# inscribed hyperbolic polygons

 Many problems in hyperbolic geometry can be solved using the Poincare disk model. In particular it is often useful to apply a hyperbolic transformation so that one of the points is at the centre of the disk. Then any hyperbolic line through this point is also a euclidean diameter of the disk. We can relate the distances in each geometry. some notation Suppose that A,B lie in the disk. Then we define s(AB) = sinh(½d(A,B)) c(AB) = cosh(½d(A,B)) t(AB) = tanh(½d(A,B)) |AB| = euclidean distance between A and B. the hybrid distances theorem Suppose that A and B are points in the disk, and O is the centre of the disk. Then |AB| = s(AB)/c(OA)c(OB). Since we know more about the euclidean geometry of the disk, this is useful. For example, we know that a hyperbolic circle is a euclidean circle which lies entirely in the disk. We can characterize the hyperbolic triangles which have hyperbolic circumcircles algebraically. This involves the quantity Δ which we first met in the sine rule. This can be expressed in terms of s(AB), s(BC), and s(CA), but this involves a root. the hyperbolic circumcircle theorem The hyperbolic triangle ΔABC has a hyperbolic circumcircle if and only if 4s(AB)s(BC)s(CA) < Δ. If the condition is satisfied, then the hyperbolic radius of the circumcircle is given by r, where tanh(r) = 4s(AB)s(BC)s(CA)/Δ. Since a hyperbolic triangle has Δ > 0, we may restate the condition as "if and only if Δ2 - (4s(AB)s(BC)s(CA))2 > 0". If we now use the formula for Δ2, and the identity cosh(2x) = 2sinh2(x)+1, we find (after some algebraic simplification) that Δ2 - (4s(AB)s(BC)s(CA))2 = 4H(s(AB),s(BC),s(CA)), where H(x,y,z) = (x+y+z)(x+y-z)(y+z-x)(z+x-y). The condition now becomes "if and only if H(s(AB),s(BC),s(CA)) > 0." We can also use identities to get a formula for the hyperbolic radius in terms of H. Summing up, we get the hyperbolic circumcircle theorem - version 2 The hyperbolic triangle ΔABC has a hyperbolic circumcircle if and only if H(s(AB),s(BC),s(CA)) > 0. If the condition is satisfied, then the hyperbolic radius of the circumcircle is r, where sinh2(r) = 4s2(AB)s2(BC)s2(CA)/H(s(AB),s(BC),s(CA)). Note that the expression for sinh2(r) is equal to (2R)2, where R is the radius of the euclidean circumcircle of the triangle with sides of length s(AB),s(BC),s(CA). This triangle exists as H(s(AB),s(BC),s(CA)) > 0. proof Only the second part remains to be proved. It follows from the identity sinh2(x) = tanh2(x)/(1-tanh2(x)), and earlier calculations. The proof of the first version also yields a result on euclidean circumcircles If A,B,C lie in the disk, then the euclidean circumcircle of ABC is (1) a circle within D if H(s(AB),s(BC),s(CA)) > 0, (2) a circle touching C if H(s(AB),s(BC),s(CA)) = 0, and (3) an i-line cutting C twice if H(s(AB),s(BC),s(CA)) < 0. As we already know, case (1) corresponds to a hyperbolic circle, and case (2) to a horocycle. Later, we shall discuss the third category as hypercircles. We shall refer to H as Heron's Polynomial, since it occurs in Heron's formula. Here, we note merely that it also characterizes the triples (k,l,m) of positive numbers for which there is a triangle with sides of length k,l,m. Such a triangle exists if and only if the sum of any two exceeds the third, and this is equivalent to H(k,l,m) > 0. We give a proof here. It works equally well in euclidean and hyperbolic geometry. If (k,l,m) satisfy the condition, we will refer to a corresponding triangle as a (k,l,m) triangle. It is, of course, unique up to conjugacy. We may now restate our result as the hybrid circumcircle theorem The hyperbolic triangle ΔABC has a hyperbolic circumcircle if and only if there exists a euclidean (s(AB),s(BC),s(CA)) triangle. When the circumcircle exists, its hyperbolic radius r is given by sinh(r) = 2R, where R is the euclidean radius of the second triangle. We use the term hybrid to describe a theorem which relates hyperbolic and euclidean properties. At this stage, we can actually prove a form of ptolemy's theorem.