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19751Re: Cubics

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  • Francisco Javier
    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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