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[EMHL] Re: A Triangle Hexagon

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  • jpehrmfr
    Dear Antreas ... [JPE] ... at ... As the circumcenter of OaObOc is obviously the NPcenter of ABC, the rectangular hyperbola above should be the Jerabek
    Message 1 of 7 , Jan 18, 2004
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      Dear Antreas
      > [APH]
      > > Parametrization:
      > >
      > > Let ABC be a triangle, H, O its orthocenter, circumcenter, resp.
      > > and Pa,Pb,Pc points on AH,BH,CH, resp. such that
      > >
      > > APa/ AH = BPa / BH = CPa / CH = t
      > >
      > > Fa, Fb, Fc = the orthogonal projections of Pa,Pb,Pc on OA,OB,OC
      > resp.
      > >
      > > Oa, Ob, Oc = the orthogonal projections of O on HA,HB,HC resp.
      > >
      > > Are the lines FaOa, FbOb, FcOc concurrent for every t?
      > >
      > > If yes, the which is the locus of the point of concurrence, as
      > > t varies?
      [JPE]
      > Yes they allways concur; after some computations, it appears that
      > the locus of the common point is a rectangular hyperbola centered
      at
      > the center of the Taylor circle. The hyperbola goes through H, the
      > NPCenter, the Kosnita point X(54), X(185), X(1147)...

      As the circumcenter of OaObOc is obviously the NPcenter of ABC, the
      rectangular hyperbola above should be the Jerabek hyperbola of
      OaObOc.
      Friendly. Jean-Pierre
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