Given a triangle ABC, let I be the incenter. The internal bisectors of angles A, B, C meet the opposite sides in A', B', C' respectively. Prove that:
1/4 < AI·BI·CI/(AA'·BB'·CC') ≤ 8/27.
Solution
Consider the areas of the three triangles ABI, BCI, CAI. Taking base BC we conclude that (area ABI + area CAI)/area ABC = AI/AA'. On the other hand, if r is the radius of the in-circle, then area ABI = AB.r/2 and similarly for the other two triangles. Hence AI/AA' = (CA + AB)/p, where p is the perimeter. Similarly BI/BB' = (AB + BC)/p and CI/CC' = (BC + CA)/p. But the arithmetic mean of (CA + AB)/p, (AB + BC)/p and (BC + CA)/p is 2/3. Hence their product is at most (2/3)3 = 8/27.
Let AB + BC - CA = 2z, BC + CA - AB = 2x, CA + AB - BC = 2y. Then x, y, z are all positive and we have AB = y + z, BC = z + x, CA = x + y. Hence (AI/AA')(BI/BB')(CI/CC') = (1/2 + y/p)(1/2 + z/p)(1/2 + x/p) > 1/8 + (x+y+z)/(4p) = 1/8 + 1/8 = 1/4.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
2 Nov 1998