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Re: [EMHL] Points on the MacBeath ellipse

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  • jpehrmfr
    Dear Antreas and Milorad [MS] ... [APH] ... (339) ... The vertices of the focal axis are the common points of the Euler line and the NP-circle; they are
    Message 1 of 14 , Nov 11, 2004
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      Dear Antreas and Milorad
      [MS]
      > >Are some other easy constructible points on the
      > >MacBeath ellipse are known?
      [APH]
      > "The only Kimberling center lying on the MacBeath inconic is X
      (339)
      > (Weisstein, Oct. 16, 2004)."
      >
      > http://mathworld.wolfram.com/MacBeathInconic.html

      The vertices of the focal axis are the common points of the Euler
      line and the NP-circle; they are X(1312) and X(1313) in ETC.
      Friendly. Jean-Pierre
    • Milorad Stevanovic
      Dear Antreas and Jean-Pierre, [MS] ... [APH] ... http://mathworld.wolfram.com/MacBeathInconic.html Does anyone know some geometric property or geometric
      Message 2 of 14 , Nov 11, 2004
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        Dear Antreas and Jean-Pierre,

        [MS]
        >Are some other easy constructible points on the
        >MacBeath ellipse are known?

        [APH]
        >"The only Kimberling center lying on the MacBeath inconic is X(339)
        >(Weisstein, Oct. 16, 2004)."

        http://mathworld.wolfram.com/MacBeathInconic.html

        Does anyone know some geometric property or geometric
        caracterization of point X(339)(beside that it is the isotomic
        conjugate of X(225))?

        [JPE]
        >The vertices of the focal axis are the common points of the Euler
        >line and the NP-circle; they are X(1312) and X(1313) in ETC.

        Thanks to both of you.
        I had this information about points X(339),X(1312),X(1313).
        It could be of interest to find some other triangle center on
        the MacBeath ellipse.
        If it is too hard then to find some other point as the point of
        tangency of some perpendicular bisector of PH, for some
        triangle center P on the circumcircle,with the MacBeath
        ellipse.
        I think that it could be easier way then to use some
        Droz-Farny line.
        But,the open problem could be,for which points P on the
        circumcircle,by the help of perpendicular bisector of PH
        we get the triangle center on the MacBeath ellipse?

        Friendly

        Milorad R.Stevanovic



        [Non-text portions of this message have been removed]
      • Milorad Stevanovic
        Dear Antreas and Jean-Pierre, [MS] ... [APH] ... http://mathworld.wolfram.com/MacBeathInconic.html The point X(339) is the point of tangency of the MacBeath
        Message 3 of 14 , Nov 12, 2004
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          Dear Antreas and Jean-Pierre,

          [MS]
          >Are some other easy constructible points on the
          >MacBeath ellipse are known?

          [APH]
          >"The only Kimberling center lying on the MacBeath inconic is X(339)
          >(Weisstein, Oct. 16, 2004)."

          http://mathworld.wolfram.com/MacBeathInconic.html

          The point X(339) is the point of tangency of the MacBeath ellipse
          and of perpendicular bisector of HTa where Ta=X(98)-Tarry point.
          The midpoint of HTa is the center of Kiepert hyperbola,or the
          midpoint of F1F2,where F1 and F2 are Fermat points.
          Antipode of Tarry point on the circumcircle is Steiner point St.
          If we take perpendicular bisector of HSt then we have the
          following triangle center on the MacBeath ellipse.

          X=(a cosA)/u^2, y=(b cosB)/v^2, z=(c cosC)/w^2,
          where
          u=(4R^2-a^2)m^2-(a^2)(b^2-c^2)^2,
          v=(4R^2-b^2)m^2-(b^2)(c^2-a^2)^2,
          w=(4R^2-c^2)m^2-(c^2)(a^2-b^2)^2,
          where
          m^2=a^4+b^4+c^4-(ab)^2-(bc)^2-(ca)^2.

          I hope that my calculations are good.

          This new triangle center could be of interest for Clark and for Edward.

