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Re: Some properties of the Euler reflection point (X110)

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  • Francisco Javier
    Dear Luis, 1) can be generalized in this way: The Miquel Point of the complete quadrangle bounded by the sidelines of ABC and the trilinear polar of a point P
    Message 1 of 8 , Nov 5, 2012
      Dear Luis,

      1) can be generalized in this way:

      The Miquel Point of the complete quadrangle bounded by the sidelines of ABC and the trilinear polar of a point P is the isogonal conjugate of the infinite point of the polar trilineal of the isotomic conjugate of P.

      Best regards,

      Francisco Javier.

      --- In Hyacinthos@yahoogroups.com, "luis240985" <luis240985@...> wrote:
      >
      > 1) X(110) of ABC is a Miquel point of the complete quadrangle bounded by the sidelines of ABC and its Lemoine axis.
      >
      > 2) If A'B'C' is the orthic triangle of ABC, then the Euler reflection points of AB'C',BC'A' and CA'B' lie on the Euler line of ABC.
      >
      > References?, previous research?
      >
      > Sincerely
      > Luis
      >
    • Chris Van Tienhoven
      Der Luis, ... Nice result! I only think you should say that X(110) of ABC is the Miquel point of the complete QUADRILATERAL bounded by the sidelines of ABC and
      Message 2 of 8 , Nov 5, 2012
        Der Luis,

        > 1) X(110) of ABC is a Miquel point of the complete quadrangle bounded by the sidelines of ABC and its Lemoine axis. References?, previous research?

        Nice result!
        I only think you should say that X(110) of ABC is the Miquel point of the complete QUADRILATERAL bounded by the sidelines of ABC and its Lemoine axis.
        I don't know about earlier research regarding a complete quadrilateral derived from a reference triangle and a special line wrt to this reference triangle.
        However there has been done some research by Randy Hutson and Eckart Schmidt regarding a complete quadrangle derived from a reference triangle and a special point wrt to this reference triangle.

        I did do some quick rearch of myself:
        * X(5) of ABC is the Morley Point of the complete QUADRILATERAL bounded by the sidelines of ABC and its Lemoine axis.
        (Morley Point see: http://www.chrisvantienhoven.nl/quadrilateral-objects/17-mathematics/quadrilateral-objects/104-ql-p2.html )

        * X(6) of ABC is the Least Squares point of the complete QUADRILATERAL bounded by the sidelines of ABC and its EULER LINE.
        (Least Squares Point see: http://chrisvantienhoven.nl/quadrilateral-objects/17-mathematics/quadrilateral-objects/131-ql-p26.html )

        * X(691) of ABC is the Miquel point of the complete QUADRILATERAL bounded by the sidelines of ABC and its EULER LINE.

        * Also new Triangle Centers can be found this way. There is a new point that is the Miquel Circumcenter of the complete QUADRILATERAL bounded by the sidelines of ABC and its Lemoine axis
        I think there is a lot more that can be found in this way.
        Barycentric coordinates: a^4 (b^2 - c^2) (4 SA^2 - b^2 c^2)
        Search: -6.120759622
        It is a point on the Lemoine Axis and {3,690}, {32,2491}, {526,1511}, {878,3455}.
        (Miquel Circumcenter see: http://www.chrisvantienhoven.nl/quadrilateral-objects/17-mathematics/quadrilateral-objects/107-ql-p4.html )

        Best regards,

        Chris van Tienhoven
      • Chris Van Tienhoven
        Dear friends, I made a nasty mistake! In my former message I wrote Euler Line , but the line I calculated with was the line X(2).X(6). So actually I should
        Message 3 of 8 , Nov 5, 2012
          Dear friends,

          I made a nasty mistake!
          In my former message I wrote "Euler Line", but the line I calculated with was the line X(2).X(6).
          So actually I should have written:
          * X(6) of ABC is the Least Squares point of the complete QUADRILATERAL bounded by the sidelines of ABC and its X(2).X(6) LINE.
          (Least Squares Point see: http://chrisvantienhoven.nl/quadrilateral-objects/17-mathematics/quadrilateral-objects/131-ql-p26.html )

          * X(691) of ABC is the Miquel point of the complete QUADRILATERAL bounded by the sidelines of ABC and its X(2).X(6) LINE.

