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[EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)

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  • Antreas
    Dear Randy Note that the conjecture is not true if one of the six triangles A B C ,...., A B C is degenerated (if A ,B ,C etc are collinear). Figures here
    Message 1 of 9 , Jan 6, 2013
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      Dear Randy

      Note that the conjecture is not true if one of the six triangles A"B'C',...., A'B"C" is degenerated (if A",B',C' etc are collinear).

      Figures here
      http://tech.groups.yahoo.com/group/Hyacinthos/message/21357

      APH

      --- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
      >
      > Antreas,
      >
      > I was able to confirm this computationally.  Of course, not the same as a proof.
      >
      > Randy Hutson
      >
      >
      >
      >
      >
      > >________________________________
      > > From: Antreas
      > >To: Hyacinthos@yahoogroups.com
      > >Sent: Friday, January 4, 2013 12:56 PM
      > >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
      > >
      > >
      > > 
      > >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
      > >
      > >Conjecture:
      > >
      > >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
      > >(and of A"B'C', B"C'A', C"A'B')
      > >
      > >True?
      > >
      > >APH
    • Randy Hutson
      Dear Antreas, The conjecture still holds if we extend the definition of Euler line to include degenerate triangles (e.g. A ,B ,C collinear) as: the line
      Message 2 of 9 , Jan 6, 2013
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        Dear Antreas,

        The conjecture still holds if we extend the definition of Euler line to include degenerate triangles (e.g. A",B',C' collinear) as: the line perpendicular to the line of vertices and passing through the centroid of the vertices.  This line meets the line at infinity at a point which is the 'circumcenter', 'orthocenter', and 'nine-point center' of the degenerate triangle.  In this case, the three Euler lines still concur.

        Best regards,
        Randy





        >________________________________
        > From: Antreas <anopolis72@...>
        >To: Hyacinthos@yahoogroups.com
        >Sent: Sunday, January 6, 2013 5:05 AM
        >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
        >
        >

        >Dear Randy
        >
        >Note that the conjecture is not true if one of the six triangles A"B'C',...., A'B"C" is degenerated (if A",B',C' etc are collinear).
        >
        >Figures here
        >http://tech.groups.yahoo.com/group/Hyacinthos/message/21357
        >
        >APH
        >
        >--- In Hyacinthos@yahoogroups.com, Randy Hutson wrote:
        >>
        >> Antreas,
        >>
        >> I was able to confirm this computationally.  Of course, not the same as a proof.
        >>
        >> Randy Hutson
        >>
        >>
        >>
        >>
        >>
        >> >________________________________
        >> > From: Antreas
        >> >To: Hyacinthos@yahoogroups.com
        >> >Sent: Friday, January 4, 2013 12:56 PM
        >> >Subject: [EMHL] GENERALIZATION (Re: ANOTHER MORLEY TRIANGLE CONJECTURE.....)
        >> >
        >> >
        >> > 
        >> >Let A'B'C', A"B"C" be two homothetic Equilateral triangles.
        >> >
        >> >Conjecture:
        >> >
        >> >The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
        >> >(and of A"B'C', B"C'A', C"A'B')
        >> >
        >> >True?
        >> >
        >> >APH
        >
        >
        >
        >
        >

