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Parallels through P

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  • Antreas P. Hatzipolakis
    Let ABC be a triangle, and P a point. The parallel through P to BC intersects AB, AC at Ab, Ac, resp. The parallel through P to CA intersects BC, BA at Bc, Ba,
    Message 1 of 2 , Apr 29, 2004
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      Let ABC be a triangle, and P a point.

      The parallel through P to BC intersects AB, AC at Ab, Ac, resp.

      The parallel through P to CA intersects BC, BA at Bc, Ba, resp.

      The parallel through P to AB intersects CA, CB at Ca, Cb, resp.

      1. For which P we have that BcCb = CaAc = AbBa ?

      2. Which is the locus of P such that

      k * (BcCb + CaAc + AbBa) = (BC + CA + AB) ?

      where k = given real


      APH
      --
    • Darij Grinberg
      Dear Antreas, ... For any point P with homogeneous trilinears P ( x : y : z ), we have BcCb = a^2 x / (ax + by + cz), and similarly we can find CaAc and AbBa.
      Message 2 of 2 , Apr 29, 2004
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        Dear Antreas,

        In Hyacinthos message #9756, you wrote:

        >> The parallel through P to BC intersects AB,
        >> AC at Ab, Ac, resp.
        >>
        >> The parallel through P to CA intersects BC,
        >> BA at Bc, Ba, resp.
        >>
        >> The parallel through P to AB intersects CA,
        >> CB at Ca, Cb, resp.

        For any point P with homogeneous trilinears
        P ( x : y : z ), we have

        BcCb = a^2 x / (ax + by + cz),

        and similarly we can find CaAc and AbBa.

        >> 1. For which P we have that BcCb = CaAc = AbBa ?

        This is for P = X(75), the isotomic conjugate
        of the incenter of triangle ABC. In this case,

        BcCb = CaAc = AbBa = 1 / (1/a + 1/b + 1/c).

        >> 2. Which is the locus of P such that
        >>
        >> k * (BcCb + CaAc + AbBa) = (BC + CA + AB) ?
        >>
        >> where k = given real

        This locus is the line with the trilinears

        ( ka^2 - a(a+b+c) : kb^2 - b(a+b+c) : kc^2 - c(a+b+c) ).

        Especially, for k = 1, we get the line with
        the trilinears

        ( a(b+c) : b(c+a) : c(a+b) ).

        And for k = infinity, i. e. for
        BcCb + CaAc + AbBa = 0, we get the line with
        the trilinears

        ( a^2 : b^2 : c^2 ),

        i. e. the isogonal tripolar of X(75).

        Nice problem, Antreas!

        Friendly,
        Darij Grinberg
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