# Romania 2006 Problem 1

Problem 1 of the Romanian Mathematical Olympiad 2006

Let ABC be a triangle and the points M and N on the sides AB and BC, respectively, such that 2CN/BC = AM/AB. Let P be a point on the line AC. Prove that the lines MN and NP are perpendicular if and only if PN is the interior angle bisector of ∠MPC.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to thi si Tuong Vi)

1) Assume MN and NP are perpendicular.

Since 2CN/BC = AM/AB, pick point Q on BC such that QN = CN. Then MQ || AC. Let E and F be the midpoints of MP and MA, respectively. We have EF || AC but N is also the midpoint of QC and MQ || AC, therefore, FN || AC and F, E and N are collinear. We then have ∠ENP = ∠NPC.

But EN = EP = EM (E is midpoint of MP and ∠MNP is right angle) causes ∠ENP = ∠EPN or ∠EPN = ∠NPC and PN is the interior bisector of ∠MPC.

2) Assume PN is interior bisector of ∠MPC.

∠EPN = ∠NPC and since MQ || AC and F and N are the midpoints of MA and QC, respectively, we have FN || AC. Therefore, ∠FNP = ∠NPC and E is the midpoint of MP. It follows that ∠EPN = ∠ENP and EN = EP = EM or ∠MNP = 90° and MN is orthogonal to NP.