        [Non-text portions of this message have been removed]
      • Nikolaos Dergiades
        Dear Antreas, If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics and A B C is the homothetic of ABC in the homothety PA = t.PA then the Euler
        Message 3 of 9 , Jan 6, 2013
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          Dear Antreas,
          If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics
          and A'B'C' is the homothetic of ABC in the homothety PA' = t.PA
          then the Euler lines of A'BC, B'CA, C'AB are concurrent at
          Q = {2 p^3 q^2 + p^2 q^3 + 5 p^3 q r + 8 p^2 q^2 r + 2 p q^3 r +
          2 p^3 r^2 + 8 p^2 q r^2 + 5 p q^2 r^2 + p^2 r^3 + 2 p q r^3 +
          2 p^4 q t + 2 p^2 q^3 t + p q^4 t + 2 p^4 r t + 4 p^3 q r t +
          4 p^2 q^2 r t + 4 p q^3 r t + 4 p^2 q r^2 t + 4 p q^2 r^2 t +
          q^3 r^2 t + 2 p^2 r^3 t + 4 p q r^3 t + q^2 r^3 t + p r^4 t -
          p^4 q t^2 + 2 p^3 q^2 t^2 - p^4 r t^2 - 2 p^3 q r t^2 +
          p q^3 r t^2 + q^4 r t^2 + 2 p^3 r^2 t^2 + 3 p q^2 r^2 t^2 +
          q^3 r^2 t^2 + p q r^3 t^2 + q^2 r^3 t^2 + q r^4 t^2 - p^3 q^2 t^3 -
          p^3 r^2 t^3 + q^3 r^2 t^3 + q^2 r^3 t^3,
          p^3 q^2 + 2 p^2 q^3 + 2 p^3 q r + 8 p^2 q^2 r + 5 p q^3 r +
          5 p^2 q r^2 + 8 p q^2 r^2 + 2 q^3 r^2 + 2 p q r^3 + q^2 r^3 +
          p^4 q t + 2 p^3 q^2 t + 2 p q^4 t + 4 p^3 q r t + 4 p^2 q^2 r t +
          4 p q^3 r t + 2 q^4 r t + p^3 r^2 t + 4 p^2 q r^2 t +
          4 p q^2 r^2 t + p^2 r^3 t + 4 p q r^3 t + 2 q^2 r^3 t + q r^4 t +
          2 p^2 q^3 t^2 - p q^4 t^2 + p^4 r t^2 + p^3 q r t^2 -
          2 p q^3 r t^2 - q^4 r t^2 + p^3 r^2 t^2 + 3 p^2 q r^2 t^2 +
          2 q^3 r^2 t^2 + p^2 r^3 t^2 + p q r^3 t^2 + p r^4 t^2 -
          p^2 q^3 t^3 + p^3 r^2 t^3 - q^3 r^2 t^3 + p^2 r^3 t^3,
          2 p^3 q r + 5 p^2 q^2 r + 2 p q^3 r + p^3 r^2 + 8 p^2 q r^2 +
          8 p q^2 r^2 + q^3 r^2 + 2 p^2 r^3 + 5 p q r^3 + 2 q^2 r^3 +
          p^3 q^2 t + p^2 q^3 t + p^4 r t + 4 p^3 q r t + 4 p^2 q^2 r t +
          4 p q^3 r t + q^4 r t + 2 p^3 r^2 t + 4 p^2 q r^2 t +
          4 p q^2 r^2 t + 2 q^3 r^2 t + 4 p q r^3 t + 2 p r^4 t + 2 q r^4 t +
          p^4 q t^2 + p^3 q^2 t^2 + p^2 q^3 t^2 + p q^4 t^2 + p^3 q r t^2 +
          3 p^2 q^2 r t^2 + p q^3 r t^2 + 2 p^2 r^3 t^2 - 2 p q r^3 t^2 +
          2 q^2 r^3 t^2 - p r^4 t^2 - q r^4 t^2 + p^3 q^2 t^3 + p^2 q^3 t^3 -
          p^2 r^3 t^3 - q^2 r^3 t^3}

