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Gergonne's point property

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  • alex
    Dear friends! Is the following property of Gergonne s point well-known? ABC is given triangle. A1 is the point on ray BA, so that AA1=BC, and A2 is the point
    Message 1 of 3 , Feb 1, 2003
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      Dear friends!
      Is the following property of Gergonne's point well-known?
      ABC is given triangle.
      A1 is the point on ray BA, so that AA1=BC, and A2 is the point on ray CA, so
      that AA2=BC. In the same way we can define B1,B2, so that BB1=BB2=AC and
      C1C2, so that CC1=CC2=BA.
      Triangle, formed by (A1A2),(B1B2),(C1C2) is perespective with ABC triangle
      and Gergonne point is the center.
      Also that triangle is homothetic to Gergonne triangle of ABC (with the
      ratio...? - I don't know)
      Can somebody give easy synthethical proof of it?
      Best regards,
      Yours sincerely,
      Alex
    • Darij Grinberg <darij_grinberg@web.de>
      Dear Alex, My first remark is that A1, A2, B1, B2, C1, C2 lie on one circle, called the Conway circle of triangle ABC. The center of this circle is the
      Message 2 of 3 , Feb 1, 2003
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        Dear Alex,

        My first remark is that A1, A2, B1, B2, C1, C2 lie on one circle,
        called the Conway circle of triangle ABC. The center of this circle
        is the incenter of ABC.

        See http://mathforum.org/epigone/geometry-puzzles/wilyangdimp/

        What concerns the question why the perspector is the Gergonne point -
        this is a special case of theorems on cevian hexagons, and I can't
        give an elementary argument right off.

        Sincerely, Darij Grinberg

        >> ABC is given triangle.
        >> A1 is the point on ray BA, so that AA1=BC,
        >> and A2 is the point on ray CA, so
        >> that AA2=BC. In the same way we can define
        >> B1,B2, so that BB1=BB2=AC and C1C2, so
        >> that CC1=CC2=BA.
        >> Triangle, formed by (A1A2),(B1B2),(C1C2)
        >> is perespective with ABC triangle and
        >> Gergonne point is the center.
        >> Also that triangle is homothetic to
        >> Gergonne triangle of ABC (with the
        >> ratio...? - I don't know)
      • Emelyanov
        Dear Alex, Let A1 and C1 be on AC, B be the vertix of the intouch triangle of ABC on AC. Then B is the midpoint of A1C1. Let triangle, formed by
        Message 3 of 3 , Feb 3, 2003
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          Dear Alex,
          Let A1 and C1 be on AC, B' be the vertix of the intouch triangle of ABC on
          AC. Then B' is the midpoint of A1C1. Let triangle, formed by
          (A1A2),(B1B2),(C1C2), be A"B"C". Then A1C1 is antiparallel to A"C" wrt
          angle A"B"C". Hence B"B' is the symmedian of A"B"C". Since BB' is the
          symmedian of A'B'C' (intouch), then B, B' and B" are collinear. The
          gomothetic ratio of A"B"C" and A'B'C' is
          ((a+b+c)*(ab+bc+ca)-a^3-b^3-c^3-abc)/(2*(a+b+c)*r).
          Best regards,
          Lev

          >> ABC is given triangle.
          >> A1 is the point on ray BA, so that AA1=BC,
          >> and A2 is the point on ray CA, so
          >> that AA2=BC. In the same way we can define
          >> B1,B2, so that BB1=BB2=AC and C1C2, so
          >> that CC1=CC2=BA.
          >> Triangle, formed by (A1A2),(B1B2),(C1C2)
          >> is perespective with ABC triangle and
          >> Gergonne point is the center.
          >> Also that triangle is homothetic to
          >> Gergonne triangle of ABC (with the
          >> ratio...? - I don't know)





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