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Re: [EMHL] Re: Pediac triangles

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  • xpolakis@otenet.gr
    Dear Floor and Jean-Pierre, ... Aren t these triangles what I called P-ac triangles (as generalizations of H-ac = orthiac triangles)? ... (Re: [EMHL]
    Message 1 of 3 , Sep 3, 2001
      Dear Floor and Jean-Pierre,

      [FvL]:
      >> Let P be a point, A'B'C' its pedal triangles. From A' drop
      >> perpendicular
      >> altitudes to the sides AC and AB, ending in Ab and Ac respectively.
      >> Then we could call A'AbAc pediac and AAbAc synpediac triangles.

      Aren't these triangles what I called P-ac triangles (as generalizations
      of H-ac = orthiac triangles)?

      | Let's now replace H with another point P:
      | Let ABC be a triangle and A'B'C' the pedal triangle of P.
      | We define the P-ac triangles A'AbAc, B'BcBa, C'CaCb
      | and Syn-P-ac triangles AAbAc, BBcBa, CCaCb.
      (Re: [EMHL] Synorthiac triangles)

      If they are the same, then I prefer Floor's term Pediac than the
      abbreviated mine (P-ac).

      [JPE]:
      >Here is a little problem related to those triangles :
      >Construct P such as the three circumcircles of the pediac triangles
      >of P (or synpediac triangles, they are the same ones) have the same
      >radius.
      >In other word, construct P such as AA' = BB' = CC'.

      I remember this variation:

      Draw three CONCURRENT cevians AA', BB', CC' such that
      AA' = BB' = CC'.

      I will look at F.G.-M., and give the precise reference at
      the end of this message.

      [JPE]:
      >I suppose that the solution this problem is well known, but not by me.
      >I'm pretty sure that there are exactly two solutions P, P', that H is
      >the midpoint of PP' and that the line PP' is parallel to the major
      >axis of a Steiner ellipse (thus this line is the Steiner line of a
      >common point of the circumcircle and the line OK).
      >Starting from that, it should not be very difficult to finish.

      I found the reference:

      On peut se proposer de determiner trois ceviennes concourantes qui
      aient meme longueur totale.

      Si F est leur point de concours, ce point est le foyer de la conique
      circonscrite au triangle donne, et qui a pour centre de centre de
      gravite du triangle (L. BICKART, I. M., 1910, p. 170, no 3724) Voir comme
      reponse, p. 256 a 260, art. par MM. WELSCH, E. MALO, etc.

      F.G.-M.: Exercices de geometrie. 5e ed. (1912), paragraph 1242. l, #4


      Antreas
    • Floor van Lamoen
      Dear Antreas, ... [APH] ... [APH] ... Yes, they are the same, but I didn t seem to notice that. Let s call them Pediac and SynPediac from now. Kind regards,
      Message 2 of 3 , Sep 3, 2001
        Dear Antreas,

        > [FvL]:
        > >> Let P be a point, A'B'C' its pedal triangles. From A' drop
        > >> perpendicular
        > >> altitudes to the sides AC and AB, ending in Ab and Ac respectively.
        > >> Then we could call A'AbAc pediac and AAbAc synpediac triangles.

        [APH]
        > Aren't these triangles what I called P-ac triangles (as generalizations
        > of H-ac = orthiac triangles)?

        [Quote of APH]:
        > | Let's now replace H with another point P:
        > | Let ABC be a triangle and A'B'C' the pedal triangle of P.
        > | We define the P-ac triangles A'AbAc, B'BcBa, C'CaCb
        > | and Syn-P-ac triangles AAbAc, BBcBa, CCaCb.
        > (Re: [EMHL] Synorthiac triangles)

        [APH]
        > If they are the same, then I prefer Floor's term Pediac than the
        > abbreviated mine (P-ac).

        Yes, they are the same, but I didn't seem to notice that. Let's call
        them Pediac and SynPediac from now.

        Kind regards,
        Sincerely,
        Floor.
      • jean-pierre.ehrmann@wanadoo.fr
        Dear Antreas and other Hyacinthists, ... (where A B C is the pedal triangle of P) ... me. ... is ... [APH] ... comme ... #4 Translation : There are two points
        Message 3 of 3 , Sep 3, 2001
          Dear Antreas and other Hyacinthists,

          > [JPE]:
          > >Here is a little problem related to those triangles :
          > >Construct P such as the three circumcircles of the pediac triangles
          > >of P (or synpediac triangles, they are the same ones) have the same
          > >radius.
          > >In other word, construct P such as AA' = BB' = CC'.
          (where A'B'C' is the pedal triangle of P)
          > >I suppose that the solution this problem is well known, but not by
          me.
          > >I'm pretty sure that there are exactly two solutions P, P', that H
          is
          > >the midpoint of PP' and that the line PP' is parallel to the major
          > >axis of a Steiner ellipse (thus this line is the Steiner line of a
          > >common point of the circumcircle and the line OK).
          > >Starting from that, it should not be very difficult to finish.

          [APH]
          > I found the reference:
          >
          > On peut se proposer de determiner trois ceviennes concourantes qui
          > aient meme longueur totale.
          >
          > Si F est leur point de concours, ce point est le foyer de la conique
          > circonscrite au triangle donne, et qui a pour centre de centre de
          > gravite du triangle (L. BICKART, I. M., 1910, p. 170, no 3724) Voir
          comme
          > reponse, p. 256 a 260, art. par MM. WELSCH, E. MALO, etc.
          >
          > F.G.-M.: Exercices de geometrie. 5e ed. (1912), paragraph 1242. l,
          #4

          Translation : There are two points with equal cevians : the focii F,
          F' of the Steiner circumellipse.
          Many thanks, Antreas for the reference.

          So, there are two points P such as AA' = BB' = CC' where A'B'C' is
          the pedal triangle of P; those points are homothetic of F, F' in the
          homothecy (L, 3/2) - L = de Longchamps point - and, of course, we can
          now construct those points.

          Friendly. Jean-Pierre
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