          The locus of Q is

          -p^2 (p - q) q^2 (p - r) (q - r) r^2 (p + q + r)^11 (p q + p r +
          q r) x^3 (p^6 q^2 x^3 - 5 p^4 q^4 x^3 + 4 p^2 q^6 x^3 -
          2 p^6 q r x^3 + 5 p^4 q^3 r x^3 + 2 p^3 q^4 r x^3 + 4 p q^6 r x^3 +
          p^6 r^2 x^3 - 2 p^3 q^3 r^2 x^3 + q^6 r^2 x^3 + 5 p^4 q r^3 x^3 -
          2 p^3 q^2 r^3 x^3 - 8 p^2 q^3 r^3 x^3 - 4 p q^4 r^3 x^3 -
          5 p^4 r^4 x^3 + 2 p^3 q r^4 x^3 - 4 p q^3 r^4 x^3 -
          2 q^4 r^4 x^3 + 4 p^2 r^6 x^3 + 4 p q r^6 x^3 + q^2 r^6 x^3 -
          3 p^6 q^2 x^2 y + 9 p^5 q^3 x^2 y + 12 p^4 q^4 x^2 y -
          18 p^3 q^5 x^2 y - 9 p^5 q^2 r x^2 y + 3 p^4 q^3 r x^2 y -
          9 p^2 q^5 r x^2 y + 6 p q^6 r x^2 y + 3 p^6 r^2 x^2 y -
          6 p^3 q^3 r^2 x^2 y + 3 q^6 r^2 x^2 y - 12 p^4 q r^3 x^2 y +
          24 p^3 q^2 r^3 x^2 y - 9 p^2 q^3 r^3 x^2 y + 6 p q^4 r^3 x^2 y -
          3 p^4 r^4 x^2 y + 18 p^2 q^2 r^4 x^2 y - 6 p q^3 r^4 x^2 y -
          6 p q r^6 x^2 y - 3 q^2 r^6 x^2 y - 18 p^5 q^3 x y^2 +
          12 p^4 q^4 x y^2 + 9 p^3 q^5 x y^2 - 3 p^2 q^6 x y^2 +
          6 p^6 q r x y^2 - 9 p^5 q^2 r x y^2 + 3 p^3 q^4 r x y^2 -
          9 p^2 q^5 r x y^2 + 3 p^6 r^2 x y^2 - 6 p^3 q^3 r^2 x y^2 +
          3 q^6 r^2 x y^2 + 6 p^4 q r^3 x y^2 - 9 p^3 q^2 r^3 x y^2 +
          24 p^2 q^3 r^3 x y^2 - 12 p q^4 r^3 x y^2 - 6 p^3 q r^4 x y^2 +
          18 p^2 q^2 r^4 x y^2 - 3 q^4 r^4 x y^2 - 3 p^2 r^6 x y^2 -
          6 p q r^6 x y^2 + 4 p^6 q^2 y^3 - 5 p^4 q^4 y^3 + p^2 q^6 y^3 +
          4 p^6 q r y^3 + 2 p^4 q^3 r y^3 + 5 p^3 q^4 r y^3 - 2 p q^6 r y^3 +
          p^6 r^2 y^3 - 2 p^3 q^3 r^2 y^3 + q^6 r^2 y^3 - 4 p^4 q r^3 y^3 -
          8 p^3 q^2 r^3 y^3 - 2 p^2 q^3 r^3 y^3 + 5 p q^4 r^3 y^3 -
          2 p^4 r^4 y^3 - 4 p^3 q r^4 y^3 + 2 p q^3 r^4 y^3 - 5 q^4 r^4 y^3 +
          p^2 r^6 y^3 + 4 p q r^6 y^3 + 4 q^2 r^6 y^3 + 3 p^6 q^2 x^2 z -
          3 p^4 q^4 x^2 z - 12 p^4 q^3 r x^2 z - 6 p q^6 r x^2 z -
          3 p^6 r^2 x^2 z - 9 p^5 q r^2 x^2 z + 24 p^3 q^3 r^2 x^2 z +
          18 p^2 q^4 r^2 x^2 z - 3 q^6 r^2 x^2 z + 9 p^5 r^3 x^2 z +
          3 p^4 q r^3 x^2 z - 6 p^3 q^2 r^3 x^2 z - 9 p^2 q^3 r^3 x^2 z -
          6 p q^4 r^3 x^2 z + 12 p^4 r^4 x^2 z + 6 p q^3 r^4 x^2 z -
          18 p^3 r^5 x^2 z - 9 p^2 q r^5 x^2 z + 6 p q r^6 x^2 z +
          3 q^2 r^6 x^2 z - 6 p^6 q^2 x y z + 9 p^5 q^3 x y z -
          6 p^4 q^4 x y z + 9 p^3 q^5 x y z - 6 p^2 q^6 x y z -
          6 p^6 q r x y z + 18 p^5 q^2 r x y z + 6 p^4 q^3 r x y z +
