The angle bisectors of the triangle ABC meet the circumcircle again at A'B'C'. Show that 16 (area A'B'C')3 ≥ 27 area ABC R4, where R is the circumradius of ABC.
Solution
Let R be the circumradius and O the circumcircle. ∠BOC = 2 ∠A, so area BOC = ½ R2 sin 2A. Hence area ABC = ½ R2(sin 2A + sin 2B + sin 2C).
∠A'C'C = ∠A'AC = ∠A/2, and ∠B'C'C = ∠B'BC = ∠B/2, so ∠A'C'B' = ∠A/2 + ∠B/2. The circumradius is the same, so area A'B'C' = ½ R2(sin(A+B) + sin(B+C) + sin(C+A) ). Thus we have to show that 4 (sin(A+B) + sin(B+C) + sin(C+A) )3 ≥ 27 (sin 2A + sin 2B + sin 2C).
Applying AM/GM to sin(A+B), sin(B+C), sin(C+A) we get (sin(A+B) + sin(B+C) + sin(C+A) )3 ≥ 27 sin(A+B) sin(B+C) sin(C+A).
We now need some straightforward applications of the forumulae for sin(x+y), cos(x+y) etc. We have sin(B+C) sin(C+A) = (cos(B-A) - cos(C+180) )/2 = (cos(B-A) + cos C)/2, so sin(A+B) sin(B+C) sin(C+A) = ½ sin(A+B) cos(B-A) + ½ sin(A+B) cos C. But sin 2A + sin 2B = sin( (A+B) + (B-A) ) + sin( (A+B) - (B--A) ) = 2 sin(A+B) cos(B-A), and sin( (A+B) + C) + sin( (A+B) - C) = 2 sin(A+B) cos C, so sin(A+B) sin(B+C) sin(C+A) = ¼ (sin 2A + sin 2B + sin(A+B-C) ) = ¼ (sin 2A + sin 2B + sin 2C). Hence (sin(A+B) + sin(B+C) + sin(C+A) )3 ≥ (27/4) (sin 2A + sin 2B + sin 2C), as required.
© John Scholes
jscholes@kalva.demon.co.uk
28 Dec 2002
Last corrected/updated 28 Dec 02