horocyclic polygons - addendum

There is an alternative to using the analytic arguments embodied in the monotonicity theorem.
We do need a geometrical version, such as that of expanding euclidean arcs to demonstrate
the existence of some hyperbolic polygon with sides of given length in the general case.

Suppose that l(1),..,l(n) are positive numbers such that B(l(1),..,l(n)) > 0. By changing the
starting point of the list, but preserving the cyclic order, we may assume that l(n) = max{l(i)}.

From result 4 on the strange mapping, we deduce

Lemma 1
Suppose that l(1),..,l(n) are positive numbers such that B(l(1),..,l(n)) > 0.
There exists a cyclic hyperbolic polygon with sides of length l(1),..,l(n) if and only if
B(s(l(1)),..,s(l(n)) > 0. The hyperbolic radius of the circumcircle is unique.

proof
From the quoted result, the existence is equivalent to the existence of a cyclic euclidean
polygon with sides of lengths s(l(1)),..,s(l(n)). From the euclidean polygon theorem,
The euclidean radius e of the circumcircle is determined by the lengths s(l(i)), and hence
by the l(i). Again by result 4, the hyperbolic radius must be h, where e = sinh(½h).

The horocyclic polygon theorem and its converse may be stated as

Lemma 2
Suppose that l(1),..,l(n) are positive numbers such that B(l(1),..,l(n)) > 0.
There exists a horocyclic polygon with sides of length l(1),..,l(n) if and only if
B(s(l(1)),..,s(l(n)) = 0.

There is no problem of uniqueness as all horocycles are hyperbolic congruent.

In the remaining case, the geometric argument gives an existence result:

Lemma 3
Suppose that l(1),..,l(n) are positive numbers such that B(l(1),..,l(n)) > 0.
If B(s(l(1)),..,s(l(n)) < 0, there is a hypercyclic polygon with sides of length l(i),..,l(n).

proof
By the hyperbolic polygon theorem, there is a hyperbolic polygon with sides
of the given lengths and vertices on a hyperbolic circle, horocycle or hypercircle.
After Lemmas 1 and 2, the vertices must lie on a hypercircle.

It remains to be shown that, in the third case, the hypercircle is unique up to congruence.

Lemma 4

proof

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