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RE: [EMHL] Re: Orthocenters

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  • Nikolaos Dergiades
    Dear Antreas and Paul, [APH]: Let ABC be a triangle P a point and PaPbPc the pedal triangle of P. Let Ha, Hb, Hc be the orthocenters of the triangles APbPc,
    Message 1 of 6 , Feb 4, 2004
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      Dear Antreas and Paul,

      [APH]: Let ABC be a triangle P a point and PaPbPc the pedal
      triangle of P.

      Let Ha, Hb, Hc be the orthocenters of the triangles
      APbPc, BPcPa, CPaPb, resp.

      Which is the locus of P such that PaPbPc, HaHbHc
      are perspective?

      If the locus is the whole plane, then what is the locus
      of the perspectors as P moves on the Euler line?


      [PY]: It is indeed the whole plane. The two triangles are
      oppositely congruent. If P = (u:v:w) in homogeneous barycentric
      coordinates, then the homothetic center is the point

      (a^2(b^2c^2u + c^2S_C v + b^2S_B w) : ... : ...).

      For P = O, this is the nine-point center, and
      for P = H, this is the center of the Taylor circle X(389).

      ***********
      A little synthetic.
      If Ma, Mb, Mc are the mid points of the sides of the pedal triangle
      and Pg is the centroid of the pedal triangle then Ma is the mid point
      of PHa and the mid point Q of PaHa is the complement of P in the
      pedal triangle or Q is produced from P in the homothety (Pg, -1/2).
      So PaHa, PbHb, PcHc are condurrent at Q and the triangles
      pedal of P and HaHbHc are oppositely congruent.
      For P = O then Pg = G and Q = X(5)
      for P = H then Pg = X(51) and Q = X(389)

      Best regards
      Nikolaos Dergiades
    • xpolakis
      Let ABC be a triangle and A B C , A B C the pedal triangles of two isogonal conjugate points P,P*. Denote: Ha := The orthocenter of A B C Hb := The
      Message 2 of 6 , Apr 1 1:28 AM
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        Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles
        of two isogonal conjugate points P,P*.

        Denote:

        Ha := The orthocenter of A'B"C"

        Hb := The orthocenter of B'C"A"

        Hc := The orthocenter of C'A"B"

        Are the triangles ABC, HaHbHc perspective + parallelogic + orthologic
        for all P's? (If no, then for which P's are they?)

        Perspector, and parallelo/orthologic centers ?

        Antreas
      • jpehrmfr
        Dear Antreas ... ABC and HaHbHc are indirectly similar; hence, they are parallelogic and orthologic (two of the centers are upon the circumcircle of ABC, the
        Message 3 of 6 , Apr 1 2:56 AM
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          Dear Antreas
          > Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles
          > of two isogonal conjugate points P,P*.
          >
          > Denote:
          >
          > Ha := The orthocenter of A'B"C"
          >
          > Hb := The orthocenter of B'C"A"
          >
          > Hc := The orthocenter of C'A"B"
          >
          > Are the triangles ABC, HaHbHc perspective + parallelogic + orthologic
          > for all P's? (If no, then for which P's are they?)
          >
          > Perspector, and parallelo/orthologic centers ?

          ABC and HaHbHc are indirectly similar; hence, they are parallelogic and orthologic (two of the centers are upon the circumcircle of ABC, the two other ones upon the circumcircle of HaHbHc)
          I think that they are perspective only when P lies on the Neuberg cubic.
          Friendly. Jean-Pierre
        • xpolakis
          Let ABC be a triangle, P = (x:y:z) a point and A B C the cevian triangle of P. Denote: Ab := BC / (perpendicular from A to BB ) Ac:= BC / (perpendicular
          Message 4 of 6 , Jun 13, 2009
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            Let ABC be a triangle, P = (x:y:z) a point and A'B'C' the
            cevian triangle of P.

            Denote:

            Ab := BC /\ (perpendicular from A to BB')

            Ac:= BC /\ (perpendicular from A to CC')

            Similarly Bc, Ba and Ca, Cb.

            Let A*, B*, C* be the orthocenters of AAbAc, BBcBa, CCaCb resp.

            ABC, A*B*C* are orthologic by construction.

            Which is the other than H orthologic center?

            If P = I, then the orth. center lies on the circumcircle of ABC.

            Which is the locus of P such that the other orthologic
            center is on the circumcircle of ABC?


            Antreas
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