Question

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.

Answer 1

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?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

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 \(\sin(\alpha_{1}) + \sin(\alpha_{2}) + ... + \sin(\alpha_{n}) = ?\)

Answer 7

can we use complex numbers?

Answer 8

Thats allowed, yes.

Answer 9

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

Answer 10

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

Answer 11

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
\[4R^{2}\sum_{k=1}^{n}\sin^{2}(\frac{ \alpha_{k} }{ 2 }+\frac{ n\pi }{ 2 })\]