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New property of X(1358) + a new related point

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  • Eric Danneels
    Dear Hyacinthians, X(1358) is defined in the current ETC as the Brisse-transform of X(101). I would like to present a new remarkable property of this point Let
    Message 1 of 4 , Jul 1, 2004
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      Dear Hyacinthians,


      X(1358) is defined in the current ETC as the Brisse-transform
      of X(101).

      I would like to present a new remarkable property of this point

      Let A'B'C' be the medial triangle of triangle ABC.
      Consider the circle, different from the ninepointcircle, through B'
      and C' touching the incircle. Let A* be the point where both circles
      touch. Define B* and C* similarly.

      X(1358) is the perspector of the triangles ABC and A*B*C*
      Its barycentrics are ( (b-c)^2/(b+c-a) : cyclic.....)

      If we replace the medial triangle by the orthic triangle we get a
      new point S with similar barycentrics

      S = ( (b-c)^2.(b+c-a)^3 : cyclic.....)

      Greetings from Bruges

      Eric Danneels
    • Antreas P. Hatzipolakis
      ... Dear Eric How about if we replace the medial triangle with the Euler triangle? (ie the triangle formed with the midpoints of AH, BH, CH) ? Also, how about
      Message 2 of 4 , Jul 1, 2004
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        [Eric Danneels]:
        >X(1358) is defined in the current ETC as the Brisse-transform
        >of X(101).
        >
        >I would like to present a new remarkable property of this point
        >
        >Let A'B'C' be the medial triangle of triangle ABC.
        >Consider the circle, different from the ninepointcircle, through B'
        >and C' touching the incircle. Let A* be the point where both circles
        >touch. Define B* and C* similarly.
        >
        >X(1358) is the perspector of the triangles ABC and A*B*C*
        >Its barycentrics are ( (b-c)^2/(b+c-a) : cyclic.....)
        >
        >If we replace the medial triangle by the orthic triangle we get a
        >new point S with similar barycentrics
        >
        >S = ( (b-c)^2.(b+c-a)^3 : cyclic.....)

        Dear Eric

        How about if we replace the medial triangle with the Euler triangle?
        (ie the triangle formed with the midpoints of AH, BH, CH) ?

        Also, how about if we replace the incircle with the three excircles?


        Antreas (waiting the great victory of Greece !)


        --
      • Eric Danneels
        Dear Antreas, you wrote ... triangle? ... My sketches show that the Euler triangle does not produce a perspector. However I think I ve found another
        Message 3 of 4 , Jul 3, 2004
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          Dear Antreas,

          you wrote

          > How about if we replace the medial triangle with the Euler
          triangle?
          > (ie the triangle formed with the midpoints of AH, BH, CH) ?

          My sketches show that the Euler triangle does not produce a
          perspector.

          However I think I've found another generalization leading to a new
          perspector.

          Let A'B'C' be the cevian triangle of a point P wrt triangle ABC.
          Through B' and C' we can draw 2 circles tangent to the incircle.
          Let A1 and A2 be the touchpoints.
          Similarly define the touchpoints B1-B2, and C1-C2 corresponding to
          C'A' and A'B'.
          It seems that A1A2, B1B2 and C1C2 go through 1 point.
          I could not find a proof of it yet.

          For the orthic triangle and the medial triangle this new perspector
          if of course the Feuerbach point.

          Greetings from Bruges & good luck on sunday

          Eric Danneels
        • Antreas P. Hatzipolakis
          Dear Eric ... Thank you. Your wishes are very much appreciated. After your nice problems with incircle, I have this locus problem: Let ABC be a triangle, P, P*
          Message 4 of 4 , Jul 3, 2004
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            Dear Eric

            >Greetings from Bruges & good luck on sunday
            >
            >Eric Danneels

            Thank you. Your wishes are very much appreciated.


            After your nice problems with incircle,
            I have this locus problem:

            Let ABC be a triangle, P, P* two isogonal points, and
            A'B'C', A"B"C" their pedal triangles.

            Consider the [finite] circle passing through A',A"
            and touching the incircle at A*.

            Similarly B*,C*.

            Which is the locus of P such that ABC, A*B*C* are perspective?


            Greetings from Athens

            Antreas


            SHKWSE TO TO G...MENO
            DEN MPORW NA PERIMENW !
            --
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