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Re: [EMHL] Cubics

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  • Fred Lang
    Dear Andreas, your generalization is exactly the one of Guido M. Pinkernell in Journal of Geometry Vol 55 (1996) Cubics Curves in the Triangle Plane . Your
    Message 1 of 42 , Oct 31, 2000
      Dear Andreas,
      your generalization is exactly the one of Guido M. Pinkernell in Journal of
      Geometry Vol 55 (1996) " Cubics Curves in the Triangle Plane".
      Your cubic is denoted as the t-pedal cubic, he defines a (2,1)
      correspondance from the t-pedal to the t-cevian cubic:

      See my message of the 8 oct: "Cubic Kn is an identified object", I wrote:


      *******BEGIN******

      I have read the interesting paper of Pinkernell, Journal of Geometry, vol
      55
      (1996) pp 141-161 and another about general quadratic transformation and I
      arrive
      to the follswing:

      The cubic Kn is the 2-cevian cubic of Pinkernell.

      It's definition could be:

      "The set of points P such that F, P, P' are collinear points",

      where F = X265 and P->P' is the quadratic transformation given by

      (x,y,z) -> ( 1/(x(1-4cosA^2)) , 1/(y(1-4cosB^2)) , 1/(z(1-4cosC^2)) )

      In Pinkernell's paper, you can see the definition of d-pedal cubic, they
      form the
      Euler-pencil (locus of the pivot = Euler line), Pinkernell give a geometric
      construction.

      He defines, for each d-pedal cubic, a corresponding d-cevian cubic, (the
      locus of
      the pivot is jerabek'hyperbola )

      and a 2-1 correspondence, the d-pedal and the (-d)-pedal corresponds to the
      same
      d-cevian.

      Example:

      1-pedal is Darboux.
      2-pedal is Neuberg.
      0-pedal is MacKay.
      (-1)-pedal is Thomson (= 17-point)
      infinite-pedal is ortho-cubic.

      0-cevian = 0-pedal.
      1-cevian = Lucas.
      2-cevian = Kn.
      infinite-cevian = threee altitudes of ABC.

      Is this not funny ?

      ******END*******

      You have a high production of problems.

      Regards Fred.


      xpolakis@... wrote :

      > I wrote:
      >
      > >Darboux Cubic:
      > >Let ABC be a triangle, and A'B'C' the pedal triangle of a point P.
      > >The locus of the points P such that A'B'C' is in perspective with ABC
      > >(or in other words, A'B'C' is the cevian triangle of some point Q)
      > >is a cubic: Darboux cubic.
      > >
      > >Generalization:
      > [...]
      >
      > Another one:
      > Let ABC be a triangle, and A'B'C' the pedal triangle of a point P.
      > Let Pa, Pb, Pc be three points on the lines PA', PB', PC' respectively
      > such that PPa, PPb, PPc be proportional to line segments PA', PB', PC',
      > that is PPa / PA' = PPb / PB' = PPc / PC' := t.
      > For t fixed, the locus of P such that PaPbPc be in perspective with
      > ABC is an isogonal cubic with pivot (cosA - tcosBcosC ::), a point
      > lying on the Euler line.
      > [or to be precise: that's what I found by calculations made by hand]:
      >
      > Equation:
      > x(y^2 - z^2)(cosA - tcosBcosC) + [cyclically] = 0
      >
      > For t = 1 (that is Pa = A', Pb = B', Pc = C') we have the Darboux cubic.
      > [the pivot is the de Longchamps point].
      >
      > For t = -2 we have the Napoleon cubic
      > [the pivot is the 9PC center N = (cos(B-C) ::)]
      >
      > and so on!
      >
      > Antreas
      >
    • Francisco Javier
      Here is an old unanswered Hyacinthos message: APH: Let P be a point on the plane of triangle ABC. The parallels to AB, AC through P intersect BC at Ab, Ac
      Message 42 of 42 , Jan 16, 2011
        Here is an old unanswered Hyacinthos message:

        APH:

        Let P be a point on the plane of triangle ABC.
        The parallels to AB, AC through P intersect BC at Ab, Ac resp.
        Let A1,A2 be the traces of the circumcenter,incenter of PAbAc on AbAc.

        Similarly we define the points B1,B2; C1,C2.

        Which are the loci of P such that (1) A1B1C1 (2) A2B2C2 be in perspective with ABC?

        ---

        I found that the loci are the following cubics:

        (1) K044 - Euler Central Cubic
        (2) K033 - Spieker Central Cubic

        Changing circumcenter, incenter by an arbitrary point Q=(u:v:w), the locus is the cubic

        (cyclic sum) (u(v+w)y z((u+v-w)y -(u-v+w)z))) = 0

        Central cubic with center the complement of Q.

        --- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:
        >
        > Let P be a point on the plane of triangle ABC.
        > The parallels to AB, AC through P intersect BC at Ab, Ac resp.
        > Let A1,A2 be the traces of the circumcenter,incenter of PAbAc on AbAc.
        > Similarly we define the points B1,B2; C1,C2.
        >
        > Which are the loci of P such that (1) A1B1C1 (2) A2B2C2 be in perspective
        > with ABC?
        >
        > I found that the loci are these cubics:
        >
        > xcos(B+u) + zcosu ycos(C+v) + xcosv zcos(A+w) + ycosw
        > ----------------- * ------------------ * ------------------ = 1
        > xcos(C-u) + ycosu ycos(A-v) + zcosv zcos(B-w) + xcosw
        >
        > where u := B-C, v := C-A, w := A-B for the (1), and
        > 2u := B-C, 2v := C-A, 2w := A-B for the (2).
        >
        > But maybe these are Darboux cubics of some triangles of the reference
        > triangle ABC...
        >
        > Antreas
        >
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