          Best regards

          Milorad R.Stevanovic

          [Non-text portions of this message have been removed]
        • Antreas P. Hatzipolakis
          Dear Milorad and Jean-Pierre I am wondering what geometrical properties [locus, envelope] has the dual of the MacBeath inconic ie the MacBeath circumconic [=
          Message 4 of 14 , Nov 12, 2004
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            Dear Milorad and Jean-Pierre

            I am wondering what geometrical properties [locus, envelope] has
            the "dual" of the MacBeath inconic ie the MacBeath circumconic
            [= the conic centered at N, and passing through A,B,C.]

            Antreas
            --
          • peter_mows
            Dear Antreas and everyone, [APH] ... My limited understanding of a dual comes from here http://www.math.fau.edu/yiu/GeometryNotes020402.ps page 120. By my
            Message 5 of 14 , Nov 12, 2004
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              Dear Antreas and everyone,

              [APH]
              >I am wondering what geometrical properties [locus, envelope] has
              >the "dual" of the MacBeath inconic ie the MacBeath circumconic
              >[= the conic centered at N, and passing through A,B,C.]

              My limited understanding of a dual comes from here
              http://www.math.fau.edu/yiu/GeometryNotes020402.ps page 120.
              By my reckoning, I think the dual is the circumconic,
              a^2 SA y z + cyclic = 0, centered on X(6) passing through
              110, 287, 648, 651, 677, 895, 1331, 1332, 1797, 1813, 1814, 1815

              If it is of interest, here are a few centers, other than X(399,
              1312, 1313), that I think are on the MacBeath conic,
              a^4 SA^2 x^2 - 2 b^2 c^2 SB SC y z + Cyclic = 0.

              rfl X(339) in X(5)
              a^2 SB SC (SB SC - SA^2)^2
              on lines {3,112},{4,147},{25,110},{114,132}

              rfl X(339) in Euler line.
              a^2 SA (b^2 - c^2)^2 (S^2 + SA^2 - 4 SB SC)^2

              (b - c)^2 (b + c - a)^2 SA
              on lines {3, 8}, {11, 123}, {116, 122}

              (b - c)^2 SB SC
              on lines {4, 145}, {25,105},{124,136}

              b^2 c^2 (b^2 - c^2)^2 SB SC
              on lines {4,94},{25,98},{115,135},{125,136}

              a^2 (b^2 - c^2) SB SC (SB - SC)
              on lines {4,147},{25,111},{127,136}

              b^2 c^2 SA (SB - SC)^2
              on lines {3,76},{115,127}

              a^2 SA^3 (SB - SC)^2
              on lines {3,74},{0,122,125},{0,127,136}

              b^2 c^2 (b - c)^2 SB SC
              on line {4, 150}

              a^2 SA (a^2 (SB SC - SA^2) + SA (SB - SC)^2)^2
              on lines {3,74},{113,131}

              b^2 c^2 SB SC (S^2 - 3 SB SC)^2
              on lines {3,107},{113,133}

              SA (a b c (b + c) - 2 S^2 + a (b + c) (b c - 2 SA) - 2 SB SC)^2
              on lines {3,100},{117,131}

              note that many of these are intersections of {Focus, NP},{NP,NP}

              In general, a point on the conic has 3 friends, its reflection in X
              (5) and their reflection in the Euler line.

              Also, I find here
              http://www.genealogy.ams.org/html/id.phtml?id=24339
              an entry for Alexander Murray MacBeath. The same one?


              Best regards,
              Peter.
            • Milorad Stevanovic
              Dear Peter, and all other friends. I have found the following result: For any point P(x:y:z) on the circumcircle, point Q((a cosA)/m^2:(b cosB)/n^2:(c
              Message 6 of 14 , Nov 13, 2004
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                Dear Peter,
                and all other friends.

                I have found the following result:

                For any point P(x:y:z) on the circumcircle,
                point Q((a cosA)/m^2:(b cosB)/n^2:(c cosC)/p^2)
                is on the MacBeath ellipse for
                m=x(cosA)^2-ycosBcos(C-A)-zcosCcos(A-B),
                n=y(cosB)^2-zcosCcos(A-B)-xcosAcos(B-C),
                p=z(cosC)^2-xcosAcos(B-C)-ycosBcos(C-A).

                This could be useful as input for good
                computer program to find a lot of new points
                on the MacBeath ellipse.

