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The 55th International Mathematical Olympiad (IMO) was held from 3-13 July 2014 in Cape Town. Here is the 1st problem written on the 2nd day of the competition (problem no. 4):
"The points P and Q are chosen on the side BC of a triangle ABC so that ∠PAB = ∠ACB and ∠QAC = ∠CBA. The points M and N are taken on the rays AP and AQ, respectively, so that AP = PM and AQ = QN. Prove that the lines BM and CN intersect on the circumcircle ABC".
Drag any of the vertices of ABC. Can you prove the result? Can you prove it in several different ways?
Also note that the products of the opposite sides of the cyclic quadrilateral ABLC, namely, AB*CL and BL*AC are equal to half the product of the diagonals AL and BC. Can you also prove this result?
(In the original IMO version, the problem was restricted to an acute-angled triangle to limit the cases for students to consider. Also note that students are not provided with a sketch and have to draw their own.)
IMO 2014 Problem 4 - Geometry
Only if stuck, or to compare your proof, click here for one possible proof.
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Modified by Michael de Villiers, 16 July 2014.