Yeah, I was afraid of that. But anyways, I have an angle-oriented approach that I'm not sure how to fix so that the distribution matches.
Our setup is to embed the cyclic polygon into the complex plain, mainly for convenience. (All results can be shown without complex numbers, but this turns most results here into just algebra.) Specifically, let the points of the polygon be
with the circumcenter as the origin and
as the circumradius. Now, define the set of reals
such that:
(This implies that
)
To proceed, we will need the following rather straightforward lemmas.
Lemma 1(a): The length
for integer
. Proof.The length is:
where the last equality holds because
for all integer
Lemma 1(b): The length
. Proof.The length is:
where the last equality holds because
Now, we can find the circumradius in terms of those angles.
Proof.Notice that the perimter of the cylic polygon is just:
but using our lemmas gives us:
Setting
allows us to find the circumradius:
( has disappeared, as we would expect: It only rotates the polygon.)
With the circumradius, we can bash out the area
using only anglesWe divide the polygon into triangles and use
with
which is probably the ugliest formula I've ever seen.
or with
sides and the circumradius.Wea again divide the polygon into triangles, but this time we will use
where
and
Notice that
is occasionally negative, but using
makes it always positive. Specifically, whenever
we should have a negative area. With this in mind, we put a
in front of that area.
Not only is the above formula ugly, but it also does just about nothing because we don't have a formula for the circumradius in terms of the sides, only the angles.
As a final tidbit, we
can write the sides in terms of the angles in a rather straightforward way. But of course, it's ugly.
Lemma 2(a): The length
can be expressed fully in terms of angles for integer
ProofWe use
Lemma 1(a) and plug in for the circumradius.
Lemma 2(b): The length
can be expressed fully in terms of angles.
ProofWe use
Lemma 1(b)[/b] and plug in for the circumradius.
So that's why I can't find a way to fix my approach. It's possible that it can't be fixed here, but we may be able to transform the distribution after calculating the expected value. However, I don't know how to do any of that. Maybe someone else?
But anyways, I don't think we're going to find a direct formula for the area because other sources have only found that seven-degree polynomial, so I'd be surprised if we can do better than that (for the general case).
Without an area formula, the only ways (that come to mind) to find the expected value are bijections or using a different distribution and fixing it later. (Because these are just thoughts, I'm going to get hand-wavy.) I have no idea about bijection, and often that requires knowing the answer in advance, and we have no pattern so far. For a distribution, we may be able to use a suitable matrix to transform the polygon into something useful and then reverse the transformation at the end.
I don't know; maybe someone will find this useful.