                Best regards

                Milorad R.Stevanovic

                [Non-text portions of this message have been removed]
              • peter_mows
                Dear Milorad, [MS] ... What a beautiful result !! a^2 SA / (a^2 SA^2 x - SB y (b^2 SB + 2 SA SC) - SC z (c^2 SC + 2 SA SB))^2 X(339) is your transform of X(98)
                Message 7 of 14 , Nov 15, 2004
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                  Dear Milorad,

                  [MS]
                  >I have found the following result:

                  >For any point P(x:y:z) on the circumcircle,
                  >point Q((a cosA)/m^2:(b cosB)/n^2:(c cosC)/p^2)
                  >is on the MacBeath ellipse for
                  >m=x(cosA)^2-ycosBcos(C-A)-zcosCcos(A-B),
                  >n=y(cosB)^2-zcosCcos(A-B)-xcosAcos(B-C),
                  >p=z(cosC)^2-xcosAcos(B-C)-ycosBcos(C-A).

                  >This could be useful as input for good
                  >computer program to find a lot of new points
                  >on the MacBeath ellipse.

                  What a beautiful result !!

                  a^2 SA / (a^2 SA^2 x - SB y (b^2 SB + 2 SA SC) -
                  SC z (c^2 SC + 2 SA SB))^2

                  X(339) is your transform of X(98)

                  Best regards
                  Peter.
                • Eric Danneels
                  Dear Milorad and Peter you wrote, ... I would like to add 2 other pairs X(1113) transforms to X(1312) X(1114) transforms to X(1313) I couldn t find other
                  Message 8 of 14 , Nov 15, 2004
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                    Dear Milorad and Peter

                    you wrote,
                    > [MS]
                    > >I have found the following result:
                    >
                    > >For any point P(x:y:z) on the circumcircle,
                    > >point Q((a cosA)/m^2:(b cosB)/n^2:(c cosC)/p^2)
                    > >is on the MacBeath ellipse for
                    > >m=x(cosA)^2-ycosBcos(C-A)-zcosCcos(A-B),
                    > >n=y(cosB)^2-zcosCcos(A-B)-xcosAcos(B-C),
                    > >p=z(cosC)^2-xcosAcos(B-C)-ycosBcos(C-A).
                    >
                    > >This could be useful as input for good
                    > >computer program to find a lot of new points
                    > >on the MacBeath ellipse.
                    >
                    > What a beautiful result !!
                    >
                    > a^2 SA / (a^2 SA^2 x - SB y (b^2 SB + 2 SA SC) -
                    > SC z (c^2 SC + 2 SA SB))^2
                    >
                    > X(339) is your transform of X(98)

                    I would like to add 2 other pairs

                    X(1113) transforms to X(1312)
                    X(1114) transforms to X(1313)

                    I couldn't find other points in the ETC

                    Greetings from Bruges

                    Eric

                    PS

                    Sorry if this message appears twice
                  • Bernard Gibert
                    Dear Milorad and Peter, ... Let M = u:v:w be a point and M2 = u^2:v^2:w^2 its barycentric square. for any M at infinity, the M2 isoconjugate of O lies on the
                    Message 9 of 14 , Nov 15, 2004
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                      Dear Milorad and Peter,

                      > [MS]
                      > >I have found the following result:
                      >
                      > >For any point P(x:y:z) on the circumcircle,
                      > >point Q((a cosA)/m^2:(b cosB)/n^2:(c cosC)/p^2)
                      > >is on the MacBeath ellipse for
                      > >m=x(cosA)^2-ycosBcos(C-A)-zcosCcos(A-B),
                      > >n=y(cosB)^2-zcosCcos(A-B)-xcosAcos(B-C),
                      > >p=z(cosC)^2-xcosAcos(B-C)-ycosBcos(C-A).
                      >
                      > [PM] This could be useful as input for good
                      > >computer program to find a lot of new points
                      > >on the MacBeath ellipse.

