PROVE THAT r/R = cosA + cosB + cosC - 1?

Question

ABC is a plane triangle

r is the radius of incircle.

R is the radius of circumcircle .

Answer 1

Here is a proof using a well-known trigonometric identity that is valid in every triangle with internal angles α, β, γ:

(***) cos α + cos β + cos γ = 4 sin(α/2) sin(β/2) sin(γ/2) + 1

It is easily derived using α + β + γ = π.

Let I is the incenter (the common point of angle bisectors) of the triangle ABC, C' - the common point of the angle bisector with the circumscribed circle, C' is of course mid-point of the arc AB, not containing C - follow the link below to see a picture:

http://farm4.static.flickr.com/3106/2813...

Triangle AC'I is isosceles (angles IAC' and AIC' are both α/2+ γ/2), let C'D is perpendicular to AI, D - mid-point of AI, let IT is perpendicular to AC, |IT| = r. Triangle AIT yields

r = |IT| = |AI| * sin(α/2) = 2*|AD| sin(α/2)

Triangle AC'D yields |AD| = |AC'| sin(β/2), so

r = 2*|AD| sin(α/2) = 2*|AC'| sin(α/2) sin(β/2)

Finally the Law of Sines, applied to triangle ACC' yields

|AC'| = 2R*sin(γ/2), hence

r = 4R sin(α/2) sin(β/2) sin(γ/2) and according (***)

r/R = cos α + cos β + cos γ - 1

By the way, using the Euler's Theorem about the distance between the incenter and circumcenter, according which

R ≥ 2r we obtain the following well--known inequalities:

sin(α/2) sin(β/2) sin(γ/2) ≤ 1/8,

cos α + cos β + cos γ ≤ 3/2,

equalities exactly when the triangle is equilateral.