          6 p^3 q^4 r x y z + 18 p^2 q^5 r x y z - 6 p q^6 r x y z -
          6 p^6 r^2 x y z + 18 p^5 q r^2 x y z - 36 p^4 q^2 r^2 x y z -
          6 p^3 q^3 r^2 x y z - 36 p^2 q^4 r^2 x y z + 18 p q^5 r^2 x y z -
          6 q^6 r^2 x y z + 9 p^5 r^3 x y z + 6 p^4 q r^3 x y z -
          6 p^3 q^2 r^3 x y z - 6 p^2 q^3 r^3 x y z + 6 p q^4 r^3 x y z +
          9 q^5 r^3 x y z - 6 p^4 r^4 x y z + 6 p^3 q r^4 x y z -
          36 p^2 q^2 r^4 x y z + 6 p q^3 r^4 x y z - 6 q^4 r^4 x y z +
          9 p^3 r^5 x y z + 18 p^2 q r^5 x y z + 18 p q^2 r^5 x y z +
          9 q^3 r^5 x y z - 6 p^2 r^6 x y z - 6 p q r^6 x y z -
          6 q^2 r^6 x y z - 3 p^4 q^4 y^2 z + 3 p^2 q^6 y^2 z -
          6 p^6 q r y^2 z - 12 p^3 q^4 r y^2 z - 3 p^6 r^2 y^2 z +
          18 p^4 q^2 r^2 y^2 z + 24 p^3 q^3 r^2 y^2 z - 9 p q^5 r^2 y^2 z -
          3 q^6 r^2 y^2 z - 6 p^4 q r^3 y^2 z - 9 p^3 q^2 r^3 y^2 z -
          6 p^2 q^3 r^3 y^2 z + 3 p q^4 r^3 y^2 z + 9 q^5 r^3 y^2 z +
          6 p^3 q r^4 y^2 z + 12 q^4 r^4 y^2 z - 9 p q^2 r^5 y^2 z -
          18 q^3 r^5 y^2 z + 3 p^2 r^6 y^2 z + 6 p q r^6 y^2 z +
          3 p^6 q^2 x z^2 - 3 p^2 q^6 x z^2 + 6 p^6 q r x z^2 +
          6 p^4 q^3 r x z^2 - 6 p^3 q^4 r x z^2 - 6 p q^6 r x z^2 -
          9 p^5 q r^2 x z^2 - 9 p^3 q^3 r^2 x z^2 + 18 p^2 q^4 r^2 x z^2 -
          18 p^5 r^3 x z^2 - 6 p^3 q^2 r^3 x z^2 + 24 p^2 q^3 r^3 x z^2 +
          12 p^4 r^4 x z^2 + 3 p^3 q r^4 x z^2 - 12 p q^3 r^4 x z^2 -
          3 q^4 r^4 x z^2 + 9 p^3 r^5 x z^2 - 9 p^2 q r^5 x z^2 -
          3 p^2 r^6 x z^2 + 3 q^2 r^6 x z^2 - 3 p^6 q^2 y z^2 +
          3 p^2 q^6 y z^2 - 6 p^6 q r y z^2 - 6 p^4 q^3 r y z^2 +
          6 p^3 q^4 r y z^2 + 6 p q^6 r y z^2 + 18 p^4 q^2 r^2 y z^2 -
          9 p^3 q^3 r^2 y z^2 - 9 p q^5 r^2 y z^2 + 24 p^3 q^2 r^3 y z^2 -
          6 p^2 q^3 r^3 y z^2 - 18 q^5 r^3 y z^2 - 3 p^4 r^4 y z^2 -
          12 p^3 q r^4 y z^2 + 3 p q^3 r^4 y z^2 + 12 q^4 r^4 y z^2 -
          9 p q^2 r^5 y z^2 + 9 q^3 r^5 y z^2 + 3 p^2 r^6 y z^2 -
          3 q^2 r^6 y z^2 + p^6 q^2 z^3 - 2 p^4 q^4 z^3 + p^2 q^6 z^3 +
          4 p^6 q r z^3 - 4 p^4 q^3 r z^3 - 4 p^3 q^4 r z^3 +
          4 p q^6 r z^3 + 4 p^6 r^2 z^3 - 8 p^3 q^3 r^2 z^3 +
          4 q^6 r^2 z^3 + 2 p^4 q r^3 z^3 - 2 p^3 q^2 r^3 z^3 -
          2 p^2 q^3 r^3 z^3 + 2 p q^4 r^3 z^3 - 5 p^4 r^4 z^3 +
          5 p^3 q r^4 z^3 + 5 p q^3 r^4 z^3 - 5 q^4 r^4 z^3 + p^2 r^6 z^3 -
          2 p q r^6 z^3 + q^2 r^6 z^3)

