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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".
Investigate: Drag any of the vertices of ABC.
Challenge: Can you prove the result? Can you prove it in several different ways?
Further Challenge: 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 IMO students are not provided with a sketch and have to draw their own.)
IMO 2014 Problem 4 - Geometry
Solution: Only if stuck, or to compare your proof, click here for one possible proof, as well as download links to two PDF files with more proofs.
Further Generalization
1) Can you generalize the result further to hexagons, etc. by using your observation 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?
2) Think about it before going to: IMO 2014 extension to compare.
Note
A cyclic quadrilateral with the products of its opposite sides equal as in the case above is called a harmonic quadrilateral. Read more about these quadrilaterals, their properties & applications to problems in this paper, About the Harmonic Quadrilateral, presented at a conference in Bulgaria in 2012 by Truong, Khanh & Quang from Vietnam.
Created by Michael de Villiers, 16 July 2014; modified & updated to WebSketchpad on 4 July 2021; updated 29 Dec 2021; 3 Feb 2023.