Loading ...
Sorry, an error occurred while loading the content.

Re: Happy New Year and Nik's locus for 2013

Expand Messages
  • yiuatfauedu
    Dear Nik and Francisco Javier, ND: If A B C is the cevian triangle of P and A BA1, CA A2 are equilateral triangles and A* is the mid point of A1, A2 in the
    Message 1 of 14 , Jan 1, 2013
    • 0 Attachment
      Dear Nik and Francisco Javier,


      ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2
      are equilateral triangles and A* is the mid point of A1, A2
      in the opposite side of BC ralative to A, and similarly define B*,
      C*, what is the locus of P such that ABC, A*B*C* are perspective?

      PY: If A_0BC, B_0CA, C_0AB are the equilateral triangles
      constructed externally on the sides BC, CA, AB respectively,
      ... A*B*C* is perspective with ABC if and only if P [lies on]
      the isocubic K(Q,X(13)), where
      Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
      has (6-9-13) search number 0.245203113787…

      ***

      I think we can go a little bit further and get a Euclidean constructible example. The point $Q$ can be regarded as the isogonal conjugate of the barycentric product of $X(15)$ (isodynamic point) and $X(17)$ (Napoleon point):
      $$
      Q = (X(15)*X(17))^*.
      $$
      It is an interior point of triangle $ABC$ if all angles are less than $120^\circ$.

      In this case, the barycentric square root $\sqrt Q$ is a point on the locus. With $P = \sqrt Q$, the perspector is the barycentric square root of
      $$
      (1/((S_A+S \sqrt 3)(S_A+S/\sqrt 3)) : ... : ...)
      = (1/(S_A+S \sqrt 3)^2 : ... : ...)*
      ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
      = (1/(S_A+S \sqrt 3)^2 : ... : ...)*Q,
      $$
      and is the barycentric product $X(17)*P$.

      Best regards
      Sincerely
      Paul
    • Antreas Hatzipolakis
      Dear Nikos, Paul Happy New Year!!! 1. We can take A B C as the pedal triangle of P and ask for the locus. 2. We can take A*,B*,C* as triangle centers
      Message 2 of 14 , Jan 1, 2013
      • 0 Attachment
        Dear Nikos, Paul

        Happy New Year!!!

        1. We can take A'B'C' as the pedal triangle of P and ask for the locus.

        2. We can take A*,B*,C* as triangle centers (Circumcenters, Orthocenters
        etc) of
        the triangles A'A1A2, B'B1B2, C'C1C2 and ask of the loci of P such taht
        ABC,A*B*C* be perspective.

        I think that interesting is the case of the circumcenters.

        Season's Greetings

        Antreas

        On Tue, Jan 1, 2013 at 2:15 AM, Nikolaos Dergiades <ndergiades@...>wrote:

        > **
        >
        >
        > Dear Antreas
        > Happy New Year to you
        > and all friends of Hyacinthos.
        >
        > A locus for the begining of 2013.
        > If A'B'C' is the cevian triangle of P
        > and A'BA1, CA'A2 are equilateral triangles
        > and A* is the mid point of A1, A2 in the
        > opposite side of BC ralative to A,
        > and similarly define B*, C* what is the locus
        > of P such that ABC, A*B*C* are perspective.
        > Best regards
        > Nikos Dergiades
        >
        >


        [Non-text portions of this message have been removed]
      • Nikolaos Dergiades
        Dear Paul, A little generalization: If A B C is the cevian triangle of a point P and A BC, B CA, C AB are similar isosceles triangles outside of ABC with base
        Message 3 of 14 , Jan 2, 2013
        • 0 Attachment
          Dear Paul,
          A little generalization:
          If A'B'C' is the cevian triangle of a point P
          and A"BC, B"CA, C"AB are similar isosceles triangles
          outside of ABC with base agnle w then the triangles
          ABC, A*B*C* where A* is the mid point of A'A" . . .
          are perspective if and only if the locus of P
          is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
          and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
          The point S lies on the Kiepert hyperbola
          and corresponds to A", B", C".
          The point S'= ((1/(S_A+3S_w):..:..)) lies also
          on the Kiepert hyperbola and corresponds to the
          Centroids of triangles A"BC, B"CA, C"AB.
          The point Q is the barycentric product of S
          and the isotomic of S'.

