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19047[EMHL] Re: Harmonic Conic

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  • Barry Wolk
    Jul 2, 2010
      > From: jpehrmfr <jean-pierre.ehrmann@...>
      >
      > 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

      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.
      --
      Barry Wolk
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