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).
Unfortunately the above 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.
Can you explain why (prove) the result is true? If stuck, have a look at Michael Fox's paper 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 Proof of BMO conic general.
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Michael de Villiers, 30 Oct 2010.