          Best regards
          Nik



          > I think we can go a little bit further and get a Euclidean
          > constructible example. The point $Q$ can be regarded as the
          > isogonal conjugate of the barycentric product of $X(15)$
          > (isodynamic point) and $X(17)$ (Napoleon point):
          > $$
          > Q = (X(15)*X(17))^*.
          > $$
          > It is an interior point of triangle $ABC$ if all angles are
          > less than $120^\circ$.
          >
          > In this case, the barycentric square root $\sqrt Q$ is a
          > point on the locus. With $P = \sqrt Q$, the perspector is
          > the  barycentric square root of
          > $$
          >   (1/((S_A+S \sqrt 3)(S_A+S/\sqrt 3)) : ... : ...)
          > = (1/(S_A+S \sqrt 3)^2 : ... : ...)*
          >   ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3])  : ... : ...)
          > = (1/(S_A+S \sqrt 3)^2 : ... : ...)*Q,
          > $$
          > and is the barycentric product $X(17)*P$.
          >
          > Best regards
          > Sincerely
          > Paul
          >
          >
          >
          >
          > ------------------------------------
          >
          > Yahoo! Groups Links
          >
          >
          >     Hyacinthos-fullfeatured@yahoogroups.com
          >
          >
        • Nikolaos Dergiades
          Dear Antreas, if A B C is the pedal triangle and not the cevian then the locus is a cubic but I think complicated. The same holds if we take circumenters. I
          Message 4 of 14 , Jan 2, 2013
          • 0 Attachment
            Dear Antreas,
            if A'B'C' is the pedal triangle and not
            the cevian then the locus is a cubic
            but I think complicated. The same holds
            if we take circumenters.
            I think that the most interesting case is:

            If A'B'C' is the cevian triangle of a point P,
            A1A'B, A2CA'are similar isosceles triangles
            outside of ABC with base angle w,
            (the bases are opposite of A1 and A2)
            A* is the centroid of triangle A'A1A2
            and similarly define B*, C*
            then the triangles ABC, A*B*C* are perspective
            if and only if the locus of P
            is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
            and pole Q = ((S_A+5S_w)/(S_A+S_w):..:..).

            Best regards
            Nikos


            > Dear Nikos, Paul
            >
            > Happy New Year!!!
            >
            > 1. We can take A'B'C' as the pedal triangle of P and ask for
            > the locus.
            >
            > 2. We can take A*,B*,C* as triangle centers (Circumcenters,
            > Orthocenters
            > etc) of
            > the triangles A'A1A2, B'B1B2, C'C1C2 and ask of the loci of
            > P such taht
            > ABC,A*B*C* be perspective.
            >
            > I think that interesting is the case of the circumcenters.
            >
            > Season's Greetings
            >
            > Antreas
            >
            > On Tue, Jan 1, 2013 at 2:15 AM, Nikolaos Dergiades <ndergiades@...>wrote:
            >
            > > **
            > >
            > >
            > > Dear Antreas
            > > Happy New Year to you
            > > and all friends of Hyacinthos.
            > >
            > > A locus for the begining of 2013.
            > > If A'B'C' is the cevian triangle of P
            > > and A'BA1, CA'A2 are equilateral triangles
            > > and A* is the mid point of A1, A2 in the
            > > opposite side of BC ralative to A,
            > > and similarly define B*, C* what is the locus
            > > of P such that ABC, A*B*C* are perspective.
            > > Best regards
            > > Nikos Dergiades
            > >
            > >
            >
            >
            > [Non-text portions of this message have been removed]
            >
            >
            >
            > ------------------------------------
            >
            > Yahoo! Groups Links
            >
            >
            >     Hyacinthos-fullfeatured@yahoogroups.com
            >
            >
          • Bernard Gibert
            Dear Nikos, Paul and friends, ... If A* is the homothetic of A under h(A , k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are
            Message 5 of 14 , Jan 2, 2013
            • 0 Attachment
              Dear Nikos, Paul and friends,

              > [ND] A little generalization:
              > If A'B'C' is the cevian triangle of a point P
              > and A"BC, B"CA, C"AB are similar isosceles triangles
              > outside of ABC with base agnle w then the triangles
              > ABC, A*B*C* where A* is the mid point of A'A" . . .
              > are perspective if and only if the locus of P
              > is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
              > and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
              > The point S lies on the Kiepert hyperbola
              > and corresponds to A", B", C".
              > The point S'= ((1/(S_A+3S_w):..:..)) lies also
              > on the Kiepert hyperbola and corresponds to the
              > Centroids of triangles A"BC, B"CA, C"AB.
              > The point Q is the barycentric product of S
              > and the isotomic of S'.

