## 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).

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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.

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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to "A British Mathematical Olympiad Problem and its dual"*

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