frac{n}{2} \sum_{k=0}^{n-1} 2 -\frac{n}{2} \sum_{k=0}^{n-1} e^{i\frac{ 2 \pi }{n}k}+e^{-i\frac{ 2 \pi }{n}k}\] That first term is \(n^2\) so now we really just have to show that this equals zero and we're home: \[ 0=\sum_{k=0}^{n-1} e^{i\frac{ 2 \pi }{n}k}+e^{-i\frac{ 2 \pi }{n}k}\] Which it does because you can either think of it as cosine terms all added up on a complete rotation so the positive part is symmetric to the negative part and cancel. The second one is a bit trickier, I haven't gotten to doing that one yet.
There are \(\left\lfloor \frac{n}{2} \right \rfloor\) edges with distinct length in a regular n-gon. For odd n, there are n edges of a particular length. For even n, there are n chords for all but one length. There are n/2 chords for that length. In total, there are \(\binom{n}{2}\) edges. \[ \begin{align*} &\phantom{=} \sum_{i=1}^{n-1}\sum_{j=i+1}^n\frac{2}{\left(c_{ij}\right)^2}\\ &=n\sum_{j=2}^{\left\lfloor \frac{n}{2}\right \rfloor+1}\frac{1}{\left(c_{1j}\right)^2} \text{ for odd n}\\ &=n\sum_{j=2}^{\frac{n}{2}}\frac{1}{\left(c_{1j}\right)^2} +\frac{n}{2}\frac{1}{\left(c_{1\frac{n}{2}}\right)^2}=n\sum_{j=2}^{\frac{n}{2}}\frac{1}{\left(c_{1j}\right)^2} +\frac{n}{8}\text{ for even since }c_{1\frac{n}{2}}=\text{diameter}=2\\ \end{align*} \] \[ \text{For odd }n\text{:}\\ \begin{align*} &\phantom{=}n\sum_{j=2}^{\left\lfloor \frac{n}{2}\right \rfloor+1}\frac{1}{\left(c_{1j}\right)^2}\\ &=n\sum_{j=2}^{\left\lfloor \frac{n}{2}\right \rfloor+1}\frac{1}{\left(\operatorname{crd}\left(\frac{2\pi j}{n}\right)\right)^2}\\ &=n\sum_{j=2}^{\left\lfloor \frac{n}{2}\right \rfloor+1}\frac{1}{\left(1-e^{\frac{2\pi ij}{n}}\right)\left(1-e^{\frac{-2\pi ij}{n}}\right)}\\ &=n\sum_{j=2}^{\left\lfloor \frac{n}{2}\right \rfloor+1}\frac{1}{2-2\operatorname{Re}\left(e^{\frac{2\pi ij}{n}}\right)}\\ &=n\sum_{j=2}^{\left\lfloor \frac{n}{2}\right \rfloor+1}\frac{1}{2-2\cos\left(\frac{2\pi ij}{n}\right)} \end{align*} \] Can't evaluate the sum.
If you look at it geometrically and rearrange your indices you'll see because \(\csc(\frac{\pi}{2})=1\) it looks like the sums will work out by symmetry.
How does that work? The problem is that it is squared so it won't cancel out.

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