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It follows from the law of cosines that if $a,b,c$ are the lengths of the sides of a triangle with respective opposite angles $\alpha,\beta,\gamma$, then $$ a^2+b^2+c^2 = 2ab\cos\gamma + 2ac\cos\beta + 2bc\cos\alpha. $$

For a cyclic (i.e. inscribed in a circle) polygon, consider the angle "opposite" a side to be the angle between adjacent diagonals whose endpoints are those of that side (it doesn't matter which vertex of the polygon serves as the vertex of that angle because of the Inscribed Angle Theorem). Then for a cyclic quadrilateral with sides $a,b,c,d$ and opposite angles $\alpha,\beta,\gamma,\delta$, one can show that $$ \begin{align} a^2+b^2+c^2+d^2 & = 2ab\cos\gamma\cos\delta + 2ac\cos\beta\cos\delta + 2ad\cos\beta\cos\gamma \\ & {} +2bc\cos\alpha\cos\delta+2bd\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta \\ & {}-4\frac{abcd}{(\text{diameter})^2} \end{align} $$ And for a cyclic pentagon, with sides $a,b,c,d,e$ and respective opposite angles $\alpha,\beta,\gamma,\delta,\varepsilon$, $$ \begin{align} a^2 + \cdots + e^2 & = 2ab\cos\gamma\cos\delta\cos\varepsilon+\text{9 more terms} \\& {} - 4\frac{abcd}{(\text{diameter})^2}\cos\varepsilon+ \text{4 more terms} \end{align} $$ And for a cyclic $n$-gon with sides $a_i$ and opposite angles $\alpha_i$, $$ \begin{align} \sum_{i=1}^n a_i^2 & = \text{a sum of }\binom{n}{2}\text{ terms each with coefficient 2} \\ & {} - \text{a sum of }\binom{n}{4}\text{ terms each with coefficient 4} \\ & {} + \text{a sum of }\binom{n}{6}\text{ terms each with coefficient 6} \\ & {} - \cdots \end{align} $$ The number of terms depends on $n$ and the power of the diameter on the bottom is in each case what is needed to make the term homogeneous of degree 2 in the side lengths ("dimensional correctness" if you like physicists' language), and the alternation of signs continues.

I showed this by induction. It should work for infinitely many sides, too, by taking limits. Each term would then have a product of infinitely many cosines.

My question is: Is there some reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle?

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This was crossposted at MO. Michael, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts (there is already an answer on the MO one by Noam Elkies). –  Zev Chonoles Jul 31 '11 at 6:11
I mildly object to the notion that the first identity follows from the law of cosines: in fact, the usual trigonometric proof proceeds by multiplying $c = a\cos{\beta} + b\cos{\alpha}$ by $c$. Doing this for each side and writing out $a^2 + b^2 - c^2$ gives you the law of cosines, writing out $a^2 + b^2 + c^2$ gives you the identity you ask about. –  t.b. Jul 31 '11 at 19:48
No geometric interpretation, but an observation: Assuming the polygon convex (a requirement for each $\alpha_i$ to be uniquely determined), we have $a_i = 2 r \sin{\alpha_i}$, where $r$ is the radius of the circle. Moreover, the arcs determined by the polygon's sides (obviously) comprise the full circle; hence the inscribed angles sum to $\pi$. So, dividing your formula by $4 r^2$ leaves a trig identity for the sum of squares of sines of $n$ (non-negative) angles that total $\pi$. –  Blue Jul 31 '11 at 23:04
@Theo: Here's a picture. :) math.stackexchange.com/questions/803/… –  Blue Jul 31 '11 at 23:42

2 Answers 2

I haven't worked it out up to arbitrary $n$, but I've managed the cases $n=3$ and $n=4$ to follow your lead and it becomes tempting to write "+4 terms" and so on.

I believe geometric interpretation of that ultimately goes down to interpreting geometrically the cosine law, since it is the main tool in your arguments.

EDIT : I just thought of something : suppose the sides of the inscribed polygon are almost are equal, or even better, that they form a regular polygon. Consider the following image :

enter image description here

You can consider the squares as a prism that rose over the inscribed polygon and then got unfolded. The sum of the squares can then be thought of an approximation of the lateral area of a cylinder with height being the average of the $a_i$'s, i.e. if you let $$ \mu = \frac{\sum_{i=1}^n a_i}n, $$ we can intuitively argue that $$ \sum_{i=1}^n a_i^2 \approx 2 \pi r \mu. $$ Hope that helps,

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Well, instead of saying "approximately", one could just put the perimeter of the polygon in place of $2\pi r$ and they'd be exactly equal. (And the approximation would be good only when all the side lengths are small.) –  Michael Hardy Aug 1 '11 at 20:06
@Michael Hardy : I never said I couldn't make it right, I just didn't want to, so I intuitively argued. There is obviously a way to define what approximately means here, and get a bound for the error term, but OP asks for intuition, so I didn't bother. =) –  Patrick Da Silva Aug 2 '11 at 4:57

Instead of an answer to the question you asked, let me give several observations.

Firstly, by homogeneity, we can consider the case where the diameter of the circle is 2, or the radius being 1. In this case, then, basic trigonometry gives you that $a = 2 \sin \alpha$, $b = 2\sin \beta$, etc.

Thus the right hand side consists of terms of the form $$\sin \times \sin \times \cos \times \cos\times \cos + \sin \times\sin \times\sin\times \sin \times \cos + \dots $$ which should be reminiscent of the angle addition rule for cosine.

Indeed, we have that

$$ \cos \sum \theta_k = \Re \exp \sum i\theta_k = \Re \prod \left( \cos \theta_k + i \sin\theta_k\right) $$

which is similar to the RHS that you've written down, except that on your RHS you do not have the term $\prod \cos\theta_k$. This term can be recovered by what you have on the LHS.

Note that if you have an inscribed polygon, $\sum \theta_k = \pi$. So $\cos \sum\theta_k = -1$. Notice that the LHS of your equation is proportional to $\sum \sin^2\theta_k$, and observe the following:

$$\begin{align} \sum 2\sin^2\theta_k & = \sum (1 - \cos(2\theta_k))\\ & = N - \sum_{k\neq 1} \cos(2\theta_k) - \cos 2\theta_1 \end{align}$$

Now using that $\sum \theta_k = \pi$, you get that $\cos2\theta_j = \cos(2\pi - 2\theta_j) = \cos(\sum_{k\neq j} 2\theta_k)$. And we observe that

$$ \cos(\sum_{k\neq j} 2\theta_k) = 2\cos(\sum_{k\neq j}\theta_j)\cos(\pi - \theta_k) - 1$$

where for $\cos(\sum_{k\neq j}\theta_j)$ we can inductively use the polynomial expression we have already derived. Doing this for all terms you end up with

$$ \sum \sin^2 \theta_k = 2N(1+\prod_k\cos\theta_k) + P \quad\quad (\sharp)$$

where $P$ is some polynomial in $\cos\theta_k$ and $\sin\theta_k$ that contains terms with positive, even number of factors in $\sin$.

So perhaps a better interpretation of your identity is actually as the identity for the angle addition formula for cosine, coupled with identity (#) above, which holds for a list of angles $\theta_k$ that sums to $\pi$.

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"Coupled with". (To be continued.....) –  Michael Hardy Aug 3 '11 at 23:19

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