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Re: [EMHL] Soddy spheres touching a plane

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  • Vladimir Dubrovsky
    Dear Frank, using the fact that the distance between the points of contact of any two of your spheres with the plane is 2sqrt(Rr), where R and r are there
    Message 1 of 14 , Dec 30, 2012
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      Dear Frank,

      using the fact that the distance between the points of contact of any two
      of your spheres with the plane is 2sqrt(Rr), where R and r are there radii,
      we see that P and Q are isodynamic points of ABC, whose trilinears are well
      known.

      Happy New Year!

      Vladimir


      2012/12/30 Frank Jackson <fjackson@...>

      > **
      >
      >
      > Dear All
      >
      > Place 3 spheres of different radii onto a flat plane such that they
      > touch each other and let their contact points with the plane be the
      > triangle ABC.
      >
      > Now introduce a 4th sphere that also touches the other 3 spheres and
      > the plane (Soddy touching spheres with one radius infinite).
      >
      > There are 2 possible solutions for the radius of the 4th sphere and
      > 2 solutions for its contact point with the plane, say P and Q.
      >
      > What are the trilinears of P and Q with respect to ABC?
      >
      > Happy New Year
      >
      > Frank Jackson
      >
      >
      >


      [Non-text portions of this message have been removed]
    • 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 2 of 14 , Dec 31, 2012
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        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 3 of 14 , Jan 1, 2013
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          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 4 of 14 , Jan 1, 2013
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            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 5 of 14 , Jan 1, 2013
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              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 6 of 14 , Jan 1, 2013
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                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 7 of 14 , Jan 1, 2013
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                  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 8 of 14 , Jan 2, 2013
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                    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 9 of 14 , Jan 2, 2013
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                      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 10 of 14 , Jan 2, 2013
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                        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 11 of 14 , Jan 2, 2013
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                          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 12 of 14 , Jan 2, 2013
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                            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 13 of 14 , Jan 2, 2013
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                              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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