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Circumcenter of N1N2N3 [EMHL] Re: Radical axes, NPCs, Os

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  • rhutson2
    For P = O, Op = complement of X(157). In general, Op = complement of the nine-point-center of the antipedal triangle of P. Randy
    Message 1 of 18 , Apr 15, 2013
      For P = O, Op = complement of X(157).

      In general, Op = complement of the nine-point-center of the antipedal triangle of P.

      Randy

      --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
      >
      > Let ABC be a triangle, P a point and N1,N2,N3 the NPC centers
      > of PBC, PCA, PAB, resp.
      >
      > Which is the circumcenter Op of N1N2N3 ?
      >
      > P = H. Op = N [N1 = N2 = N3 = N (N1N2N3 is degenerated)]
      >
      > P = I. Op = N
      >
      > P = O. Op lies on the line NU, where U is the Poncelet point of O
      > wrt ABC (ie the point where concur the NPCs of OBC,OCA,OAB and ABC)
      > Coordinates???
      >
      > P = N. Op lies on the Euler line of ABC.
      > Coordinates???
      >
      > If P describes a line (Euler line, etc) which is the locus of Op?
      >
      > If P describes the circumcircle, Op is the Poncelet point of P
      > wrt ABC. If P describes other curves?
      >
      > Figures:
      > http://anthrakitis.blogspot.gr/2013/04/circumcenter-of-n1n2n3.html
      >
      > APH
      >
      > [APH]
      > > Let ABC be a triangle.
      > >
      > > Denote:
      > >
      > > N1,N2,N3 = The NPC centers of IBC, ICA, IAB, resp.
      >
      > >
      > > 2. The Circumcenter of N1N2N3 is N, the NPC center of ABC.
      >
    • Antreas Hatzipolakis
      ... Euler line (where A1,B1,C1 are the NPC centers of PBC, PCA, PAB)? ... I think it is the same with Bernard s Q038:
      Message 2 of 18 , Apr 16, 2013
        > [APH]
        >
        > Which is the locus of P such that the circumcenter of A1B1C1 is on the
        Euler line (where A1,B1,C1 are the NPC centers of PBC, PCA, PAB)?

        [Francisco]:

        > It is a circular quintic through I, H and N.
        > The other three intersection points with Euler line are X30
        > (the infinite point) and the instersections with the
        > circumcircle X1113 and X1114.
        > Equation:
        >
        > a^6 c^2 x^3 y^2 - 3 a^4 b^2 c^2 x^3 y^2 + 3 a^2 b^4 c^2 x^3 y^2 -
        > b^6 c^2 x^3 y^2 - a^4 c^4 x^3 y^2 + a^2 b^2 c^4 x^3 y^2 -
        > a^2 c^6 x^3 y^2 + c^8 x^3 y^2 + a^6 c^2 x^2 y^3 -
        > 3 a^4 b^2 c^2 x^2 y^3 + 3 a^2 b^4 c^2 x^2 y^3 - b^6 c^2 x^2 y^3 -
        > a^2 b^2 c^4 x^2 y^3 + b^4 c^4 x^2 y^3 + b^2 c^6 x^2 y^3 -
        > c^8 x^2 y^3 + 3 a^2 b^4 c^2 x^3 y z - 3 b^6 c^2 x^3 y z -
        > 3 a^2 b^2 c^4 x^3 y z + 3 b^2 c^6 x^3 y z + 2 a^6 c^2 x^2 y^2 z -
        > 2 b^6 c^2 x^2 y^2 z - 4 a^4 c^4 x^2 y^2 z + 4 b^4 c^4 x^2 y^2 z +
        > 2 a^2 c^6 x^2 y^2 z - 2 b^2 c^6 x^2 y^2 z + 3 a^6 c^2 x y^3 z -
        > 3 a^4 b^2 c^2 x y^3 z + 3 a^2 b^2 c^4 x y^3 z - 3 a^2 c^6 x y^3 z -
        > a^6 b^2 x^3 z^2 + a^4 b^4 x^3 z^2 + a^2 b^6 x^3 z^2 - b^8 x^3 z^2 +
        > 3 a^4 b^2 c^2 x^3 z^2 - a^2 b^4 c^2 x^3 z^2 - 3 a^2 b^2 c^4 x^3 z^2 +
        > b^2 c^6 x^3 z^2 - 2 a^6 b^2 x^2 y z^2 + 4 a^4 b^4 x^2 y z^2 -
        > 2 a^2 b^6 x^2 y z^2 + 2 b^6 c^2 x^2 y z^2 - 4 b^4 c^4 x^2 y z^2 +
        > 2 b^2 c^6 x^2 y z^2 + 2 a^6 b^2 x y^2 z^2 - 4 a^4 b^4 x y^2 z^2 +
        > 2 a^2 b^6 x y^2 z^2 - 2 a^6 c^2 x y^2 z^2 + 4 a^4 c^4 x y^2 z^2 -
        > 2 a^2 c^6 x y^2 z^2 + a^8 y^3 z^2 - a^6 b^2 y^3 z^2 -
        > a^4 b^4 y^3 z^2 + a^2 b^6 y^3 z^2 + a^4 b^2 c^2 y^3 z^2 -
        > 3 a^2 b^4 c^2 y^3 z^2 + 3 a^2 b^2 c^4 y^3 z^2 - a^2 c^6 y^3 z^2 -
        > a^6 b^2 x^2 z^3 + b^8 x^2 z^3 + 3 a^4 b^2 c^2 x^2 z^3 +
        > a^2 b^4 c^2 x^2 z^3 - b^6 c^2 x^2 z^3 - 3 a^2 b^2 c^4 x^2 z^3 -
        > b^4 c^4 x^2 z^3 + b^2 c^6 x^2 z^3 - 3 a^6 b^2 x y z^3 +
        > 3 a^2 b^6 x y z^3 + 3 a^4 b^2 c^2 x y z^3 - 3 a^2 b^4 c^2 x y z^3 -
        > a^8 y^2 z^3 + a^2 b^6 y^2 z^3 + a^6 c^2 y^2 z^3 -
        > a^4 b^2 c^2 y^2 z^3 - 3 a^2 b^4 c^2 y^2 z^3 + a^4 c^4 y^2 z^3 +
        > 3 a^2 b^2 c^4 y^2 z^3 - a^2 c^6 y^2 z^3 = 0


        I think it is the same with Bernard's Q038:
        http://bernard.gibert.pagesperso-orange.fr/curves/q038.html

        If yes, Bernard may add that property of the quintic.

        APH


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