horocyclic polygons

The work we have done so far allows us to describe the hyperbolic polygons with vertices
on a horocycle. The result on euclidean circumcircles shows that A,B,C lies on a
horocycle if and only if H(s(AB),s(AC),s(BC)) = 0, where s(PQ) = sinh(½d(P,Q)), and
H(x,y,z) is the Heron polynomial. Thus one of s(AB),s(AC),s(BC) is the sum of the others.
Observe that, if A,B,C lie on a horocycle K, then one of the points lies on the finite arc
defined by the other two. The arcs theorem shows that, if C lies on the finite arc AB,
then E(A,B,C) < 0, so s2(AB) > s2(AC)+s2(CB). We must then have s(AB)=s(AC)+s(CB).
This generalizes by induction on m to

the horocycle polygon theorem
Suppose that A(0),..,A(m) lie in order on a horocycle, then
s(A(0)A(m)) = s(A(0)A(1))+s(A(1)A(2))+..+s(A(m-1)A(m)).

Saying that the points lie in order means that, for each k, A(k) is on the finite arc defined
by A(0)A(k+1). The induction step is the application of the result for a hyperbolic triangle
to A(0)A(k)A(k+1).

This has a valid converse :

the converse horocycle polygon theorem
Suppose that L,l(1),..,l(m) are positive numbers satisfying
sinh(½L) = sinh(½l(1))+..+sinh(½l(m)). Then there is a
hyperbolic (L,l(1),..,l(m)) polygon with vertices on a horocycle.

This is proved by choosing A(0) on a horocycle K, and choosing A(i), i=1,..,m
in order along K so that d(A(i-1),A(i))=l(i). By the theorem, d(A(0),A(m)) = L.

arc length along a horocycle
If A,B lie on a horocycle K and x = d(A,B), then the length of the finite arc AB
of K is L = 2sinh(½x).

As you might expect, we approximate L by n polygonal arcs each of length ln.
Let Ln = nln. As n tends to ∞, Ln tends to L. By the theorem, sinh(½x) = n sinh(½Ln/n).
Taking the limit as n tends to ∞, we get the result as the limit of nsinh(K/n) is K.

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