IMO 1998 SL 8

Origin: IND

Problem

Let ABC be a triangle such that \angleA = 90◦and \angleB < \angleC. The tangent at A to its circumcircle \omega meets the line BC at D. Let E be the reflection of A across BC, X the foot of the perpendicular from A to BE, and Y the midpoint of AX. Let the line BY meet \omega again at Z. Prove that the line BD is tangent to the circumcircle of triangle ADZ.

Solution

Let M be the point of intersection of AE and BC, and let N be the point on \omega diametrically opposite A. Since \angleB < \angleC, points N and B are on the same side of AE. Furthermore, \angleNAE = \angleBAX = 90◦−\angleABE; hence the triangles NAE and BAX are similar. Con- sequently, \triangleBAY and \triangleNAM are also similar, since M is the midpoint A B C D E M N X Y Z \omega of AE. Thus \angleANZ = \angleABZ = \angleABY = \angleANM, implying that N, M, Z are collinear. Now we have \angleZMD = 90◦−\angleZMA = \angleEAZ = \angleZED (the last equality because ED is tangent to \omega); hence ZMED is a cyclic quadrilateral. It follows that \angleZDM = \angleZEA = \angleZAD, which is enough to conclude that MD is tangent to the circumcircle of AZD. Remark. The statement remains valid if \angleB \geq\angleC.