IMO 1990 SL 9

Origin: HUN

Problem

The incenter of the triangle ABC is K. The midpoint of AB is C1 and that of AC is B1. The lines C1K and AC meet at B2, the lines B1K and AB at C2. If the areas of the triangles AB2C2 and ABC are equal, what is the measure of angle \angleCAB?

Solution

Let a, b, c be the lengths of the sides of \triangleABC, s = a+b+c , r the inradius of the triangle, and c1 and b1 the lengths of AB2 and AC2 respectively. As usual we will denote by S(XY Z) the area of \triangleXY Z. We have S(AC1B2) = AC1 \cdot AB2 AC \cdot AB S(ABC) = c1rs 2b , S(AKB2) = c1r 2 , S(AC1K) = cr 4 . From S(AC1B2) = S(AKB2) + S(AC1K) we get c1rs 2b = c1r 2 + cr 4 ; there- fore (a −b + c)c1 = bc. By looking at the area of \triangleAB1C2 we sim- ilarly obtain (a + b −c)b1 = bc. From these two equations and from S(ABC) = S(AB2C2), from which we have b1c1 = bc, we obtain a2 −(b −c)2 = bc ⇒b2 + c2 −a2 2bc = cos(\angleBAC) = 1 2 ⇒\angleBAC = 60◦.