British Mathematical Olympiad problem: Conic Generalization

If a hexagon ABCDEF is circumscribed around a conic and the midpoints P, Q and R of the alternate sides AB, CD & EF touch the conic, and the other three points where the remaining sides of the hexagon touches the conic are labelled X, Y and Z as shown, then the lines PY, XR & QZ connecting opposite tangential points are concurrent at the Brianchon point.

Corollary 1: For the hexagon above, sin A * sin C * sin E = sin B * sin D * sin F.

Corollary 2: The hexagon ABCDEF is inscribed in a conic.

(Drag any of the 'bright' red points).


Please enable Java for an interactive construction (with Cinderella).


Unfortunately the above Cinderella Java applet no longer runs on some newer browsers, so below is a picture illustrating the result for those who can't see the applet loading.

BMO conic general

Download the dynamic geometry software Cinderella 2 for FREE from here, and use it (after unzipping) to view & manipulate the Cinderella BMO conic general dynamic sketch illustrating the above result.

Challenge: Can you explain why (prove) the result is true? If stuck, have a look at the 2005 paper of my friend & colleague, Michael Fox, from the United Kingdom at Proof of BMO conic general which provides not only a proof, but a further generalization. A dynamic sketch by Michael Fox that illustrates the proof and generalization is at Michael Fox theorem.

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Created by Michael de Villiers, 30 Oct 2010 with Cinderella; updated 7 May 2021.