          Best regards,

          Chris van Tienhoven

          --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
          >
          > Der Luis,
          >
          > > 1) X(110) of ABC is a Miquel point of the complete quadrangle bounded by the sidelines of ABC and its Lemoine axis. References?, previous research?
          >
          > Nice result!
          > I only think you should say that X(110) of ABC is the Miquel point of the complete QUADRILATERAL bounded by the sidelines of ABC and its Lemoine axis.
          > I don't know about earlier research regarding a complete quadrilateral derived from a reference triangle and a special line wrt to this reference triangle.
          > However there has been done some research by Randy Hutson and Eckart Schmidt regarding a complete quadrangle derived from a reference triangle and a special point wrt to this reference triangle.
          >
          > I did do some quick rearch of myself:
          > * X(5) of ABC is the Morley Point of the complete QUADRILATERAL bounded by the sidelines of ABC and its Lemoine axis.
          > (Morley Point see: http://www.chrisvantienhoven.nl/quadrilateral-objects/17-mathematics/quadrilateral-objects/104-ql-p2.html )
          >
          > * X(6) of ABC is the Least Squares point of the complete QUADRILATERAL bounded by the sidelines of ABC and its EULER LINE.
          > (Least Squares Point see: http://chrisvantienhoven.nl/quadrilateral-objects/17-mathematics/quadrilateral-objects/131-ql-p26.html )
          >
          > * X(691) of ABC is the Miquel point of the complete QUADRILATERAL bounded by the sidelines of ABC and its EULER LINE.
          >
          > * Also new Triangle Centers can be found this way. There is a new point that is the Miquel Circumcenter of the complete QUADRILATERAL bounded by the sidelines of ABC and its Lemoine axis
          > I think there is a lot more that can be found in this way.
          > Barycentric coordinates: a^4 (b^2 - c^2) (4 SA^2 - b^2 c^2)
          > Search: -6.120759622
          > It is a point on the Lemoine Axis and {3,690}, {32,2491}, {526,1511}, {878,3455}.
          > (Miquel Circumcenter see: http://www.chrisvantienhoven.nl/quadrilateral-objects/17-mathematics/quadrilateral-objects/107-ql-p4.html )
          >
          > Best regards,
          >
          > Chris van Tienhoven
          >
        • Nikolaos Dergiades
          Dear friends, Let A B C be the cevian triangle of a point P. Let L1 be the conic with axis the perpendicular bisector of AP and that passes through A,P,B ,C .
          Message 4 of 8 , Nov 6, 2012
            Dear friends,
            Let A'B'C' be the cevian triangle of a point P.
            Let L1 be the conic with axis the perpendicular
            bisector of AP and that passes through A,P,B',C'.
            Let O1 be the center of L1. Similarly define the
            centers O2, O3. What is the locus of P such that
            the triangles ABC, O1O2O3 are perspective.
            One component of the locus is a cubic.
            Is this cubic known?

            Best regards
            Nikos Dergiades
          • Bernard Gibert
            Dear Nikolaos, ... I should say yes since it is the Neuberg cubic... best regards Bernard [Non-text portions of this message have been removed]
            Message 5 of 8 , Nov 7, 2012
              Dear Nikolaos,

              > [ND] Let A'B'C' be the cevian triangle of a point P.
              > Let L1 be the conic with axis the perpendicular
              > bisector of AP and that passes through A,P,B',C'.
              > Let O1 be the center of L1. Similarly define the
              > centers O2, O3. What is the locus of P such that
              > the triangles ABC, O1O2O3 are perspective.
              > One component of the locus is a cubic.
              > Is this cubic known?

              I should say yes since it is the Neuberg cubic...

              best regards

              Bernard

              [Non-text portions of this message have been removed]
            • Antreas Hatzipolakis
              ... ..... and the other components?? aph ... [Non-text portions of this message have been removed]
              Message 6 of 8 , Nov 7, 2012
                On Wed, Nov 7, 2012 at 10:07 AM, Bernard Gibert <bg42@...> wrote:

                > **
                >
                >
                > Dear Nikolaos,
                >
                > > [ND] Let A'B'C' be the cevian triangle of a point P.
                >
                > > Let L1 be the conic with axis the perpendicular
                > > bisector of AP and that passes through A,P,B',C'.
                > > Let O1 be the center of L1. Similarly define the
                > > centers O2, O3. What is the locus of P such that
                > > the triangles ABC, O1O2O3 are perspective.
                > > One component of the locus is a cubic.
                >


                ..... and the other components??

                aph



                > > Is this cubic known?
                >
                > I should say yes since it is the Neuberg cubic...
                >
                > best regards
                >
                > Bernard
                >


                [Non-text portions of this message have been removed]
              • Bernard Gibert
                Dear Antreas, ... the circles with diameters BC, CA, AB the line at infinity the sidelines of ABC not very interesting... best regards Bernard ... [Non-text
                Message 7 of 8 , Nov 7, 2012
                  Dear Antreas,

                  > > Dear Nikolaos,
                  > >
                  > > > [ND] Let A'B'C' be the cevian triangle of a point P.
                  > >
                  > > > Let L1 be the conic with axis the perpendicular
                  > > > bisector of AP and that passes through A,P,B',C'.
                  > > > Let O1 be the center of L1. Similarly define the
                  > > > centers O2, O3. What is the locus of P such that
                  > > > the triangles ABC, O1O2O3 are perspective.
                  > > > One component of the locus is a cubic.
                  > >
                  >
                  > [APH] ..... and the other components??


                  the circles with diameters BC, CA, AB
                  the line at infinity
                  the sidelines of ABC

                  not very interesting...

                  best regards

                  Bernard


                  >
                  > > >[ND] Is this cubic known?
                  > >
                  > >[BG] I should say yes since it is the Neuberg cubic...



                  [Non-text portions of this message have been removed]
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