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9173Re: [EMHL] Orthologic Triangles

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  • jpehrmfr
    Feb 1, 2004
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      Dear Fred, Darij and Antreas
      > Dear Jean-Pierre, Antreas and Darij
      > > [FL]
      > > > Let ABC be a triangle, P a point, and (Oa, Ob, Oc)
      > > > the Circumcenters of the triangles PBC, PCA, PAB, resp.
      > > >
      > > > The locus of P such that the perpendiculars
      > > > from A,B,C to POa, POb, POc are concurrent is:
      > > > infinity line*circumcircle*KjP cubic.
      > > [JPE]
      > > Consider three lines La, Lb, Lc going respectively through A, B,
      C
      > > and such as (La,POa) = (Lb,POb) = (Lc,POc) = phi where (L1,L2)
      is
      > > the measure modulo Pi of the oriented angle of lines L1, L2.
      > > Then La, Lb, Lc concur if and only if
      > > (AP,BC)+(BP,CA)+(CP,AB) = Pi/2-phi
      > > Thus, we get the MacCay cubic for phi = 0, the Kjp cubic for phi
      =
      > > Pi/2 and, for any theta, a member of the pencil generated by Kjp
      and
      > > MacCay.
      > > Friendly. Jean-Pierre
      > [FL]
      > That's fine.
      > A link between nK0 and pK.
      >
      > I have the pencil MacCay -4s tan(phi) Kjp.
      >
      > This cubic is an isocubic iff phi = 0 or phi = Pi/2
      >
      > Remark: MacCay and Kjp are isogonal cubic but this cubic is not
      > (if phi non 0 or Pi/2).
      >
      > Question (probably difficult) to JPE: Locus of perspector?

      This is the same cubic with -phi instead of phi - ie the isogonal
      conjugate of the locus of P -.
      Of course, in the case of Mac Cay or Kjp, the locus of the common
      point of La, Lb, Lc and the locus of P are the same ones.
      Friendly. Jean-Pierre
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