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Re: [EMHL] Re: Mccay ?

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  • Bernard Gibert
    Dear Francisco, ... Your S is probably the actual area of ABC and not Conway s S i.e. twice the area. Anyway, your cubic is a nK(X6, X6, ?) without known
    Message 1 of 10 , Apr 5, 2009
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      Dear Francisco,

      > No, I see that the cubic with equation
      >
      > 4 S^2 xyz + CyclicSum(a^2 y z(c^2y+b^2z)) = 0 (*)
      >
      > also passes through the centers of Apollonian circles, but it is not
      > K192.

      Your "S" is probably the actual area of ABC and not Conway's S i.e.
      twice the area.

      Anyway, your cubic is a nK(X6, X6, ?) without known center on it and
      without any remarkable other property as far as I can see.

      Best regards

      Bernard




      [Non-text portions of this message have been removed]
    • garciacapitan
      Yes, S was the area of ABC, like in your K192 page.
      Message 2 of 10 , Apr 5, 2009
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        Yes, S was the area of ABC, like in your K192 page.

        --- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
        >
        > Dear Francisco,
        >
        > > No, I see that the cubic with equation
        > >
        > > 4 S^2 xyz + CyclicSum(a^2 y z(c^2y+b^2z)) = 0 (*)
        > >
        > > also passes through the centers of Apollonian circles, but it is not
        > > K192.
        >
        > Your "S" is probably the actual area of ABC and not Conway's S i.e.
        > twice the area.
        >
        > Anyway, your cubic is a nK(X6, X6, ?) without known center on it and
        > without any remarkable other property as far as I can see.
        >
        > Best regards
        >
        > Bernard
        >
        >
        >
        >
        > [Non-text portions of this message have been removed]
        >
      • xpolakis
        Dear Francisco Thanks! Here is a similar problem: Let ABC be a triangle, P a point and A B C , A B C its pedal, circumcevian triangles, resp. Let O be the
        Message 3 of 10 , Apr 6, 2009
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          Dear Francisco

          Thanks!

          Here is a similar problem:

          Let ABC be a triangle, P a point and A'B'C', A"B"C"
          its pedal, circumcevian triangles, resp.

          Let O' be the center of the pedal circle of P, and
          A*B*C* the circumcevian of O' wrt A'B'C' in the pedal
          circle of P. (ie A*,B*,C* are the antipodes of A',B',C'
          in the pedal circle).

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

          Antreas

          [APH]
          > > Let ABC be a triangle, P,P* two isogonal conjugate points
          > > and A*B*C* the pedal triangle of P*.
          > >
          > > Denote:
          > >
          > > A'B'C' := the circumcevian triangle of P wrt ABC
          > > (in the circumcircle of ABC)
          > >
          > > A"B"C" := the circumcevian triangle of P* wrt A*B*C* in
          > > the pedal circle of P*
          > >
          > > (ie A',B',C' are the second intersections of AP,BP,CP
          > > with
          > > the circumcircle of ABC, and A",B",C" are the second
          > > intersections
          > > of P*A*,PB*,P*C* with the circumcircle of A*B*C*).
          > >
          > > Which is the locus of P such that A'B'C', A"B"C" are perspective?
          > >
          > > (For P = O or H the perspector is G)

          [Francisco Javier]
          > Yes, I get the McCay cubic, also with the line at infinity,
          > the circumcircle and the cubic with equation
          >
          > 4 S^2 xyz + CyclicSum(a^2 yz(c^2y+b^2z)) = 0
          >
          > Surely Bernard can give this cubic a name.
        • garciacapitan
          In this case we get the sidelines, the line at infinity, the circumcircle and a septic trhough the orthocenter, the incenter, the excenters and the feet of the
          Message 4 of 10 , Apr 6, 2009
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            In this case we get the sidelines, the line at infinity, the circumcircle and a septic trhough the orthocenter, the incenter, the excenters and the feet of the altitudes

