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Re: New point related to Feuerbach-point

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  • jpehrmfr
    Dear Eric ... and ... triangle ... Your point is the reflection of X(40) (reflection of I wrt O) wrt X(100) (anticomplement of F); he lies on the line
    Message 1 of 8 , Nov 3, 2003
      Dear Eric
      > I wonder if the following property is known
      >
      > -------------------------------------------------
      > Let A', B' and C' be the midpoints of BC, CA and AB
      > Let Ia, Ib and Ic be the A-, B- and C-excenter
      > Let F be the Feuerbach-point
      > Consider the perpendiculars Pa from Ia to FA', Pb from Ib to FB'
      and
      > Pc from Ic to FC'
      >
      > ==> the lines Pa, Pb and Pc are concurrent
      >
      > or in words
      >
      > The perpendiculars from each excenter to the line joining the
      > Feuerbach point with the corresponding vertex of the medial
      triangle
      > are concurrent.

      Your point is the reflection of X(40) (reflection of I wrt O)
      wrt X(100) (anticomplement of F); he lies on the line I-F-NPcenter
      and on the circle (I,2R).
      Friendly. Jean-Pierre
    • Antreas P. Hatzipolakis
      ... Dear Eric Your theorem rephrased: The perpendiculars from each vertex of the antipedal triangle of I to the line joining the Feuerbach point with the
      Message 2 of 8 , Nov 5, 2003
        [Eric Danneels]:

        >The perpendiculars from each excenter to the line joining the
        >Feuerbach point with the corresponding vertex of the medial triangle
        >are concurrent.

        Dear Eric

        Your theorem rephrased:

        The perpendiculars from each vertex of the antipedal triangle of I
        to the line joining the Feuerbach point with the corresponding vertex
        of the pedal triangle of O are concurrent.

        The "dual" (by replacing the points I,O):

        The perpendiculars from each vertex of the antipedal triangle of O
        to the line joining the Feuerbach point with the corresponding vertex
        of the pedal triangle of I are concurrent.

        Is it true??

        Antreas
        --
      • Eric Danneels
        Antreas wrote ... vertex ... vertex ... I don t think so (if I may believe my sketches) I also tried anticevian triangles instead of antipedal triangles
        Message 3 of 8 , Nov 5, 2003
          Antreas wrote

          > Your theorem rephrased:
          >
          > The perpendiculars from each vertex of the antipedal triangle of I
          > to the line joining the Feuerbach point with the corresponding
          vertex
          > of the pedal triangle of O are concurrent.
          >
          > The "dual" (by replacing the points I,O):
          >
          > The perpendiculars from each vertex of the antipedal triangle of O
          > to the line joining the Feuerbach point with the corresponding
          vertex
          > of the pedal triangle of I are concurrent.
          >
          > Is it true??
          >
          ----------------------------

          I don't think so (if I may believe my sketches)

          I also tried anticevian triangles instead of antipedal triangles
          (because IaIbIc is the antipedal and anticevian triangle of I) but
          it doesn't work either

          I also tried replacing one of the excenters by the incenter and
          replacing the Feuerbach point by the tangent point of the
          ninepointcircle and the selected excircle but without success


          PS

          I would like to thank Paul for the coordinates of this point
          and Jean-Pierre for some nice properties of it and I'm sorry for
          the typing errors in the formulae in my original mail

          Greetings from Bruges

          Eric Danneels
        • Antreas P. Hatzipolakis
          ... [...] ... What the extraversions of this problem are? Antreas --
          Message 4 of 8 , Nov 5, 2003
            [ED]:
            >Antreas wrote
            >
            >> Your theorem rephrased:
            >>
            >> The perpendiculars from each vertex of the antipedal triangle of I
            >> to the line joining the Feuerbach point with the corresponding
            >> vertex of the pedal triangle of O are concurrent.
            [...]
            >I also tried replacing one of the excenters by the incenter and
            >replacing the Feuerbach point by the tangent point of the
            >ninepointcircle and the selected excircle but without success

            What the extraversions of this problem are?

            Antreas
            --
          • Eric Danneels
            ... of I ... Sorry, I don t understand your last question Meanwhile I discovered that using my original statement we get a new concurrency if we replace the
            Message 5 of 8 , Nov 5, 2003
              --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"
              <xpolakis@o...> wrote:
              > [ED]:
              > >Antreas wrote
              > >
              > >> Your theorem rephrased:
              > >>
              > >> The perpendiculars from each vertex of the antipedal triangle
              of I
              > >> to the line joining the Feuerbach point with the corresponding
              > >> vertex of the pedal triangle of O are concurrent.
              > [...]
              > >I also tried replacing one of the excenters by the incenter and
              > >replacing the Feuerbach point by the tangent point of the
              > >ninepointcircle and the selected excircle but without success
              >
              > What the extraversions of this problem are?
              >
              -------------------
              Sorry, I don't understand your last question

              Meanwhile I discovered that using my original statement we get a new
              concurrency if we replace the excenters by the vertices of the pedal
              triangle of I

              so

              The perpendiculars from the vertices of the intouch triangle on the
              lines joining the Feuerbach-point with the corresponding vertices of
              the medial triangle are concurrent


              Greetings from Bruges

              Eric Danneels
            • Antreas P. Hatzipolakis
              ... Extraversists (JHC et al) understand. JHC? Antreas --
              Message 6 of 8 , Nov 5, 2003
                >> [ED]:
                >> >Antreas wrote
                >> >
                >> >> Your theorem rephrased:
                >> >>
                >> >> The perpendiculars from each vertex of the antipedal triangle
                >of I
                >> >> to the line joining the Feuerbach point with the corresponding
                >> >> vertex of the pedal triangle of O are concurrent.
                >> [...]
                >> >I also tried replacing one of the excenters by the incenter and
                >> >replacing the Feuerbach point by the tangent point of the
                >> >ninepointcircle and the selected excircle but without success
                >>
                >> What the extraversions of this problem are?
                >>
                >-------------------
                >Sorry, I don't understand your last question


                Extraversists (JHC et al) understand.

                JHC?

                Antreas



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