An extension of the IMO 2014 Problem 4

imo logo 2014

Problem no. 4 of the IMO 2014 in Cape Town involves the following construction and associated theorem:
"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. Then the intersection L of the lines BM and CN lies on the circumcircle ABC". Go here for a dynamic sketch of the problem and links to several proofs.

An interesting corollary is that the products of the opposite (alternate) sides of the cyclic quadrilateral ABLC are equal, and hence according to Ptolemy’s theorem, equal to half the product of the diagonals. Thus, AB⋅CL = BL⋅AC = ½AL⋅BC.

Extension to cyclic hexagon
If points N and M are constructed on sides AB and AC, respectively, in the same way as L as shown below, we analogously obtain that the two products of the alternate sides of the cyclic hexagon ANBLCM are equal, as well as 1/8 of the product of the main diagonals. In other words, NB⋅LC⋅MA = BL⋅CM⋅AN = AL⋅BM⋅CN/8.

An extension of the IMO 2014 Problem 4

Explore: Drag either of the red vertices A or B to explore dynamically.

Challenge: Can you prove the hexagon result above?

Observation: Also note that the main diagonals AL, CN and BM are concurrent. This result follows immediately from the lovely theorem by Cartensen (2000-2001): The main diagonals of a cyclic hexagon are concurrent, if and only if, the two products of alternate sides are equal; i.e. NB⋅LC⋅MA = BL⋅CM⋅AN. This theorem also appears in Gardiner & Bradley (2002, p. 96; 99) and also on the Math Stack Exchange (2013).
Cartensen, J. (2000-2001). About hexagons. Mathematical Spectrum, 33(2), pp. 37–40.
Gardiner, A.D. & Bradley, C.J. (2005). Plane Euclidean Geometry: Theory and Problems. Leeds, University of Leeds: The United Kingdom Mathematics Trust.
Math Stack Exchange. (2013). Accessed on 31 July 2021 at: diagonals of cyclic hexagon.

Extension to cyclic octagon
Click on the 'Link to Octagon' button. If we now similarly construct points O and P, respectively in relation to triangles BAN and BCL and on opposite arcs AN and LC as shown, we obtain: AO⋅NB⋅LP⋅CM = ON⋅BL⋅PC⋅MA = OP⋅NC⋅BM⋅LA/64.

Extension to cyclic decagon
Click on the 'Link to Decagon' button. If we similarly construct points Q and R respectively on opposite arcs AO and PL of the preceding octagon as shown, we obtain: AQ⋅ON⋅BL⋅RP⋅CM = QO⋅NB⋅LR⋅PC⋅MA = QR⋅OP⋅NC⋅BM⋅LA/1152AQ.

Continuing constructions in this way, the result can be generalized to cyclic 2n-gons in different ways. The reader is invited to also explore other variations of these interesting constructions.

Read my Dec 2021 paper in the Learning & Teaching Mathematics journal at: An Extension of an IMO 2014 Geometry Problem .

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Created by Michael de Villiers, 29 July 2014; updated 1 August & 29 Dec 2021.