IMO 1989/A2 solution

IMO 1989

Problem A2

In an acute-angled triangle ABC, the internal bisector of angle A meets the circumcircle again at A₁. Points B₁ and C₁ are defined similarly. Let A₀ be the point of intersection of the line AA₁ with the external bisectors of angles B and C. Points B₀ and C₀ are defined similarly. Prove that the area of the triangle A₀B₀C₀ is twice the area of the hexagon AC₁BA₁CB₁ and at least four times the area of the triangle ABC.

Solution

By Marcin Mazur, University of Illinois at Urbana-Champaign

Let I be the point of intersection of AA₀, BB₀, CC₀ (the in-center). BIC = 180 - 1/2 ABC - 1/2 BCA = 180 - 1/2 (180 - CAB) = 90 + 1/2 CAB. Hence CA₁B = 180 - CAB [BA₁CA is cyclic] = 2(180 - BIC) = 2CA₀B. But A₁B = A₁C, so A₁ is the center of the circumcircle of BCA₀. But I lies on this circumcircle (IBA₀ = ICA₀ = 90), and hence A₁A₀ = A₁I.

Hence area IBA₁ = area A₀BA₁ and area ICA₁ = area A₀CA₁. Hence area IBA₀C = 2 area IBA₁C. Similarly, area ICB₀A = 2 area ICB₁A and area IAC₀B = 2 area IAC₁B. Hence area A₀B₀C₀ = 2 area hexagon AB₁CA₁BC₁.

A neat solution for the rest is as follows by Thomas Jäger

Let H be the orthocentre of ABC. Let H₁ be the reflection of H in BC, so H₁ lies on the circumcircle. So area BCH = area BCH₁ <= area BCA₁. Adding to the two similar inequalities gives area ABC <= area hexagon - area ABC.

Mazur's solution was as follows

CAB = 180^o - CA₁B and A₁B = A₁C, so A₁BC = 90^o - 1/2 CA₁B = 1/2 CAB. Hence the perpendicular from A₁ to BC has length 1/2 BC tan(CAB/2) and area CA₁B = 1/4 BC² tan(CAB/2).

Put r = radius of in-circle of ABC, x = cot(CAB/2), y = cot(ABC/2), z = cot(BCA/2). Then BC = r(y + z) and area CA₁B = r²(y + z)²/(4x). Also area BIC = 1/2 r BC. Similarly for the other triangles, so area ABC = area BIC + area CIA + area AIB = r²(x + y + z). We have to show that area ABC ≤ area CA₁B + area AB₁C + area BC₁A, or (x + y + z) ≤ (y + z)²/(4x) + (z + x)²/(4y) + (x + y)²/(4z).

Putting s = x + y + z, this is equivalent to: 4s ≤ (s - x)²/x + (s - y)²/y + (s - z)²/z, or 9s ≤ s²(1/x + 1/y + 1/z), but this is just the statement that the arithmetic mean of x, y, z is not less than the harmonic mean.

Note in passing that the requirement for ABC to be acute is unnecessary.

Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

30th IMO 1989