IMO 1989 SL 6

Origin: CZS

Problem

For a triangle ABC, let k be its circumcircle with radius r. The bisectors of the inner angles A, B, and C of the triangle intersect respec- tively the circle k again at points A′, B′, and C′. Prove the inequality 16Q3 \geq27r4P, where Q and P are the areas of the triangles A′B′C′ and ABC respec- tively.

Solution

Let us denote the measures of the inner angles of the triangle ABC by \alpha, \beta, \gamma. Then P = r2(sin 2\alpha + sin 2\beta + sin 2\gamma)/2. Since the inner angles of the triangle A′B′C′ are (\beta + \gamma)/2, (\gamma + \alpha)/2, (\alpha + \beta)/2, we also have Q = r2[sin (\beta + \gamma) + sin (\gamma + \alpha) + sin (\alpha + \beta)]/2. Applying the AM–GM mean inequality, we now obtain 16Q3 = 16 8 r6(sin (\beta + \gamma) + sin (\gamma + \alpha) + sin (\alpha + \beta))3 \geq54r6 sin (\beta + \gamma) sin (\gamma + \alpha) sin (\alpha + \beta) = 27r6[cos(\alpha −\beta) −cos(\alpha + \beta + 2\gamma)] sin(\alpha + \beta) = 27r6[cos(\alpha −\beta) + cos \gamma] sin(\alpha + \beta) = 27 2 r6[sin(\alpha + \beta + \gamma) + sin(\alpha + \beta −\gamma) + sin 2\alpha + sin 2\beta] = 27 2 r6[sin(2\gamma) + sin 2\alpha + sin 2\beta] = 27r4P. This completes the proof.