            2 a^4 c^6 x^4 y^3 - 2 b^4 c^6 x^4 y^3 - 4 a^2 c^8 x^4 y^3 +
            2 c^10 x^4 y^3 + 2 a^4 c^6 x^3 y^4 - 2 b^4 c^6 x^3 y^4 +
            4 b^2 c^8 x^3 y^4 - 2 c^10 x^3 y^4 - a^6 b^2 c^2 x^4 y^2 z +
            3 a^4 b^4 c^2 x^4 y^2 z - 3 a^2 b^6 c^2 x^4 y^2 z +
            b^8 c^2 x^4 y^2 z + 3 a^4 b^2 c^4 x^4 y^2 z +
            6 a^2 b^4 c^4 x^4 y^2 z - 9 b^6 c^4 x^4 y^2 z -
            7 a^2 b^2 c^6 x^4 y^2 z + 3 b^4 c^6 x^4 y^2 z + 5 b^2 c^8 x^4 y^2 z -
            a^8 c^2 x^3 y^3 z + 2 a^6 b^2 c^2 x^3 y^3 z -
            2 a^2 b^6 c^2 x^3 y^3 z + b^8 c^2 x^3 y^3 z + 7 a^6 c^4 x^3 y^3 z +
            3 a^4 b^2 c^4 x^3 y^3 z - 3 a^2 b^4 c^4 x^3 y^3 z -
            7 b^6 c^4 x^3 y^3 z - 11 a^4 c^6 x^3 y^3 z + 11 b^4 c^6 x^3 y^3 z +
            5 a^2 c^8 x^3 y^3 z - 5 b^2 c^8 x^3 y^3 z - a^8 c^2 x^2 y^4 z +
            3 a^6 b^2 c^2 x^2 y^4 z - 3 a^4 b^4 c^2 x^2 y^4 z +
            a^2 b^6 c^2 x^2 y^4 z + 9 a^6 c^4 x^2 y^4 z -
            6 a^4 b^2 c^4 x^2 y^4 z - 3 a^2 b^4 c^4 x^2 y^4 z -
            3 a^4 c^6 x^2 y^4 z + 7 a^2 b^2 c^6 x^2 y^4 z - 5 a^2 c^8 x^2 y^4 z +
            a^6 b^2 c^2 x^4 y z^2 - 3 a^4 b^4 c^2 x^4 y z^2 +
            7 a^2 b^6 c^2 x^4 y z^2 - 5 b^8 c^2 x^4 y z^2 -
            3 a^4 b^2 c^4 x^4 y z^2 - 6 a^2 b^4 c^4 x^4 y z^2 -
            3 b^6 c^4 x^4 y z^2 + 3 a^2 b^2 c^6 x^4 y z^2 + 9 b^4 c^6 x^4 y z^2 -
            b^2 c^8 x^4 y z^2 + 12 a^4 b^4 c^2 x^3 y^2 z^2 -
            12 a^2 b^6 c^2 x^3 y^2 z^2 - 12 a^4 b^2 c^4 x^3 y^2 z^2 +
            12 a^2 b^2 c^6 x^3 y^2 z^2 + 12 a^6 b^2 c^2 x^2 y^3 z^2 -
            12 a^4 b^4 c^2 x^2 y^3 z^2 + 12 a^2 b^4 c^4 x^2 y^3 z^2 -
            12 a^2 b^2 c^6 x^2 y^3 z^2 + 5 a^8 c^2 x y^4 z^2 -
            7 a^6 b^2 c^2 x y^4 z^2 + 3 a^4 b^4 c^2 x y^4 z^2 -
            a^2 b^6 c^2 x y^4 z^2 + 3 a^6 c^4 x y^4 z^2 +
            6 a^4 b^2 c^4 x y^4 z^2 + 3 a^2 b^4 c^4 x y^4 z^2 -
