IMO 1999 SL G1

Origin: ARM | Category: Geometry

Problem

Let ABC be a triangle and M an interior point. Prove that min{MA, MB, MC} + MA + MB + MC < AB + AC + BC.

Solution

We use the following simple lemma. Lemma. Suppose that M is the interior point of a convex quadrilateral ABCD. Then it follows that MA + MB < AD + DC + CB. Proof. We repeatedly make use of the triangle inequality. The line AM, in addition to A, intersects the quadrilateral in a second point N. In that case AM + MB < AN + NB < AD + DC + CB. We now apply this lemma in the following way. Let D, E, and F be median points of BC, AC, and AB. Any point M in the interior of \triangleABC is contained in at least two of the three convex quadrilaterals ABDE, BCEF, and CAFD. Let us assume without loss of generality that M is in the interior of BCEF and CAFD. In that case we apply the lemma to obtain AM + CM < AF + FD + DC and BM + CM < CE + EF + FB to obtain CM + AM + BM + CM < AF + FD + DC + CE + EF + FB = AB + AC + BC from which the required conclusion immediately follows.