IMO 1993 SL 15

Origin: MCD

Problem

For three points A, B, C in the plane we define m(ABC) to be the smallest length of the three altitudes of the triangle ABC, where in the case of A, B, C collinear, m(ABC) = 0. Let A, B, C be given points in the plane. Prove that for any point X in the plane, m(ABC) \leqm(ABX) + m(AXC) + m(XBC).

Solution

Denote by d(PQR) the diameter of a triangle PQR. It is clear that d(PQR) \cdot m(PQR) = 2SPQR. So if the point X lies inside the triangle ABC or on its boundary, we have d(ABX), d(BCX), d(CAX) \leqd(ABC), which implies

m(ABX) + m(BCX) + m(CAX) = 2SABX d(ABX) + 2SBCX d(BCX) + 2SCAX d(CAX) \geq2SABX + 2SBCX + 2SCAX d(ABC) = 2SABC d(ABC) = m(ABC). If X is outside \triangleABC but inside the angle BAC, consider the point Y of intersection of AX and BC. Then m(ABX) + m(BCX) + m(CAX) \geq m(ABY ) + m(BCY ) + m(CAY ) \geqm(ABC). Also, if X is inside the opposite angle of \angleBAC (i.e., \angleDAE, where B(D, A, B) and B(E, A, C)), then m(ABX) + m(BCX) + m(CAX) \geqm(BCX) \geqm(ABC). Since these are substantially all possible different positions of point X, we have finished the proof.