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Re: Pedal triangles

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  • jpehrmfr
    Dear Francois ... P recur ... the third ... isometric ... ratio k ... If I don t mistake,if P is inside ABC, the third pedal triangle cannot be isometric
    Message 1 of 37 , Nov 11, 2007
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      Dear Francois
      > Dear Bernard and dear friends
      > It is known that the shapes of repeated triangles wrt a fixed point
      P recur
      > with period 3 and the Mac Cay cubic is the locus for P for which
      the third
      > pedal triangle is homothetic to the reference triangle.
      > Do you know the locus of P for which this third pedal triangle is
      isometric
      > with the reference triangle or more generally similar with a given
      ratio k >
      > 0?
      > I think this is some sextic?

      If I don't mistake,if P is inside ABC, the third pedal triangle
      cannot be isometric with the reference triangle because the ratio of
      similarity k is <=1/8.
      In fact k = (PPa.PPb.PPc)/(PA.PB.PC) ( with PaPbPc = pedal triangle
      of P)and the Oppenheim inequality says that PA.PB.PC >=(Pb+Pc)(Pc+Pa)
      (Pa+Pb) and
      (PPb+PPc)(PPc+PPa)(PPa+PPb)>= 8.PPa.PPb.PPc
      Friendly. Jean-Pierre
    • Nikolaos Dergiades
      Dear Jeff, Yes. And the points A1, B1, C1 are found in this case as previously but by negative rotations. Have a Happy New Year. Best regards Nikos Dergiades
      Message 37 of 37 , Dec 30, 2008
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        Dear Jeff,
        Yes.
        And the points A1, B1, C1
        are found in this case as previously
        but by negative rotations.
        Have a Happy New Year.
        Best regards
        Nikos Dergiades

        > Dear Nikos and Francois,
        >
        > So, if instead of the triangle A'B'C', we use a
        > triangle inversely
        > similar to A'B'C', then we find the pedal
        > triangle of the point P*
        > inverse of P in the ABC circumcircle?
        >
        > Sincerely, Jeff
        >
        >
        > [ND]
        > > Dear Francois,
        > > I think you mean if we construct the point C1
        > > as we constructed the points A1, B1,
        > > if the reflections
        > > of C in A1B1,
        > > of A in B1C1
        > > of B in C1A1
        > > coincide?
        > > Yes.
        > >
        > > Best regards
        > > Nikos Dergiades
        > >
        > > > Dear Nikos
        > > > Thank you very much for your help.
        > > > Now, I will think about your construction and
        > compare it
        > > > with mines.
        > > > Have you already look at the configuration you
        > get when you
        > > > perform your
        > > > construction with other vertices?
        > > > Friendly
        > > > Francois
        > > >
        > > > 2008/11/24 Nikolaos Dergiades
        > <ndergiades@...>
        > > >
        > > > > Dear Francois,
        > > > > you are right.
        > > > > A simpler construction is the following:
        > > > >
        > > > > Let O be the circumcenter of ABC.
        > > > > The perpendicular from O to BC meets the
        > > > > rotation of line CO about C by the signed
        > angle
        > > > > B'A'C' at the point A1.
        > Similarly
        > > > > the perpendicular from O to CA meets the
        > > > > rotation of line AO about A by the signed
        > angle
        > > > > C'B'A' at the point B1.
        > > > > The reflection of C in A1B1 is the required
        > point P.
        > > > > Best regards
        > > > > Nikos Dergiades
        > > > >
        >
        >
        >
        > ------------------------------------
        >
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        >
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        >


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