IMO 1995 SL G3

Origin: TUR | Category: Geometry

Problem

The incircle of ABC touches BC, CA, and AB at D, E, and F respectively. X is a point inside ABC such that the incircle of XBC touches BC at D also, and touches CX and XB at Y and Z, respectively. Prove that EFZY is a cyclic quadrilateral.

Solution

If EF is parallel to BC, \triangleABC must be isosceles and E, Y are symmetric to F, Z with respect to AD, so the result follows. Now suppose that EF meets BC at P. By Menelaus’s theorem, BP CP = BF F A \cdot AE EC = BD DC (since BD = BF, CD = CE, AE = AF). It follows that the point P depends only on D and not on A. In particular, the same point is obtained as the intersection of ZY with BC. Therefore PE \cdotPF = PD2 = PY \cdotPZ, from which it follows that EFZY is a cyclic quadrilateral. Second solution. Since CD = CY = CE and BD = BZ = BF, all angles of EFZY can be calculated in terms of angles of ABC and Y ZBC. In fact, \angleFEY = 1 2(\angleA + \angleC + \angleBCY ) and \angleFZY = 1 2(180◦+ \angleB + \angleBCY ), which gives us \angleFEY + \angleFZY = 180◦.