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
$$\sum\limits_{i\ne j}\dfrac{1}{{c_{ij}}^2} = \dfrac{n^2-1}{12}$$

Question

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
$$\sum\limits_{i\ne j}\dfrac{1}{{c_{ij}}^2} = \dfrac{n^2-1}{12}$$

freckles · Answer

|dw:1441591152116:dw|
how to show:
$$c_{ij}=c_{jk}=c_{ki}=\sqrt{3}$$

thomas5267 · Answer

$$
\operatorname{crd}(x)=2\sin\left(\frac{x}{2}\right)\
$$

thomas5267 · Answer

What would be the sum of n=3?
$$
\begin{align*}
\sum\limits_{i\ne j} {c_{ij}}^2 &= c_{12}+c_{13}+c_{23}?\
&=c_{12}+c_{13}+c_{23}+c_{21}+c_{31}+c_{32}?
\end{align*}
$$

ganeshie8 · Answer

sry i think order doesn't matter
freckles has correct interpretation it seems..

there are only $\binom{n}{2}$ chords, not $n^2$

ganeshie8 · Answer

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...

freckles · Answer

|dw:1441592259514:dw|
well this my interpretation for a 4-gon...$$c^2_{12}+c^2_{23}+c^2_{34}+c^2_{41}=4^2 \ \text{ but this a a reg  n-gon } \ \text{ so } c_{12}=c_{23}=c_{34}=c_{41} \ \text{ so this really becomes } 4 c_{12}^2=4^2  \ \text{ and so we really just want to show } c_{12}^2=4 \text{ or that } c_{12}=2$$