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.