IMO 1988 SL 3

Origin: CAN

Problem

The triangle ABC is inscribed in a circle. The interior bi- sectors of the angles A, B, and C meet the circle again at A′, B′, and C′ respectively. Prove that the area of triangle A′B′C′ is greater than or equal to the area of triangle ABC.

Solution

Let R be the circumradius, r the inradius, s the semiperimeter, ∆the area of ABC and ∆′ the area of A′B′C′. The angles of triangle A′B′C′ are A′ = 90◦−A/2, B′ = 90◦−B/2, and C′ = 90◦−C/2, and hence ∆= 2R2 sin A sin B sin C and ∆′ = 2R2 sin A′ sin B′ sin C′ = 2R2 cos A 2 cos B 2 cos C 2 . Hence, ∆ ∆′ = sin A sin B sin C cos A 2 cos B 2 cos C = 8 sin A 2 sin B 2 sin C 2 = 2r R ,

where we have used that r = AI sin(A/2) = \cdot \cdot \cdot = 4R sin(A/2) \cdot sin(B/2) \cdot sin(C/2). Euler’s inequality 2r \leqR shows that ∆\leq∆′. Second solution. Let H be orthocenter of triangle ABC, and Ha, Hb, Hc points symmetric to H with respect to BC, CA, AB, respectively. Since \angleBHaC = \angleBHC = 180◦−\angleA, points Ha, Hb, Hc lie on the circumcircle of ABC, and the area of the hexagon AHcBHaCHb is double the area of ABC. (1) Let us apply the analogous result for the triangle A′B′C′. Since its or- thocenter is the incenter I of ABC, and the point symmetric to I with respect to B′C′ is the point A, we find by (1) that the area of the hexagon AC′BA′CB′ is double the area of A′B′C′. But it is clear that the area of ∆CHaB is less than or equal to the area of ∆CA′B etc.; hence, the area of AHcBHaCHb does not exceed the area of AC′BA′CB′. The statement follows immediately.