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a theorem on cyclic hyperbolic quadrilaterals There is a convex cyclic quadrilateral with sides of length r,s,t,u if and only if B = B(s(r),s(s),s(t),s(u)) > 0, where, for real x, s(x) denotes sinh(½x). When it exists, the hyperbolic circumcircle has radius d, where sinh2(d) = 4S(s(r),s(s),s(t),s(u))/B, S(a,b,c,d) = (ab+cd)(ac+bd)(cd+bc)
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proof Suppose that ABCD is a hyperbolic quadrilateral with d(A,B)=r, d(B,C)=s, d(C,D)=t, d(D,A)=u and d(A,C)=x. By the second hyperbolic circle theorem, this will be convex and cyclic if and only if E(A,B,C) = -E(A,D,C) and A,B,C lie on a hyperbolic circle. By the second form of the circumcircle theorem, the second condition amounts to H = H(s(r),s(s),s(x)) > 0. By tedious algebra, we can check that E(A,B,C) =-E(A,D,C) if and only if (s(r)s(s)+s(t)s(u))s(x)2=(s(r)s(t)+s(s)s(u))(s(r)s(u)+s(s)s(t)). Note that, as r,s,t,u are positive, this has a positive solution x. If we rewrite H(x,y,z) as ((x+y)2-z2)(z2-(x-y)2), and substitute, then H(s(r),s(s),s(x)) = 4s2(r)s2(s)B(s(r),s(s),s(t),s(u))/(s(r)s(s)+s(t)s(u))2 Thus, if we are given that the value of B is positive, the H value is positive. Hence we can find a (unique) convex cyclic quadrilateral, as required, precisely when the value of B is positive.
The radius of the hyperbolic circumcircle is the radius of the sinh2(d) = 4S(s(r),s(s),s(t),s(u))/B(s(r),s(s),s(t),s(u)), with S as stated.
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Note that we can also find y = d(B,D) by the method of the proof simply by interchanging s and t. We obtain (s(r)s(s)+s(t)s(u))s(x)2=(s(r)s(t)+s(s)s(u))(s(r)s(u)+s(s)s(t)). (s(r)s(t)+s(s)s(u))s(y)2=(s(r)s(s)+s(t)s(u))(s(r)s(u)+s(s)s(t)). From these, the original form of ptolemy's theorem is soon recovered.
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