              If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is pK(Ω, S) with

              pole Ω = (SA k + (2-k)Sw) / (SA + Sw) : : ,

              pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,

              isopivot : SA k + (2-k)Sw : : , on the line X2-X69.

              When k varies, the locus of Ω is a line through X2,

              When w varies, the locus of Ω is a nodal circum-cubic with node X2, the nodal tangents being X2-X6 and Euler line.

              This cubic is a member of the pencil generated by :
              the three medians (k = 1)
              Kiepert + Euler (k = 0)
              line X2-X6 + isotomic of Euler (k = 2).

              ----------------------------------------------------------

              The locus of X is another pK(Ω', S') with

              pole Ω' = 1 / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

              pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

              isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.


              Best regards and Happy New Year to you all.

              Bernard

              [Non-text portions of this message have been removed]
            • Paul Yiu
              Dear Nik, Bernard, and Francisco Javier, Thank you for these generalizations. FJ, I did not realize that the locus contains G! Best regards Sincerely Paul
              Message 6 of 14 , Jan 2, 2013
              • 0 Attachment
                Dear Nik, Bernard, and Francisco Javier,

                Thank you for these generalizations. FJ, I did not realize that the locus contains G!

                Best regards
                Sincerely
                Paul
                ________________________________________
                From: Hyacinthos@yahoogroups.com [Hyacinthos@yahoogroups.com] on behalf of Bernard Gibert [bg42@...]
                Sent: Wednesday, January 02, 2013 8:00 AM
                To: Hyacinthos@yahoogroups.com
                Subject: Re: [EMHL] Re: Happy New Year and Nik's locus for 2013

                Dear Nikos, Paul and friends,

                > [ND] A little generalization:
                > If A'B'C' is the cevian triangle of a point P
                > and A"BC, B"CA, C"AB are similar isosceles triangles
                > outside of ABC with base agnle w then the triangles
                > ABC, A*B*C* where A* is the mid point of A'A" . . .
                > are perspective if and only if the locus of P
                > is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
                > and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
                > The point S lies on the Kiepert hyperbola
                > and corresponds to A", B", C".
                > The point S'= ((1/(S_A+3S_w):..:..)) lies also
                > on the Kiepert hyperbola and corresponds to the
                > Centroids of triangles A"BC, B"CA, C"AB.
                > The point Q is the barycentric product of S
                > and the isotomic of S'.

                If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is pK(Ω, S) with

                pole Ω = (SA k + (2-k)Sw) / (SA + Sw) : : ,

                pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,

                isopivot : SA k + (2-k)Sw : : , on the line X2-X69.

                When k varies, the locus of Ω is a line through X2,

                When w varies, the locus of Ω is a nodal circum-cubic with node X2, the nodal tangents being X2-X6 and Euler line.

                This cubic is a member of the pencil generated by :
                the three medians (k = 1)
                Kiepert + Euler (k = 0)
                line X2-X6 + isotomic of Euler (k = 2).

                ----------------------------------------------------------

                The locus of X is another pK(Ω', S') with

                pole Ω' = 1 / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

                pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

                isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.


                Best regards and Happy New Year to you all.

                Bernard

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



                ------------------------------------

                Yahoo! Groups Links
              • Ricardo Barroso
                Happy New year 2013 . From Sevilla  Ricardo http://personal.us.es/rbarroso/trianguloscabri/ [Non-text portions of this message have been removed]
                Message 7 of 14 , Jan 2, 2013
                • 0 Attachment
                  Happy New year 2013 .

                  From Sevilla 

                  Ricardo

                  http://personal.us.es/rbarroso/trianguloscabri/

                  [Non-text portions of this message have been removed]
                • Bernard Gibert
                  Dear all, If A B C is the pedal triangle of P (instead of cevian triangle of P), and A , B , C as before, If A* is the homothetic of A under h(A , k), B*
                  Message 8 of 14 , Jan 2, 2013
                  • 0 Attachment
                    Dear all,

                    If A'B'C' is the pedal triangle of P (instead of cevian triangle of P), and A", B", C" as before,

                    If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is a central cubic passing through X3 with asymptotes parallel to the altitudes of ABC and concurring on the Brocard axis, very similar to the Darboux cubic which is obtained when k = 0.

                    The locus of the perspector X is pK(X2, T) where

                    T = (-a^4+b^4+c^4)k-4 SA Sw : : , on the line X4-X69.

                    If k = 0, T = X69 and this is the Lucas cubic.

                    Best regards

                    Bernard

                    [Non-text portions of this message have been removed]
                  Your message has been successfully submitted and would be delivered to recipients shortly.