IMO 2002 SL G8

Origin: ARM | Category: Geometry

Problem

Let S1 and S2 be circles meeting at the points A and B. A line through A meets S1 at C and S2 at D. Points M, N, K lie on the line segments CD, BC, BD respectively, with MN parallel to BD and MK parallel to BC. Let E and F be points on those arcs BC of S1 and BD of S2 respectively that do not contain A. Given that EN is perpendicular to BC and FK is perpendicular to BD, prove that ∡EMF = 90◦.

Solution

Let G be the other point of intersection of the line FK with the arc BAD. Since BN/NC

DK/KB and \angleCEB

\angleBGD the triangles CEB and BGD are similar. Thus BN/NE = DK/KG = FK/KB. From BN

MK and BK

MN it follows that MN/NE = FK/KM. But we also have that \angleMNE = 90◦+ \angleMNB = 90◦+ \angleMKB

\angleFKM, and hence \triangleMNE ∼\triangleFKM. A B C D M K N E F G S1 S2 Now \angleEMF = \angleNMK −\angleNME −\angleKMF = \angleNMK −\angleNME − \angleNEM = \angleNMK −90◦+ \angleBNM = 90◦as claimed.