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Variation on the Van Lamoen-Grinberg-Wolk-transform

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  • efn4900
    Dear Hyacynthians, looking for variations of the Van Lamoen-Grinberg-Wolk-transform I discovered the following properties: Consider a point P and a triangle
    Message 1 of 2 , Jun 1, 2003
      Dear Hyacynthians,

      looking for variations of the Van Lamoen-Grinberg-Wolk-transform I
      discovered the following properties:

      Consider a point P and a triangle ABC.
      Let A'B'C' be the circumcevian triangle of P with respect to ABC.
      Let A°, B° and C° be the circumcenters of the triangles PB'C', PC'A'
      and PA'B'.

      ==> The lines AA°, BB° and CC° pass through a point S on the
      circumcircle of ABC.


      I found the following relationships using the ETC

      P = Incenter X(1) ==> S = X(104)
      P = Centroid X(2) ==> S = X(98)
      P = Circumcenter X(3) ==> S = X(74)
      P = Orthocenter X(4) ==> S = undetermined
      P = Euler-center X(5) ==> S = X(1141)
      P = Lemoine point X(6) ==> S = X(74)
      P = Isogonal conjugate of Euler-center X(54) ==> S = X(74)


      If we consider ABC as the circumcevian triangle of P with respect to
      A'B'C' we have

      Let A*, B* and C* be the circumcenters of the triangles PBC, PCA and
      PAB.

      ==> The lines A'A*, B'B* and C'C* pass through a point T on the
      circumcircle of ABC,

      I found the following using the ETC

      P = Incenter X(1) ==> T = undetermined
      P = Centroid X(2) ==> T = X(1296)
      P = Circumcenter X(3) ==> T = X(110)
      P = Orthocenter X(4) ==> T = X(110)
      P = Euler-center X(5) ==> T = X(1291)
      P = Lemoine point X(6) ==> T = X(1296)
      P = Isogonal conjugate of Euler-center X(54) ==> T = X(1291)

      It seems that isogonal conjugates map to the same point.

      Are these properties known ?

      Greetings from Belgium

      Eric Danneels
    • Darij Grinberg
      Dear Eric Danneels, ... Do you have a synthetic proof? The property was established by Paul Yiu with barycentrics in Hyacinthos message #3957, but I think that
      Message 2 of 2 , Jun 1, 2003
        Dear Eric Danneels,

        You wrote:

        >> Consider a point P and a triangle ABC.
        >> Let A'B'C' be the circumcevian triangle of P
        >> with respect to ABC.
        >> Let A°, B° and C° be the circumcenters of
        >> the triangles PB'C', PC'A' and PA'B'.
        >>
        >> ==> The lines AA°, BB° and CC° pass through
        >> a point S on the circumcircle of ABC.

        Do you have a synthetic proof? The property was
        established by Paul Yiu with barycentrics in
        Hyacinthos message #3957, but I think that there
        must be some easy proof.

        >> If we consider ABC as the circumcevian
        >> triangle of P with respect to A'B'C' we have
        >>
        >> Let A*, B* and C* be the circumcenters of the
        >> triangles PBC, PCA and PAB.
        >>
        >> ==> The lines A'A*, B'B* and C'C* pass
        >> through a point T on the circumcircle of ABC,

        >> It seems that isogonal conjugates map to the
        >> same point.

        This is very interesting.

        Probably, an idea to prove that the lines A'A*,
        B'B* and C'C* concur at a point on the
        circumcircle of ABC analytically but without
        much calculation would be using the angles
        A' = angle BPC, B' = angle CPA, C' = angle APB.
        If I don't mistake, the point A* has trilinears

        A* ( -cos A : cos(C+A') : cos(B+A') ).

        But what are the trilinears of A' ?

        Sincerely,
        Darij Grinberg
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