          and the Euler lines of AB'C', BC'A', CA'B' are concurrent at

          {2 p^3 q^2 + 2 p^3 r^2 + 3 p q^2 r^2 + q^3 r^2 + q^2 r^3 + 2 p^4 q t -
          p^3 q^2 t + 3 p^2 q^3 t + 2 p^4 r t + 4 p^3 q r t +
          6 p^2 q^2 r t + 4 p q^3 r t + q^4 r t - p^3 r^2 t + 6 p^2 q r^2 t +
          q^3 r^2 t + 3 p^2 r^3 t + 4 p q r^3 t + q^2 r^3 t + q r^4 t -
          p^4 q t^2 + 3 p^3 q^2 t^2 - p^2 q^3 t^2 + p q^4 t^2 - p^4 r t^2 +
          7 p^3 q r t^2 + 7 p^2 q^2 r t^2 + p q^3 r t^2 + 3 p^3 r^2 t^2 +
          7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 + q^3 r^2 t^2 - p^2 r^3 t^2 +
          p q r^3 t^2 + q^2 r^3 t^2 + p r^4 t^2 - p^3 q^2 t^3 + p^2 q^3 t^3 -
          4 p^3 q r t^3 - p^2 q^2 r t^3 + 2 p q^3 r t^3 - p^3 r^2 t^3 -
          p^2 q r^2 t^3 + 2 p q^2 r^2 t^3 + p^2 r^3 t^3 + 2 p q r^3 t^3,
          2 p^2 q^3 + p^3 r^2 + 3 p^2 q r^2 + 2 q^3 r^2 + p^2 r^3 +
          3 p^3 q^2 t - p^2 q^3 t + 2 p q^4 t + p^4 r t + 4 p^3 q r t +
          6 p^2 q^2 r t + 4 p q^3 r t + 2 q^4 r t + p^3 r^2 t +
          6 p q^2 r^2 t - q^3 r^2 t + p^2 r^3 t + 4 p q r^3 t + 3 q^2 r^3 t +
          p r^4 t + p^4 q t^2 - p^3 q^2 t^2 + 3 p^2 q^3 t^2 - p q^4 t^2 +
          p^3 q r t^2 + 7 p^2 q^2 r t^2 + 7 p q^3 r t^2 - q^4 r t^2 +
          p^3 r^2 t^2 + 7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 + 3 q^3 r^2 t^2 +
          p^2 r^3 t^2 + p q r^3 t^2 - q^2 r^3 t^2 + q r^4 t^2 + p^3 q^2 t^3 -
          p^2 q^3 t^3 + 2 p^3 q r t^3 - p^2 q^2 r t^3 - 4 p q^3 r t^3 +
          2 p^2 q r^2 t^3 - p q^2 r^2 t^3 - q^3 r^2 t^3 + 2 p q r^3 t^3 +
          q^2 r^3 t^3,
          p^3 q^2 + p^2 q^3 + 3 p^2 q^2 r + 2 p^2 r^3 + 2 q^2 r^3 + p^4 q t +
          p^3 q^2 t + p^2 q^3 t + p q^4 t + 4 p^3 q r t + 4 p q^3 r t +
          3 p^3 r^2 t + 6 p^2 q r^2 t + 6 p q^2 r^2 t + 3 q^3 r^2 t -
          p^2 r^3 t + 4 p q r^3 t - q^2 r^3 t + 2 p r^4 t + 2 q r^4 t +
          p^3 q^2 t^2 + p^2 q^3 t^2 + p^4 r t^2 + p^3 q r t^2 +
          7 p^2 q^2 r t^2 + p q^3 r t^2 + q^4 r t^2 - p^3 r^2 t^2 +
          7 p^2 q r^2 t^2 + 7 p q^2 r^2 t^2 - q^3 r^2 t^2 + 3 p^2 r^3 t^2 +
          7 p q r^3 t^2 + 3 q^2 r^3 t^2 - p r^4 t^2 - q r^4 t^2 +
          2 p^3 q r t^3 + 2 p^2 q^2 r t^3 + 2 p q^3 r t^3 + p^3 r^2 t^3 -
          p^2 q r^2 t^3 - p q^2 r^2 t^3 + q^3 r^2 t^3 - p^2 r^3 t^3 -
          4 p q r^3 t^3 - q^2 r^3 t^3}

          ND

          > Let A'B'C', A"B"C" be two homothetic
          > Equilateral triangles.
          >
          > Conjecture:
          >
          > The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
          > (and of A"B'C', B"C'A', C"A'B')
          >
          > True?
          >
          > APH
        • Antreas
          Dear Nikos Thanks!! Using this Therorem we can find new Triangle centers, but with very complicated coordinates, I guess! A family of triangle equilateral
          Message 4 of 9 , Jan 7, 2013
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            Dear Nikos

            Thanks!!

            Using this Therorem we can find new Triangle centers, but with very
            complicated coordinates, I guess!
            A family of triangle equilateral triangles is Morley triangle and its homothetics. We have already discussed the case of the 1st Morley and Russel triangles. It will be interesting if the two centers we get are on the Euler line of the reference triangle (remarked by Frank Jackson, Hyacinthos #21354)

            APH


            --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades wrote:
            >
            > Dear Antreas,
            > If ABC is equilateral triangle P = (p : q : r) in ABC barycentrics
            > and A'B'C' is the homothetic of ABC in the homothety PA' = t.PA
            > then the Euler lines of A'BC, B'CA, C'AB are concurrent at
            > Q = ......

            >
            > The locus of Q is ...........

            >
            > and the Euler lines of AB'C', BC'A', CA'B' are concurrent at......
            >
            > ND
            >
            > > Let A'B'C', A"B"C" be two homothetic
            > > Equilateral triangles.
            > >
            > > Conjecture:
            > >
            > > The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.
            > > (and of A"B'C', B"C'A', C"A'B')
            > >
            > > True?
            > >
            > > APH
            >
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