the semi-perimeter in hyperbolic geometry

 In line with the philosophy of duality, the semi-perimeter s = ½(a+b+c) ought to be related to the quantity σ = ½(α+β+γ). From the Gauss-Bonet Formula, the area of a hyperbolic triangle is D = π-2σ, so σ = ½π-½D. Earlier, we showed that sin(½D) = Δ/4cosh(½a)cosh(½b)cosh(½c). We therefore have the σ theorem For any hyperbolic triangle, cos(σ) = Δ/4cosh(½a)cosh(½b)cosh(½c). Applying duality, we get the s theorem For any hyperbolic triangle, sinh(s) = Φ/4sin(½α)sin(½β)sin(½γ). proof Applying duality, the cos(σ) is replaced by ±i.sinh(s), Δ by i.Φ, and cosh(½a) by ±sin(½α), etc. Cancelling the i, and noting that each expression is positive, we have the result. In our work on circumcircles, we showed that, if ΔABC has a hyperbolic circumcircle, then its radius r is given by tanh(r) = 4sinh(½a)sinh(½b)sinh(½c)/Δ. If we combine this with the σ theorem, and note that 2sinh(x)cosh(x) = sinh(2x), we get cos(σ)/tanh(r) = Δ2/2sinh(a)sinh(b)sinh(c) = ½Φ, by (R2). Hence the circumradius formula If ΔABC has a hyperbolic circumcircle of radius r, then tanh(r) = 2cos(σ)/Φ. The dual of the right-hand-side is 2sinh(s)/Δ. We should expect this to be related to some length associated with ΔABC. The circumcentre, if it exists, is the intersection of the bisectors of the sides. Perhaps we should expect to encounter the bisectors of the angles. We do indeed find that the incircle and inradius are involved. We can find results about incircle and excircles. Taking the dual of (1) in the inradius theorem, we get a final set of expressions for the circumradius: other circumradius formulae If ΔABC has a hyperbolic circumcircle of radius r, then tanh(r) = tanh(½a)/cos(σ-α) = tanh(½b)/cos(σ-β) = tanh(½c)/cos(σ-γ).