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Extra Summer Laboratory of triangles

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  • Ricardo Barroso
    Friends of Hyacinthos: Four problems are offered  in  http://personal.us.es/rbarroso/trianguloscabri/ One of the inradius and circumradius, of Vcente Vicario
    Message 1 of 10 , Jun 30, 2010
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      Friends of Hyacinthos:

      Four problems are offered 

      in

       http://personal.us.es/rbarroso/trianguloscabri/

      One of the inradius and
      circumradius, of Vcente Vicario

      another on the
      equilateral, proposed for Ricard Peiró

      and two of Paul Halmos
      Greetings from
      Sevilla

      Ricardo
      Barroso
















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    • Alexey Zaslavsky
      Dear Hauke, Barry and Jean-Pierre! ... I don t agree with you. Your circles are member of the pencil with Poncelet points (circles with radius 0) A and the
      Message 2 of 10 , Jul 2, 2010
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        Dear Hauke, Barry and Jean-Pierre!

        > Some computation shows that for any circle that meets the 2 sides AB and AC harmonically, its center lies on the perpendicular bisector of BC.

        I don't agree with you. Your circles are member of the pencil with Poncelet points (circles with radius 0) A and the projection of A upon BC.

        The conic meets AB harmonically iff A lyes on the polar of B. Thus triangle ABC is autopolar wrt Hauke's conic. But in this case all six common points of this conic and the sidelines can't be real.

        Sincerely Alexey

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      • Barry Wolk
        ... I finally realized that I used an incorrect formula for harmonic conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct (BP/PC)/(BQ/QC)=-1 when P,Q
        Message 3 of 10 , Jul 2, 2010
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          > From: jpehrmfr <jean-pierre.ehrmann@...>
          >
          > Dear Barry
          > [Barry Wolk]
          > > Some computation shows that for any circle that meets
          > the 2 sides AB and AC harmonically, its center lies on the
          > perpendicular bisector of BC.
          >
          > I don't agree with you. Your circles are member of the
          > pencil with Poncelet points (circles with radius 0) A and
          > the projection of A upon BC.
          > Friendly. Jean-Pierre

          I finally realized that I used an incorrect formula for harmonic conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct (BP/PC)/(BQ/QC)=-1 when P,Q are harmonic conjugates wrt B,C. So my results apply to this strange kind of "conjugates," but not to harmonics.
          --
          Barry Wolk
        • Eisso Atzema
          Dear Hyacinthians, Finally catching up on my mail. At the risk of having missed part of the discussion, I only partially agree with Alexey (if I understand him
          Message 4 of 10 , Jul 19, 2010
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            Dear Hyacinthians,

            Finally catching up on my mail. At the risk of having missed part of the
            discussion, I only partially agree with Alexey (if I understand him
            correctly). Using his observation that ABC is autopolar with respect to
            Hauke's conic, it is a straightforward exercise in barycentric calculus
            to show that any circle with the property that Hauke is looking for
            always has its center at the orthocenter of ABC. Also, with a little
            more work it can be shown that the radius of such a circle has to be
            equal to 2*R*sqrt(-cos(aA)cos(aB)cos(aC)), with R the radius of the
            circumcircle of ABC and aA, aB, aC the angles at A, B, C. It follows
            that there is no such (real) circle for acute triangles and exactly one
            for obtuse triangles.

            Eisso

            On 7/2/10 3:42 AM, Alexey Zaslavsky wrote:
            >
            > Dear Hauke, Barry and Jean-Pierre!
            >
            > > Some computation shows that for any circle that meets the 2 sides AB
            > and AC harmonically, its center lies on the perpendicular bisector of BC.
            >
            > I don't agree with you. Your circles are member of the pencil with
            > Poncelet points (circles with radius 0) A and the projection of A upon BC.
            >
            > The conic meets AB harmonically iff A lyes on the polar of B. Thus
            > triangle ABC is autopolar wrt Hauke's conic. But in this case all six
            > common points of this conic and the sidelines can't be real.
            >
            > Sincerely Alexey
            >
            >




            --

            ========================================
            Eisso J. Atzema, Ph.D.
            Department of Mathematics& Statistics
            University of Maine
            Orono, ME 04469
            Tel.: (207) 581-3928 (office)
            (207) 866-3871 (home)
            Fax.: (207) 581-3902
            E-mail: atzema@...
            ========================================



            [Non-text portions of this message have been removed]
          • Francois Rideau
            I just remember the Faure theorem. A triangle ABC is autopolar wrt the conic Gamma iff the ABC-circumcircle is orthogonal to the orthoptic circle of Gamma.
            Message 5 of 10 , Jul 29, 2010
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              I just remember the Faure theorem.
              A triangle ABC is autopolar wrt the conic Gamma iff the ABC-circumcircle is
              orthogonal to the orthoptic circle of Gamma.
              Friendly
              Francois

              On Fri, Jul 2, 2010 at 8:15 PM, Barry Wolk <wolkbarry@...> wrote:

              >
              >
              > > From: jpehrmfr <jean-pierre.ehrmann@...<jean-pierre.ehrmann%40wanadoo.fr>
              > >
              >
              > >
              > > Dear Barry
              > > [Barry Wolk]
              > > > Some computation shows that for any circle that meets
              > > the 2 sides AB and AC harmonically, its center lies on the
              > > perpendicular bisector of BC.
              > >
              > > I don't agree with you. Your circles are member of the
              > > pencil with Poncelet points (circles with radius 0) A and
              > > the projection of A upon BC.
              > > Friendly. Jean-Pierre
              >
              > I finally realized that I used an incorrect formula for harmonic
              > conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct
              > (BP/PC)/(BQ/QC)=-1 when P,Q are harmonic conjugates wrt B,C. So my results
              > apply to this strange kind of "conjugates," but not to harmonics.
              > --
              > Barry Wolk
              >
              >
              >


              [Non-text portions of this message have been removed]
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