            9 a^4 c^6 x y^4 z^2 - 3 a^2 b^2 c^6 x y^4 z^2 + a^2 c^8 x y^4 z^2 -
            2 a^4 b^6 x^4 z^3 + 4 a^2 b^8 x^4 z^3 - 2 b^10 x^4 z^3 +
            2 b^6 c^4 x^4 z^3 + a^8 b^2 x^3 y z^3 - 7 a^6 b^4 x^3 y z^3 +
            11 a^4 b^6 x^3 y z^3 - 5 a^2 b^8 x^3 y z^3 -
            2 a^6 b^2 c^2 x^3 y z^3 - 3 a^4 b^4 c^2 x^3 y z^3 +
            5 b^8 c^2 x^3 y z^3 + 3 a^2 b^4 c^4 x^3 y z^3 -
            11 b^6 c^4 x^3 y z^3 + 2 a^2 b^2 c^6 x^3 y z^3 +
            7 b^4 c^6 x^3 y z^3 - b^2 c^8 x^3 y z^3 -
            12 a^6 b^2 c^2 x^2 y^2 z^3 + 12 a^2 b^6 c^2 x^2 y^2 z^3 +
            12 a^4 b^2 c^4 x^2 y^2 z^3 - 12 a^2 b^4 c^4 x^2 y^2 z^3 +
            5 a^8 b^2 x y^3 z^3 - 11 a^6 b^4 x y^3 z^3 + 7 a^4 b^6 x y^3 z^3 -
            a^2 b^8 x y^3 z^3 - 5 a^8 c^2 x y^3 z^3 + 3 a^4 b^4 c^2 x y^3 z^3 +
            2 a^2 b^6 c^2 x y^3 z^3 + 11 a^6 c^4 x y^3 z^3 -
            3 a^4 b^2 c^4 x y^3 z^3 - 7 a^4 c^6 x y^3 z^3 -
            2 a^2 b^2 c^6 x y^3 z^3 + a^2 c^8 x y^3 z^3 + 2 a^10 y^4 z^3 -
            4 a^8 b^2 y^4 z^3 + 2 a^6 b^4 y^4 z^3 - 2 a^6 c^4 y^4 z^3 -
            2 a^4 b^6 x^3 z^4 + 2 b^10 x^3 z^4 - 4 b^8 c^2 x^3 z^4 +
            2 b^6 c^4 x^3 z^4 + a^8 b^2 x^2 y z^4 - 9 a^6 b^4 x^2 y z^4 +
            3 a^4 b^6 x^2 y z^4 + 5 a^2 b^8 x^2 y z^4 -
            3 a^6 b^2 c^2 x^2 y z^4 + 6 a^4 b^4 c^2 x^2 y z^4 -
            7 a^2 b^6 c^2 x^2 y z^4 + 3 a^4 b^2 c^4 x^2 y z^4 +
            3 a^2 b^4 c^4 x^2 y z^4 - a^2 b^2 c^6 x^2 y z^4 -
            5 a^8 b^2 x y^2 z^4 - 3 a^6 b^4 x y^2 z^4 + 9 a^4 b^6 x y^2 z^4 -
            a^2 b^8 x y^2 z^4 + 7 a^6 b^2 c^2 x y^2 z^4 -
            6 a^4 b^4 c^2 x y^2 z^4 + 3 a^2 b^6 c^2 x y^2 z^4 -
            3 a^4 b^2 c^4 x y^2 z^4 - 3 a^2 b^4 c^4 x y^2 z^4 +
            a^2 b^2 c^6 x y^2 z^4 - 2 a^10 y^3 z^4 + 2 a^6 b^4 y^3 z^4 +
            4 a^8 c^2 y^3 z^4 - 2 a^6 c^4 y^3 z^4

