the existence theorem for excircles
Suppose that the hyperbolic triangle ABC has <CAB = α, <ABC = β, <BCA = γ. Then
there is an hyperbolic excircle opposite C if and only if cos(½γ) < sin(½α)+sin(½β),
and a horocyclic excircle opposite C if and only if cos(½γ) = sin(½α)+sin(½β).
proof
The excircle exists if and only if the exterior bisectors at A and B meet in a point X.
This X is then the centre of the excircle, and the excircle is a horocycle if X is on C.
Since the angles at A,B are α,β, these bisectors make angles ½(π-α), ½(π-β) with
AB. Let c = d(A,B). The bisectors meet if and only if there is a hyperbolic triangle
ABX with <ABX=½(π-α), <BXA = ½(π-β). From the third existence theorem,
this is equivalent to sin(½(π-α))sin(½(π-β))cosh(c)-cos(½(π-α))cos(½(π-β)) ≤ 1,
i.e. to cos(½α)cos(½β)cosh(c) - sin(½α)sin(½β) ≤ 1.
By the Second Cosine Theorem for ΔABC, sin(α)sin(β)cosh(c) = cos(γ)+cos(α)cos(β).
If we substitute for cosh(c) in the inequality and simplify, we get
cos2(½γ) ≤ (sin(½α)+sin(½β))2
Since the half-angles are in [0,½π), we can take non-negative roots and obtain
cos(½γ) ≤ sin(½α)+sin(½β).
The cases of a hyperbolic circle and horocycle occur when X is in or on C, so correspond
to inequality and equality respectively.
|