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

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  • Francisco Javier
    Dear Nikos, I also did the calculations but when I was about to post them, I felt gladly overwhelmed by the impressive answer by Paul. Happy New Year to all
    Message 1 of 14 , Jan 1, 2013
      Dear Nikos, I also did the calculations but when I was about to post them, I felt gladly overwhelmed by the impressive answer by Paul.

      Happy New Year to all Hyacinthos members!

      Francisco Javier.

      --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...> wrote:
      >
      > Dear Paul,
      > we have the same result as for A* . . .
      > but I am impressed with the factorization
      > you had in the equation of the cubic.
      > Happy New Year.
      > Best regards
      > Nik
      >
      >
      > >
      > > 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?
      > >
      > > ***
      > >
      > > If A_0BC, B_0CA, C_0AB are the equilateral triangles
      > > constructed externally on the sides BC, CA, AB respectively,
      > > your A* is the midpoint of A'A_0.  If  P = (u:v:W)
      > > in homogeneous barycentric coordinates, then
      > >
      > > A* = (-(S_B+S_C)(v+w) : (S_C+S/Sqrt[3])v + (S_C+S Sqrt[3])w
      > >
      > >         : (S_B+S/Sqrt[3])v + (S_B+S
      > > Sqrt[3])w),
      > > and  similarly for B* and C*.
      > >
      > > A*B*C* is perspective with ABC if and only if
      > >
      > > Cyclic sum [(S_B+S/Sqrt[3])(S_C+S Sqrt[3]) uv^2
      > >            - (S_B+S
      > > Sqrt[3])(S_C+S/Sqrt[3]) uw^2]= 0.
      > >
      > > Rewriting this equation as
      > >
      > > Cyclic sum  u/(S_A+S/Sqrt[3])[(S_C+S
      > > Sqrt[3])/(S_C+S/Sqrt[3]) v^2
      > >           - (S_B+S
      > > Sqrt[3])/(S_B+S/Sqrt[3])w^2] = 0 ,
      > > we identify the locus of P as 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…
      > >
      > > Best regards
      > > Sincerely
      > > Paul
      > >
      > >
      > >
      > >
      > > ------------------------------------
      > >
      > > Yahoo! Groups Links
      > >
      > >
      > >     Hyacinthos-fullfeatured@yahoogroups.com
      > >
      > >
      >
    • 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 2 of 14 , Jan 1, 2013
        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 3 of 14 , Jan 1, 2013
          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 4 of 14 , Jan 2, 2013
            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 5 of 14 , Jan 2, 2013
              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 6 of 14 , Jan 2, 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]
              • 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 7 of 14 , Jan 2, 2013
                  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]



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

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                • 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 8 of 14 , Jan 2, 2013
                    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 9 of 14 , Jan 2, 2013
                      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]
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