existence of a convex polygon in hyperbolic geometry

hyperbolic polygon theorem
If a(1),..,a(n) is a list of positive numbers, with a(n) ≥ a(i) for all i,
and a(n) < a(1)+...+a(n). then
(1) if B(s(a(1)),..,s(a(n))) > 0 there is a unique convex cyclic
hyperbolic polygon with sides of length a(1),..,a(n) in order.
(2) in any case, there is a convex cyclic. horocyclic or hypercyclic
hyperbolic polygon with sides of length a(1),..,a(n).

proof
For (1), we can either mimic the proof of the euclidean theorem
or use the euclidean theorem and result 4. In the former, we need
to note that we have hyperbolic circles of arbitrarily large hyperbolic
radius. Also, we need to use the hyperbolic sine formula in the last
part - sinh(½a(i)) = sinh(r)sin(½θ(i)).

We begin by choosing a segment AB of hyperbolic length a(n).
We may as well choose this so that the mid point is O and such that
AB is a vertical segment in the picture. We use the construction
described earlier, choosing a euclidean circle K of radius R, through
A and B. We then construct A(1) = A,..,A(n) on K with A(i)A(i+1) of
hyperbolic length a(i), i = 1,..,n-1.

Note if the circle K meets the disk boundary, then as hyperbolic
circles lie entirely within the disk, a hyperbolic circle centred on K
and of any radius will cut K twice within the disk.

Examination of the euclidean proof actually shows that, if we have
a case where all the A(i) lie on the left arc for centre O, then there
will be just a cyclic hyperbolic polygon.

For the other case, as R increases, the left arc of K approaches the
line AB. Since a(1)+..+a(n-1) > a(n), A(n) will eventually lie beyond
B. By continuity, there will be at least one R for which A(n) = B.

Note In case (2) there is no obvious way to show that the solution
is unique without some new ideas.

hyperbolic geometry