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
In our work on circumcircles, we showed that, if ΔABC has a hyperbolic circumcircle, then its radius r is given by 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.
We can find results about incircle and excircles. Taking the dual of (1) in the inradius theorem,
other circumradius formulae If ΔABC has a hyperbolic circumcircle of radius r, then
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