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Re: [EMHL] GpP // GH

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  • xpolakis@otenet.gr
    Dear Jean-Pierre, [APH] ... Hmmm... If one tries it analyticaly, he has to work a week, I am afraid! Let P = (x:y:z) in normals. 1st step: Computation of the
    Message 1 of 1 , Aug 31 4:19 PM
      Dear Jean-Pierre,

      [APH]
      >> Let ABC be a triangle and Gp the centroid of the pedal triangle of P.
      >> Which is the locus of P such that the line PGp is parallel to the
      >> Euler line of ABC?

      [JPE]:
      >P -> Gp is affine with K (symedian) as fixed point. Thus, the locus
      >of P such as PGp has a fixed direction is allways a line through K.
      >For P = O, PGp is obviously the Euler line. Thus your locus is the
      >Brocard axis OK.

      Hmmm... If one tries it analyticaly, he has to work a week, I am afraid!

      Let P = (x:y:z) in normals.
      1st step: Computation of the normals of Gp.
      They are (2x + ycosC + zcosB : xcosC + 2y + zcosA : xcosB + ycosA + 2z)

      2nd step: Formulation of the equation of the line PGp. It is:
      (y^2cosA - z^2cosA + xycosB - xzcosC)X +
      (z^2cosB - x^2cosB + yzcosC - yxcosA)Y +
      (x^2cosC - y^2cosC + zxcosA - zycosB)Z = 0

      3rd step: Finding the condition to be the lines:
      line at oo: xsinA + cyclic = 0, Euler line: xsin2Asin(B-C) + cyclic = 0,
      and PGp concurrent.

      They are concurrent iff

      | sinA sinB sinC |
      | |
      | sin2Asin(B-C) sin2Bsin(C-A) sin2Csin(A-B) | = 0
      | |
      | y^2cosA - z^2cosA + xycosB - xzcosC [no space for the rest!]

      .... and no time for more computations!

      [I believe that if one makes them, then will get the line at oo and your OK]


      Thanks !

      Antreas
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