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Happy New Year

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  • Nikolaos Dergiades
    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
    Message 1 of 14 , Dec 31, 2012
      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
    • yiuatfauedu
      Dear Nik and Antreas, Happy New Year to you and all friends of Hyacinthos. ND: If A B C is the cevian triangle of P and A BA1, CA A2 are equilateral triangles
      Message 2 of 14 , Jan 1, 2013
        Dear Nik and Antreas,

        Happy New Year to you and all friends of Hyacinthos.

        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
      • Nikolaos Dergiades
        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
        Message 3 of 14 , Jan 1, 2013
          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
          >
          >
        • 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 4 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 5 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 6 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 7 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 8 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 9 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 10 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]



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

                        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 11 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 12 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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