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Re: [EMHL] Re: incenters & squares

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  • Paul Y. Yiu
    Dear Bernard and Jean-Pierre, [BG] Let P be a point on the circumcircle of triangle ABC with incenter I denote by Pa, Pb, Pc the incenters of triangles PBC,
    Message 1 of 2 , Dec 3 9:12 AM
      Dear Bernard and Jean-Pierre,

      [BG] Let P be a point on the circumcircle of triangle ABC with incenter
      I denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB. I,
      Pa, Pb, Pc are the vertices of a rectangle. Find P such that the
      rectangle becomes a square.

      [JPE]: Suppose that P lies on the arc BC of the circumcircle opposite
      to A. The lines PaPb and PaPc go respectively through the excenters Ib
      and I. The rectangle will be a square if PaI is bisector of IbPaIc.
      Hence the construction : Pa lies on the circle (Ca) of diameter IbIc
      (center Wa) such that angle IbPaI = angle IPaIc = 45°. P lies on the
      line WaPa. Do the same for P lying on the arcs CA or AB of the
      circumcircle.

      *** Thanks for an excellent problem and a beautiful solution.
      I tried to follow up to see if PaPbPc is the circumcevian triangle of
      some point. Unfortunately it is not. Pa has homogeneous barycentric
      coordinates

      -a^2/( (b+c)(c+a-b)(a+b-c) + 2aS)
      : b^2/((c+a-b)(b(a+b-c)+S))
      : c^2/((a+b-c)(c(c+a-b)+S))

      From this we get an alternative description of Pa: it is the isogonal
      conjugate of the infinite point of the line joining the Nagel point to
      the A-trace of the isotomic conjugate of the inner Soddy center.

      Best regards
      Sincerely
      Paul
    • Bernard Gibert
      Dear Jean-Pierre and Paul, ... I arrived to the same conclusions with the use of rotated circles about I, angle +/-90°. I have found that the triangle with
      Message 2 of 2 , Dec 3 9:48 AM
        Dear Jean-Pierre and Paul,

        > [BG] Let P be a point on the circumcircle of triangle ABC with
        > incenter
        > I denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB. I,
        > Pa, Pb, Pc are the vertices of a rectangle. Find P such that the
        > rectangle becomes a square.
        >
        > [JPE]: Suppose that P lies on the arc BC of the circumcircle opposite
        > to A. The lines PaPb and PaPc go respectively through the excenters Ib
        > and I. The rectangle will be a square if PaI is bisector of IbPaIc.
        > Hence the construction : Pa lies on the circle (Ca) of diameter IbIc
        > (center Wa) such that angle IbPaI = angle IPaIc = 45°. P lies on the
        > line WaPa. Do the same for P lying on the arcs CA or AB of the
        > circumcircle.
        >
        > [PY] I tried to follow up to see if PaPbPc is the circumcevian
        > triangle of
        > some point. Unfortunately it is not. Pa has homogeneous barycentric
        > coordinates
        >
        > -a^2/( (b+c)(c+a-b)(a+b-c) + 2aS)
        > : b^2/((c+a-b)(b(a+b-c)+S))
        > : c^2/((a+b-c)(c(c+a-b)+S))
        >
        > From this we get an alternative description of Pa: it is the isogonal
        > conjugate of the infinite point of the line joining the Nagel point to
        > the A-trace of the isotomic conjugate of the inner Soddy center.

        I arrived to the same conclusions with the use of rotated circles
        about I, angle +/-90°.

        I have found that the triangle with vertices the 3 points Pa (or the
        centers of the squares ) is similar to the pedal triangle of X1124,
        hence a relation with the Paasche point and inscribed congruent circles.

        I didn't find any interesting perspectivity yet.

        Best regards

        Bernard



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