Loading ...
Sorry, an error occurred while loading the content.

Re: Problem from Vietnam Team Selection Tests 2009

Expand Messages
  • garciacapitan
    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
    Message 1 of 10 , Aug 1, 2009
    • 0 Attachment
      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.
      >
    • Moses, Peter J. C.
      Hi Francisco, Your conic is the Feuerbach circumhyperbola of the Tangential triangle, sometimes called the Stammler hyperbola. It contains X{1, 3, 6, 155,
      Message 2 of 10 , Aug 1, 2009
      • 0 Attachment
        Hi Francisco,

        Your conic is the Feuerbach circumhyperbola of the Tangential triangle, sometimes called the Stammler hyperbola. It contains X{1, 3, 6, 155, 159, 195, 399, 610, 1498, 1740, 2574, 2575, 2916, 2917, 2918, 2929, 2930, 2931, 2935, 2948, 3216, 3360, 3499, 3511}, the ex-centers & the centers of the Stammler circles.

        Best regards,
        Peter.


        ----- Original Message -----
        From: garciacapitan
        To: Hyacinthos@yahoogroups.com
        Sent: Saturday, August 01, 2009 11:11 AM
        Subject: [EMHL] Re: Problem from Vietnam Team Selection Tests 2009



        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.
        >





        [Non-text portions of this message have been removed]
      • xpolakis
        ... Let ABC be a triangle, P a point, A B C the cevian triangle of P and A ,B ,C the second intersections of the circumcircle of ABC with the circumcircles
        Message 3 of 10 , Aug 2, 2009
        • 0 Attachment
          > Another problem with the "Vietnameze" triangle.

          Let ABC be a triangle, P a point, A'B'C' the cevian triangle
          of P and A",B",C" the second intersections of the circumcircle
          of ABC with the circumcircles of AB'C', BC'A', CA'B', resp.

          Denote:

          A* := BB" /\ CC"

          B* := CC" /\ AA"

          C* := AA" /\ BB"

          Which is the locus of P such that
          ABC, A*B*C* are perspective?

          APH
        • Quang Tuan Bui
          Dear Antreas, They are always perspective at a point with barycentrics: a^2 /(q + r) : : Best regards, Bui Quang Tuan
          Message 4 of 10 , Aug 2, 2009
          • 0 Attachment
            Dear Antreas,
            They are always perspective at a point with barycentrics:

            a^2 /(q + r) : :

            Best regards,
            Bui Quang Tuan

            --- On Mon, 8/3/09, xpolakis <anopolis72@...> wrote:

            > From: xpolakis <anopolis72@...>
            > Subject: [EMHL] Re: Problem from Vietnam Team Selection Tests 2009
            > To: Hyacinthos@yahoogroups.com
            > Date: Monday, August 3, 2009, 12:47 AM
            > > Another problem with the
            > "Vietnameze" triangle.
            >
            > Let ABC be a triangle, P a point, A'B'C' the cevian
            > triangle
            > of P and A",B",C" the second intersections of the
            > circumcircle
            > of ABC with the circumcircles of AB'C', BC'A', CA'B',
            > resp.
            >
            > Denote:
            >
            > A* := BB" /\ CC"
            >
            > B* := CC" /\ AA"
            >
            > C* := AA" /\ BB"
            >
            > Which is the locus of P such that
            > ABC, A*B*C* are perspective?
            >
            > APH 
            >
          • xpolakis
            ... Let ABC be a triangle, P a point, A B C the cevian triangle of P and A ,B ,C the second intersections of the circumcircle of ABC with the circumcircles
            Message 5 of 10 , Aug 2, 2009
            • 0 Attachment
              > Another problem with the "Vietnameze" triangle.

              Let ABC be a triangle, P a point, A'B'C' the cevian triangle
              of P and A",B",C" the second intersections of the circumcircle
              of ABC with the circumcircles of AB'C', BC'A', CA'B', resp.

              For which P's the triangle A"B"C" is equilateral?

              APH
            Your message has been successfully submitted and would be delivered to recipients shortly.