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Re: NPC-related conic

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  • Francisco Javier
    Dear Randy, you can take A B C the medial triangle and A B C the cevian triangle of any point (p:q:r) and this gives a conic with center {(p + q) (p + r) (10
    Message 1 of 3 , Feb 24, 2013
      Dear Randy, you can take A'B'C' the medial triangle and A"B"C" the cevian triangle of any point (p:q:r) and this gives a conic with center

      {(p + q) (p + r) (10 p^3 q^2 + 25 p^2 q^3 + 5 p q^4 + 12 p^3 q r +
      87 p^2 q^2 r + 58 p q^3 r + 3 q^4 r + 10 p^3 r^2 + 87 p^2 q r^2 +
      98 p q^2 r^2 + 13 q^3 r^2 + 25 p^2 r^3 + 58 p q r^3 +
      13 q^2 r^3 + 5 p r^4 + 3 q r^4), (p + q) (q + r) (5 p^4 q +
      25 p^3 q^2 + 10 p^2 q^3 + 3 p^4 r + 58 p^3 q r + 87 p^2 q^2 r +
      12 p q^3 r + 13 p^3 r^2 + 98 p^2 q r^2 + 87 p q^2 r^2 +
      10 q^3 r^2 + 13 p^2 r^3 + 58 p q r^3 + 25 q^2 r^3 + 3 p r^4 +
      5 q r^4), (p + r) (q + r) (3 p^4 q + 13 p^3 q^2 + 13 p^2 q^3 +
      3 p q^4 + 5 p^4 r + 58 p^3 q r + 98 p^2 q^2 r + 58 p q^3 r +
      5 q^4 r + 25 p^3 r^2 + 87 p^2 q r^2 + 87 p q^2 r^2 + 25 q^3 r^2 +
      10 p^2 r^3 + 12 p q r^3 + 10 q^2 r^3)}

      that is homothetic to the circumconic

      (3 p + q) (p + 3 q) r (p + r) (q + r) x y +
      q (p + q) (3 p + r) (q + r) (p + 3 r) x z +
      p (p + q) (p + r) (3 q + r) (q + 3 r) y z = 0.

      Best regards,

      Francisco Javier.







      --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
      >
      > Let A'B'C' be the medial triangle and A"B"C" be the orthic triangle.
      > (So A', A" are the intersections of the NPC with BC, etc.)
      > Let Ab be the {A',A"}-Harmonic conjugate of B, and let Ac be the {A',A"}-Harmonic conjugate of C. Define Bc, Ba, Ca, Cb cyclically.
      >
      > Ab, Ac, Bc, Ba, Ca, Cb lie on a common conic.
      >
      > What are the coordinates of the conic center?
      >
      > Best regards,
      > Randy
      >
    • Randy Hutson
      Thank you, Francisco!  This could be further generalized by letting A B C and A B C be any 2 cevian triangles. Rsndy ... [Non-text portions of this message
      Message 2 of 3 , Feb 25, 2013
        Thank you, Francisco!  This could be further generalized by letting A'B'C' and A"B"C" be any 2 cevian triangles.

        Rsndy





        >________________________________
        > From: Francisco Javier <garciacapitan@...>
        >To: Hyacinthos@yahoogroups.com
        >Sent: Sunday, February 24, 2013 2:14 AM
        >Subject: [EMHL] Re: NPC-related conic
        >
        >

        >Dear Randy, you can take A'B'C' the medial triangle and A"B"C" the cevian triangle of any point (p:q:r) and this gives a conic with center
        >
        >{(p + q) (p + r) (10 p^3 q^2 + 25 p^2 q^3 + 5 p q^4 + 12 p^3 q r +
        >87 p^2 q^2 r + 58 p q^3 r + 3 q^4 r + 10 p^3 r^2 + 87 p^2 q r^2 +
        >98 p q^2 r^2 + 13 q^3 r^2 + 25 p^2 r^3 + 58 p q r^3 +
        >13 q^2 r^3 + 5 p r^4 + 3 q r^4), (p + q) (q + r) (5 p^4 q +
        >25 p^3 q^2 + 10 p^2 q^3 + 3 p^4 r + 58 p^3 q r + 87 p^2 q^2 r +
        >12 p q^3 r + 13 p^3 r^2 + 98 p^2 q r^2 + 87 p q^2 r^2 +
        >10 q^3 r^2 + 13 p^2 r^3 + 58 p q r^3 + 25 q^2 r^3 + 3 p r^4 +
        >5 q r^4), (p + r) (q + r) (3 p^4 q + 13 p^3 q^2 + 13 p^2 q^3 +
        >3 p q^4 + 5 p^4 r + 58 p^3 q r + 98 p^2 q^2 r + 58 p q^3 r +
        >5 q^4 r + 25 p^3 r^2 + 87 p^2 q r^2 + 87 p q^2 r^2 + 25 q^3 r^2 +
        >10 p^2 r^3 + 12 p q r^3 + 10 q^2 r^3)}
        >
        >that is homothetic to the circumconic
        >
        >(3 p + q) (p + 3 q) r (p + r) (q + r) x y +
        >q (p + q) (3 p + r) (q + r) (p + 3 r) x z +
        >p (p + q) (p + r) (3 q + r) (q + 3 r) y z = 0.
        >
        >Best regards,
        >
        >Francisco Javier.
        >
        >
        >--- In Hyacinthos@yahoogroups.com, "rhutson2" wrote:
        >>
        >> Let A'B'C' be the medial triangle and A"B"C" be the orthic triangle.
        >> (So A', A" are the intersections of the NPC with BC, etc.)
        >> Let Ab be the {A',A"}-Harmonic conjugate of B, and let Ac be the {A',A"}-Harmonic conjugate of C. Define Bc, Ba, Ca, Cb cyclically.
        >>
        >> Ab, Ac, Bc, Ba, Ca, Cb lie on a common conic.
        >>
        >> What are the coordinates of the conic center?
        >>
        >> Best regards,
        >> Randy
        >>
        >
        >
        >
        >
        >

        [Non-text portions of this message have been removed]
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