            Best regards,

            Francisco Javier.

            --- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
            >
            > Dear Francisco
            >
            > Thanks!
            >
            > Here is a similar problem:
            >
            > Let ABC be a triangle, P a point and A'B'C', A"B"C"
            > its pedal, circumcevian triangles, resp.
            >
            > Let O' be the center of the pedal circle of P, and
            > A*B*C* the circumcevian of O' wrt A'B'C' in the pedal
            > circle of P. (ie A*,B*,C* are the antipodes of A',B',C'
            > in the pedal circle).
            >
            > Which is the locus of P such that A"B"C", A*B*C* are
            > perspective?
            >
            > Antreas
            >
            > [APH]
            > > > Let ABC be a triangle, P,P* two isogonal conjugate points
            > > > and A*B*C* the pedal triangle of P*.
            > > >
            > > > Denote:
            > > >
            > > > A'B'C' := the circumcevian triangle of P wrt ABC
            > > > (in the circumcircle of ABC)
            > > >
            > > > A"B"C" := the circumcevian triangle of P* wrt A*B*C* in
            > > > the pedal circle of P*
            > > >
            > > > (ie A',B',C' are the second intersections of AP,BP,CP
            > > > with
            > > > the circumcircle of ABC, and A",B",C" are the second
            > > > intersections
            > > > of P*A*,PB*,P*C* with the circumcircle of A*B*C*).
            > > >
            > > > Which is the locus of P such that A'B'C', A"B"C" are perspective?
            > > >
            > > > (For P = O or H the perspector is G)
            >
            > [Francisco Javier]
            > > Yes, I get the McCay cubic, also with the line at infinity,
            > > the circumcircle and the cubic with equation
            > >
            > > 4 S^2 xyz + CyclicSum(a^2 yz(c^2y+b^2z)) = 0
            > >
            > > Surely Bernard can give this cubic a name.
            >
          • xpolakis
            Dear Francisco [APH] ... Hmmm... I don t see O in the locus. If P = O, then A*B*C* is the Euler triangle (A*,B*,C* are the midpoints of AH,BH,CH) and A ,B ,C
            Message 5 of 10 , Apr 6, 2009
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              Dear Francisco

              [APH]
              > > Here is a similar problem:
              > >
              > > Let ABC be a triangle, P a point and A'B'C', A"B"C"
              > > its pedal, circumcevian triangles, resp.
              > >
              > > Let O' be the center of the pedal circle of P, and
              > > A*B*C* the circumcevian of O' wrt A'B'C' in the pedal
              > > circle of P. (ie A*,B*,C* are the antipodes of A',B',C'
              > > in the pedal circle).
              > >
              > > Which is the locus of P such that A"B"C", A*B*C* are
              > > perspective?

              [Francisco Javier]:
              > In this case we get the sidelines, the line at infinity, the >circumcircle and a septic trhough the orthocenter, the incenter, the >excenters and the feet of the altitudes

              Hmmm... I don't see O in the locus.

              If P = O, then A*B*C* is the Euler triangle
              (A*,B*,C* are the midpoints of AH,BH,CH)
              and A",B",C" are the antipodes of A,B,C
              in the circumcircle)
              I think that A"B"C", A*B*C* are perspective
              at G (as in the first case)

              APH
            • garciacapitan
              Dear Antreas, yes, you are right, O is on the locus, but is not true the general case that if P is on the locus then the same occur with its isogonal
              Message 6 of 10 , Apr 6, 2009
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                Dear Antreas, yes, you are right, O is on the locus, but is not true the general case that if P is on the locus then the same occur with its isogonal conjugate.


                --- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
                >
                > Dear Francisco
                >
                > [APH]
                > > > Here is a similar problem:
                > > >
                > > > Let ABC be a triangle, P a point and A'B'C', A"B"C"
                > > > its pedal, circumcevian triangles, resp.
                > > >
                > > > Let O' be the center of the pedal circle of P, and
                > > > A*B*C* the circumcevian of O' wrt A'B'C' in the pedal
                > > > circle of P. (ie A*,B*,C* are the antipodes of A',B',C'
                > > > in the pedal circle).
                > > >
                > > > Which is the locus of P such that A"B"C", A*B*C* are
                > > > perspective?
                >
                > [Francisco Javier]:
                > > In this case we get the sidelines, the line at infinity, the >circumcircle and a septic trhough the orthocenter, the incenter, the >excenters and the feet of the altitudes
                >
                > Hmmm... I don't see O in the locus.
                >
                > If P = O, then A*B*C* is the Euler triangle
                > (A*,B*,C* are the midpoints of AH,BH,CH)
                > and A",B",C" are the antipodes of A,B,C
                > in the circumcircle)
                > I think that A"B"C", A*B*C* are perspective
                > at G (as in the first case)
                >
                > APH
                >
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