19047[EMHL] Re: Harmonic Conic
- Jul 2, 2010
> From: jpehrmfr <jean-pierre.ehrmann@...>I finally realized that I used an incorrect formula for harmonic conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct (BP/PC)/(BQ/QC)=-1 when P,Q are harmonic conjugates wrt B,C. So my results apply to this strange kind of "conjugates," but not to harmonics.
> Dear Barry
> [Barry Wolk]
> > Some computation shows that for any circle that meets
> the 2 sides AB and AC harmonically, its center lies on the
> perpendicular bisector of BC.
> I don't agree with you. Your circles are member of the
> pencil with Poncelet points (circles with radius 0) A and
> the projection of A upon BC.
> Friendly. Jean-Pierre
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