IMO 1994 SL G5

Origin: CYP | Category: Geometry

Problem

A line l does not meet a circle \omega with center O. E is the point on l such that OE is perpendicular to l. M is any point on l other than E. The tangents from M to \omega touch it at A and B. C is the point on MA such that EC is perpendicular to MA. D is the point on MB such that ED is perpendicular to MB. The line CD cuts OE at F. Prove that the location of F is independent of that of M.

Solution

We first prove that AB cuts OE in a fixed point H. Note that \angleOAH = \angleOMA = \angleOEA (because O, A, E, M lie on a circle); hence \triangleOAH ∼ \triangleOEA. This implies OH \cdot OE = OA2, i.e., H is fixed. Let the lines AB and CD meet at K. Since EAOBM and ECDM are cyclic, we have \angleEAK

\angleEMB = \angleECK, so ECAK is cyclic. Therefore \angleEKA = 90◦, hence EKBD is also cyclic and EK \parallel OM. Then \angleEKF

\angleEBD = \angleEOM = \angleOEK, from which we deduce that KF = FE. However, since \angleEKH = 90◦, the O E M A B C D H K F l point F is the midpoint of EH; hence it is fixed.