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Re: [EMHL] Cubics

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  • xpolakis@otenet.gr
    ... [APH] ... The circumradius of this triangle is R = R/4cosAcosBcosC. Probably some of its notable points (O,H, etc) is a notable point for the reference
    Message 1 of 42 , Nov 6, 2000
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      [BG]:

      >>2). 3 circles, one through B & C and tangent at B to OB, at C to OC
      >>the other two defined similarly.

      [APH]
      >Now, let Ka, Kb, Kc be the centers of the three circles.
      >The triangle KaKbKc in general is not in perspective with ABC

      The circumradius of this triangle is R' = R/4cosAcosBcosC.
      Probably some of its notable points (O,H, etc) is a notable
      point for the reference triangle (ABC) too.

      [APH]:
      >Let P be a point on the plane of a triangle ABC, and Ab, Ac the centers
      >of the circles passing through B, C and tangent to PB,PC at B, C resp.
      >Let A' = BAb /\ CAc. Similarly we define B', C'.
      >Which is the locus of the points P such that A'B'C' be in perspective
      >with ABC? (locus of perspectors?)
      >
      >More simply (since AAb is _|_ to PB, and CCc is _|_ to PC):
      ^^^ ^^^
      BAb CAc

      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
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        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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