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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.

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Created by Michael de Villiers, 16 July 2014; modified & updated to *WebSketchpad* on 4 July 2021; updated 29 Dec 2021.