Home > Geometry, Olympiad > Balkan Mathematical Olympiad 2012 Problem 1

Balkan Mathematical Olympiad 2012 Problem 1


Let {A}, {B} and {C} be points lying on a circle {\Gamma} with centre {O}. Assume that {\angle ABC > 90}. Let {D} be the point of intersection of the line {AB} with the line perpendicular to {AC} at {C}. Let {l} be the line through {D} which is perpendicular to { A O }. Let {E} be the point of intersection of {l} with the line {AC}, and let {F} be the point of intersection of {\Gamma} with {l} that lies between {D} and {E}. Prove that the circumcircles of triangles {BFE} and {CFD} are tangent at {F}.

Balkan Mathematical Olympiad 2012 Problem 1

Solution: Note that the circles are tangent if \angle CDF +\angle EBF=\angle CFE, and this condition is equivalent to \angle FCD=\angle FBE. Since E is the orthocenter of the triangle formed by AD, l, DC it follows that BE \perp AD, and the desired condition is equivalent to \angle FCA=\angle FBD, which is true because ABFC is cyclic.

Advertisements
Categories: Geometry, Olympiad
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: