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 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 1 of 14 , Jan 1, 2013
    • 0 Attachment
      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 2 of 14 , Jan 1, 2013
      • 0 Attachment
        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 3 of 14 , Jan 1, 2013
        • 0 Attachment
          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 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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.