An extension of the IMO 2014 Problem 4

Problem 4 of IMO 2014
Problem no. 4 of the IMO 2014 in Cape Town involved 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.

Corollary
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?

Concurrency
Also observe 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).
References
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.

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

Harmonic Quadrilaterals
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.
Harmonic quadrilaterals in turn are special cases of Apollonius quadrilaterals, which are general quadrilaterals with the two products of the lengths of their opposite sides equal.

Some External Related Links
Harmonic quadrilateral (Wikipedia)
Apollonius quadrilateral (Wikipedia)
SA Mathematics Olympiad Questions and worked solutions for past South African Mathematics Olympiad papers can be found at this link.
(Note, however, that prospective users will need to register and log in to be able to view past papers and solutions.)

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Created by Michael de Villiers, 29 July 2014; updated 1 August; 29 Dec 2021; 1 May 2025.

An extension of the IMO 2014 Problem 4

.sketch_canvas { border: medium solid lightgray; display: inline-block; } An extension of the IMO 2014 Problem 4

An extension of the IMO 2014 Problem 4