**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* = ½*AL⋅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).

*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 2*n*-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.