IMO 2001 SL G8

Origin: ISR | Category: Geometry

Problem

Let ABC be a triangle with ∡BAC = 60◦. Let AP bisect \angleBAC and let BQ bisect \angleABC, with P on BC and Q on AC. If AB + BP = AQ + QB, what are the angles of the triangle?

Solution

Let S and T respectively be the points on the extensions of AB and AQ over B and Q such that BS = BP and QT = QB. It is given that AS = AB + BP = AQ + QB = AT . Since \anglePAS = \anglePAT , the triangles APS

and APT are congruent, from which we deduce that \angleATP = \angleASP = \beta/2 = \angleQBP. Hence \angleQTP = \angleQBP. If P does not lie on BT, then the last equality implies that \triangleQBP and \triangleQTP are congruent, so P lies on the internal bisector of \angleBQT. But P also lies on the internal bisector of \angleQAB; consequently, P is an excenter of \triangleQAB, thus lying on the internal bisector of \angleQBS as well. It follows that \anglePBQ = \beta/2 = \anglePBS = 180◦−\beta, so \beta = 120◦, which is impossible. Therefore P \inBT, which means that T \equivC. Now from QC = QB we conclude that 120◦−\beta = \gamma = \beta/2, i.e., \beta = 80◦and \gamma = 40◦.