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18089Re: Problem from Vietnam Team Selection Tests 2009

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  • garciacapitan
    Aug 1, 2009
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      Dear Antreas,

      The locus is formed by the circumcircle and the quartic
      -a^2 c^2 x^2 y^2 + b^2 c^2 x^2 y^2 + a^2 b^2 x^2 z^2 - b^2 c^2 x^2 z^2 - a^2 b^2 y^2 z^2 + a^2 c^2 y^2 z^2 = 0.

      This quartic is the isogonal conjugate of the conic

      b^4 c^2 x^2 - b^2 c^4 x^2 - a^4 c^2 y^2 + a^2 c^4 y^2 + a^4 b^2 z^2 - a^2 b^4 z^2 = 0

      This conic, always an hyperbola, is centered at X110, the focus of the Kiepert parabola. Some points on this hyperbola include

      X1
      X3
      X6
      X155
      X159
      X195
      X399
      X610

      Best regards from Almería, Spain.

      --- In Hyacinthos@yahoogroups.com, "xpolakis" <anopolis72@...> wrote:
      >
      >
      > Another problem with the "Vietnameze" triangle.
      >
      > Let ABC be a triangle, P a point, A1B1C1 the circumcevian
      > triangle of P and A2B2C2 the cevian triangle of P.
      > Let A3,B3,C3 be the second intersections of the circumcircle
      > of ABC with the circumcircles of AB2C2, BC2A2, CA2B2, resp.
      >
      > Which is the locus of P such that A1B1C1, A3B3C3 are
      > perspective?
      >
      > Antreas
      >
      > [APH]
      > > Let an acute triangle ABC with curcumcircle (O). Call A_1,B_1,C_1
      > > are foots of perpendicular line from A,B,C to opposite side.
      > > A_2,B_2,C_2 are reflect points of A_1,B_1,C_1 over midpoints
      > > of BC,CA,AB respectively.
      > > Circle (AB_2C_2),(BC_2A_2),(CA_2B_2) cut (O) at A_3,B_3,C_3
      > > respectively.
      > > Prove that: A_1A_3,B_1B_3,C_1C_3 are concurent.
      > >
      > > Vietnam Team Selection Tests 2009
      > > http://www.mathlinks.ro/resources.php?c=186&cid=41&year=2009
      > >
      > >
      > > In other words:
      > >
      > > Let ABC be a triangle, A1B1C1 the cevian triangle of H, A2B2C2 the
      > > cevian triangle of the isotomic conjugate of H, and A3,B3,C3
      > > the second intersections of the circumcircle of ABC
      > > with the circumcircles of AB2C2, BC2A2, CA2B2, resp.
      > >
      > > The triangles A1B1C1, A3B3C3 are perspective.
      >
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