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Re: Radical axes - locus

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  • Angel
    Dear Antreas (Oa) = the circumcircle of the triangle bounded by the lines (AB,AC, parallel_to_B C _through_A ) and similarly (Ob), (Oc). (O1),(O2),(O3) = the
    Message 1 of 8 , Feb 19, 2013
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      Dear Antreas

      (Oa) = the circumcircle of the triangle bounded by the lines (AB,AC,
      parallel_to_B'C'_through_A') and similarly (Ob), (Oc).

      (O1),(O2),(O3) = the circles with diameters AA',BB',CC', resp.

      Are the triangles ABC, triangle bounded by the radical axes ((Oa),(O1)),((Ob),(O2)), ((Oc),(O3)), perspective?

      Yes, If P is on the medians of ABC, or Lucas cubic, or sextic:

      a^4c^2x^3y^3 - 2a^2b^2c^2x^3y^3 + b^4c^2x^3y^3 -
      c^6x^3y^3 - a^4c^2x^3y^2z + b^4c^2x^3y^2z + 2a^2c^4x^3y^2z - c^6x^3y^2z + a^4c^2x^2y^3z - b^4c^2x^2y^3z + 2b^2c^4x^2y^3z -
      c^6x^2y^3z - a^4b^2x^3yz^2 + 2a^2b^4x^3yz^2 -
      b^6x^3yz^2 + b^2c^4x^3yz^2 - a^6x^2y^2z^2 +
      a^4b^2x^2y^2z^2 + a^2b^4x^2y^2z^2 - b^6x^2y^2z^2 + a^4c^2x^2y^2z^2 - 2a^2b^2c^2x^2y^2z^2 + b^4c^2x^2y^2z^2 +
      a^2c^4x^2y^2z^2 + b^2c^4x^2y^2z^2 -
      c^6x^2y^2z^2 - a^6xy^3z^2 + 2a^4b^2xy^3z^2 -
      a^2b^4xy^3z^2 + a^2c^4xy^3z^2 + a^4b^2x^3z^3 -
      b^6x^3z^3 - 2a^2b^2c^2x^3z^3 + b^2c^4x^3z^3 +
      a^4b^2x^2yz^3 - b^6x^2yz^3 + 2b^4c^2x^2yz^3 - b^2c^4x^2yz^3 -
      a^6xy^2z^3 + a^2b^4xy^2z^3 + 2a^4c^2xy^2z^3 -
      a^2c^4xy^2z^3 - a^6y^3z^3 + a^2b^4y^3z^3 - 2a^2b^2c^2y^3z^3 + a^2c^4y^3z^3=0.

      Angel Montesdeoca
    • Angel
      The locus of P, so that the triangles ABC and triangle bounded by the radical axes ((Oa),(O1)),((Ob),(O2)), ((Oc),(O3)) are perspective is the Lucas cubic and
      Message 2 of 8 , Feb 20, 2013
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        The locus of P, so that the triangles ABC and triangle bounded by the radical axes ((Oa),(O1)),((Ob),(O2)), ((Oc),(O3)) are perspective is the Lucas cubic and sextic said. Please forget about the medians of ABC, it is my mistake.

        Angel M.

        --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
        >
        >
        > Dear Antreas
        >
        > (Oa) = the circumcircle of the triangle bounded by the lines (AB,AC,
        > parallel_to_B'C'_through_A') and similarly (Ob), (Oc).
        >
        > (O1),(O2),(O3) = the circles with diameters AA',BB',CC', resp.
        >
        > Are the triangles ABC, triangle bounded by the radical axes ((Oa),(O1)),((Ob),(O2)), ((Oc),(O3)), perspective?
        >
        > Yes, If P is on the medians of ABC, or Lucas cubic, or sextic:
        >
        > a^4c^2x^3y^3 - 2a^2b^2c^2x^3y^3 + b^4c^2x^3y^3 -
        > c^6x^3y^3 - a^4c^2x^3y^2z + b^4c^2x^3y^2z + 2a^2c^4x^3y^2z - c^6x^3y^2z + a^4c^2x^2y^3z - b^4c^2x^2y^3z + 2b^2c^4x^2y^3z -
        > c^6x^2y^3z - a^4b^2x^3yz^2 + 2a^2b^4x^3yz^2 -
        > b^6x^3yz^2 + b^2c^4x^3yz^2 - a^6x^2y^2z^2 +
        > a^4b^2x^2y^2z^2 + a^2b^4x^2y^2z^2 - b^6x^2y^2z^2 + a^4c^2x^2y^2z^2 - 2a^2b^2c^2x^2y^2z^2 + b^4c^2x^2y^2z^2 +
        > a^2c^4x^2y^2z^2 + b^2c^4x^2y^2z^2 -
        > c^6x^2y^2z^2 - a^6xy^3z^2 + 2a^4b^2xy^3z^2 -
        > a^2b^4xy^3z^2 + a^2c^4xy^3z^2 + a^4b^2x^3z^3 -
        > b^6x^3z^3 - 2a^2b^2c^2x^3z^3 + b^2c^4x^3z^3 +
        > a^4b^2x^2yz^3 - b^6x^2yz^3 + 2b^4c^2x^2yz^3 - b^2c^4x^2yz^3 -
        > a^6xy^2z^3 + a^2b^4xy^2z^3 + 2a^4c^2xy^2z^3 -
        > a^2c^4xy^2z^3 - a^6y^3z^3 + a^2b^4y^3z^3 - 2a^2b^2c^2y^3z^3 + a^2c^4y^3z^3=0.
        >
        > Angel Montesdeoca
        >
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