35th IMO 1994 shortlist

Problem G5

L is a line not meeting a circle center O. E is the foot of the perpendicular from O to L and M is a variable point on L (not E). The tangents from O to the circle meet it at A and B. The feet of the perpendiculars from E to MA and MB are C and D respectively. The lines CD and OE meet at F. Show that F is fixed.



Let AB meet OM at G and OE at H. Then ∠MEH = ∠MGH = 90o, so GHEM is cyclic, so OH·OE = OG·OM. Triangles AOG and MOA are similar, so OG·OM = OA2. Hence OH = OA2/OE, which is constant.

Let the lines AB and CD meet at K. Angles OAM, OEM, OBM are all 90o, so EAOBM is cyclic. Similarly ∠ECM = ∠EDM = 90o, so ECDM is cyclic. Hence ∠EAK = ∠EMB = ∠EMD = ∠ECK, so EKAC is cylic. Hence angle EKA = 90o. So EKBD is cyclic. So ∠EKF = ∠EKD = ∠EBD = ∠EBM = ∠EOM (EAOBM cyclic) = ∠FEK (OM parallel EK). Hence EF = FK. But ∠EKH = 90o, so EF = FH. So F is the midpoint of EH. But E and H are fixed, so F is fixed also.



35th IMO shortlist 1994

© John Scholes
22 Sep 2002