                      Let M = u:v:w be a point and M2 = u^2:v^2:w^2 its barycentric square.

                      for any M at infinity, the M2 isoconjugate of O lies on the MacBeath
                      conic.
                      this is the point u^2/(a^2SA) : : .
                      with M=X525, we find X339.

                      consequently, for any point N = p:q:r on the circumcircle, the X(25)
                      isoconjugate of M2 also lies on the MacBeath conic.
                      this is the point a^2/(p^2 SA) : : .
                      with N = X112, we find X339.

                      this gives a very simple way to find many simple points on the conic,
                      and in fact on any in-conic with given perspector or center.

                      I do not think this is very interesting...

                      Best regards

                      Bernard

                      [Non-text portions of this message have been removed]
                    • Milorad Stevanovic
                      Dear Peter,Eric,Bernard I could add a new result. For any point P which is on circumcircle and on ortho cubic of triangle ABC we have 1.Perpendicular bisector
                      Message 10 of 14 , Nov 15, 2004
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                        Dear Peter,Eric,Bernard

                        I could add a new result.
                        For any point P which is on circumcircle
                        and on ortho cubic of triangle ABC we have
                        1.Perpendicular bisector of PH is Wallace line
                        of P wrt triangle ABC.
                        2.This line is tangent to the MacBeath ellipse.

                        Are the coordinates of points P known?
                        Can we find the corresponding points on the
                        MacBeath ellipse ?

                        If the Wallace line of point P wrt triangle ABC
                        is orthogonal to line PH then P is also on the
                        ortho cubic of triangle ABC.

                        Best regards

                        Milorad R.Stevanovic

                        [Non-text portions of this message have been removed]
                      • ndergiades
                        Dear friends, ... (339) ... see also at http://pages.infinit.net/spqrsncf/ngorecent.htm#L2I-11 but I think that this inconic is known in F.G-M fifth edition
                        Message 11 of 14 , Nov 16, 2004
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                          Dear friends,

                          > "The only Kimberling center lying on the MacBeath inconic is X
                          (339)
                          > (Weisstein, Oct. 16, 2004)."

                          see also at
                          http://pages.infinit.net/spqrsncf/ngorecent.htm#L2I-11

                          but I think that this inconic is known in F.G-M fifth edition
                          problem §130. In my Greek translation there is no reference
                          but in the French edition 5 or 8 there is the reference that
                          this conic is due to Paul Seret N.A. 1865 p. 428.

                          Best regards
                          Nikolaos Dergiades
                        • Ricardo Barroso
                          Dear Nikolaos and friends: In FGM problem 130, it is the ellipse of centers H and O tangent to sides of triangle one talks about Paul Serret 1865, to M.E.
                          Message 12 of 14 , Nov 18, 2004
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                            Dear Nikolaos and friends:

                            In FGM problem 130, it is the ellipse of centers H and O
                            tangent to sides of triangle one talks about Paul Serret
                            1865, to M.E. Lemoine 1858 and Picquet 1866.



                            University of Michigan Historical Math Collection

                            Exercices de géométrie, comprenant l'esposé des méthodes
                            géométriques et 2000 questions résolues par F. G.-M. ---
                            Frère Gabriel Marie, 1820-1891.
                            5. ed.: 3 p. L., [iii]-xxiv, 1302 p. diagrs. 22 cm.
                            Tours,
                            A. Mame et fils; [etc., etc.]
                            1912.

                            In :
                            http://www.personal.us.es/rbarroso/ellipsefocusOHtangent.htm

                            there is a link to the pdf of the FGM

                            Greetings

                            Ricardo
                            --- ndergiades <ndergiades@...> escribió:
                            >
                            > Dear friends,
                            >
                            > > "The only Kimberling center lying on the MacBeath
                            > inconic is X
                            > (339)
                            > > (Weisstein, Oct. 16, 2004)."
                            >
                            > see also at
                            > http://pages.infinit.net/spqrsncf/ngorecent.htm#L2I-11
                            >
                            > but I think that this inconic is known in F.G-M fifth
                            > edition
                            > problem §130. In my Greek translation there is no
                            > reference
                            > but in the French edition 5 or 8 there is the reference
                            > that
                            > this conic is due to Paul Seret N.A. 1865 p. 428.
                            >
                            > Best regards
                            > Nikolaos Dergiades
                            >
                            >
                            >
                            >



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