existence of a convex polygon in euclidean geometry

euclidean polygon theorem
If a(1),..,a(n) is a list of positive numbers, with a(n) ≥ a(i) for all i,
then there is a unique convex cyclic polygon with sides of length
a(1),..,a(n) in order.

proof
We use the construction described earlier, choosing points A,B
on a circle K of radius R, with |AB| = a(n), and A(1) = A,..,A(n)
with |A(i)A(i+1)|=a(i), i = 1,..,n-1, in order on K.
Let θ(i) denote the angle A(i)CA(i+1), measured conterclockwise,
so θ(i) ≤ π by our choice of A(i+1). Let θ(n) be the measure of
angle BCA in the range (0,π].
Let C* be the mid-point of AB.

In the first case, for C = C*, we have θ(1)+..+θ(n) ≥ 2π.
Now sin(½θ(i)) = a(i)/2R. As C moves to the left, R
increases so all the θ(i) decrease. As R tends to ∞, each θ(i)
tends to zero, so there is exactly one choice of R for which
the sum is 2π. Hence there is one cyclic polygon.

In the second case, for C = C*, θ(1)+..+θ(n-1) < θ(n) = π.
Here, we require a C for which θ(1)+..+θ(n-1) = θ(n).
As we move C to the right, all the θ(i) decrease, so we cannot
immediately assert that there is a (unique) solution.
We can argue that, as R tends to infinity, the circle K approaches
the line AB. As a(1)+..+a(n-1) > a(n) = |AB|, the point A(n)
eventually passes B. There must be at least one value of R for
which A(n) = B. Thus there is a suitable polygon.

The problem with this geometrical approach is that it does not
establish the uniqueness of R. This can be done using a certain
trigonometric identity.

Observe that, by elementary trigonometry,
|A(1)A(n)|
= 2Rsin(½θ(1)+..+½θ(n-1))
= ΣrP(n-1,r)S(r)2Rsin(½θ(r)),
= ΣrP(n-1,r)S(r)a(r),

As R tends to infinity, each θ(i) decreases towards 0, and so
each P(n-1,r) and S(r) increase towards 1. Thus |A(1)A(n)|
increases towards Σra(r). Thus there is a unique value of R
for which |A(1)A(n)| = a(n), i.e. for which A(n) = B.

For any α(1),..,α(m),
sin(α(1)+..+α(m)) = ΣrP(m,r)S(r)sin(α(r)),
where
P(m,m) = 1,
P(m,r) = cos(α(r+1))..cos(α(m)) for r < m, and
S(1) = 1,
S(r) = cos(α(1)+..+α(r-1)) for r > 1.

There is, of course, a hyperbolic analogue.

hyperbolic geometry