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THEOREM (was : Locus and generalization)

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  • Antreas P. Hatzipolakis
    [APH] ... that is, cotB + cot(B,C) ... cotC + cot(C,B) that is, ABC, A#B#C# are perspective. So we have the THEOREM Let ABC, A#B#C# be two triangles and P a
    Message 1 of 4 , Jun 24, 2005
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      [APH]
      >GENERALIZATION
      >
      >Let ABC be a triangle P a point and A'B'C' the cevian triangle
      >of P.
      >
      >Let A#BC, B#CA, C#AB be three fixed triangles erected on the sides
      >of ABC (out / inwardly).
      >
      >Ac := AC# /\ CC'
      >
      >Ab := AB# /\ BB'
      >
      >A* := BAc /\ CAb. Similarly B*, C*
      >
      >Denote fixed angles C#AB = (A,B), B#AC = (A,C) etc
      >
      >Now, the triangles ABC, A*B*C* are perspective for all P, if
      >
      >
      > cotA + cot(A,C)
      > ------------------------ * CYCLIC = 1
      > cotA + cot(A,B)

      that is,

      cotB + cot(B,C)
      ------------------------ * CYCLIC = 1
      cotC + cot(C,B)


      that is, ABC, A#B#C# are perspective.


      So we have the

      THEOREM

      Let ABC, A#B#C# be two triangles and P a point.

      Denote

      Ac := AC# /\ CP, Ab := AB# /\ BP

      A* := AcB /\ AbC. Similarly B*,C*.


      ABC, A#B#C# are perspective ==> ABC, A*B*C* are perspective
      for all Ps


      Synthetic proof??


      Antreas













      --
    • Jeff Brooks
      ... Hey Antreas! Have you given any more thought to this theorem? Sincerely, Jeff
      Message 2 of 4 , Mar 2, 2006
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        --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"
        <xpolakis@...> wrote:
        >
        > [APH]
        > >GENERALIZATION
        > >
        > >Let ABC be a triangle P a point and A'B'C' the cevian triangle
        > >of P.
        > >
        > >Let A#BC, B#CA, C#AB be three fixed triangles erected on the sides
        > >of ABC (out / inwardly).
        > >
        > >Ac := AC# /\ CC'
        > >
        > >Ab := AB# /\ BB'
        > >
        > >A* := BAc /\ CAb. Similarly B*, C*
        > >
        > >Denote fixed angles C#AB = (A,B), B#AC = (A,C) etc
        > >
        > >Now, the triangles ABC, A*B*C* are perspective for all P, if
        > >
        > >
        > > cotA + cot(A,C)
        > > ------------------------ * CYCLIC = 1
        > > cotA + cot(A,B)
        >
        > that is,
        >
        > cotB + cot(B,C)
        > ------------------------ * CYCLIC = 1
        > cotC + cot(C,B)
        >
        >
        > that is, ABC, A#B#C# are perspective.
        >
        >
        > So we have the
        >
        > THEOREM
        >
        > Let ABC, A#B#C# be two triangles and P a point.
        >
        > Denote
        >
        > Ac := AC# /\ CP, Ab := AB# /\ BP
        >
        > A* := AcB /\ AbC. Similarly B*,C*.
        >
        >
        > ABC, A#B#C# are perspective ==> ABC, A*B*C* are perspective
        > for all Ps
        >
        >
        > Synthetic proof??
        >

        Hey Antreas!

        Have you given any more thought to this theorem?

        Sincerely, Jeff
      • Jeff Brooks
        Waiting patiently for a synthetic proof, Jeff. ... sides
        Message 3 of 4 , Mar 3, 2006
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          Waiting patiently for a synthetic proof, Jeff.
          >
          > --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"
          > <xpolakis@> wrote:
          > >
          > > [APH]
          > > >GENERALIZATION
          > > >
          > > >Let ABC be a triangle P a point and A'B'C' the cevian triangle
          > > >of P.
          > > >
          > > >Let A#BC, B#CA, C#AB be three fixed triangles erected on the
          sides
          > > >of ABC (out / inwardly).
          > > >
          > > >Ac := AC# /\ CC'
          > > >
          > > >Ab := AB# /\ BB'
          > > >
          > > >A* := BAc /\ CAb. Similarly B*, C*
          > > >
          > > >Denote fixed angles C#AB = (A,B), B#AC = (A,C) etc
          > > >
          > > >Now, the triangles ABC, A*B*C* are perspective for all P, if
          > > >
          > > >
          > > > cotA + cot(A,C)
          > > > ------------------------ * CYCLIC = 1
          > > > cotA + cot(A,B)
          > >
          > > that is,
          > >
          > > cotB + cot(B,C)
          > > ------------------------ * CYCLIC = 1
          > > cotC + cot(C,B)
          > >
          > >
          > > that is, ABC, A#B#C# are perspective.
          > >
          > >
          > > So we have the
          > >
          > > THEOREM
          > >
          > > Let ABC, A#B#C# be two triangles and P a point.
          > >
          > > Denote
          > >
          > > Ac := AC# /\ CP, Ab := AB# /\ BP
          > >
          > > A* := AcB /\ AbC. Similarly B*,C*.
          > >
          > >
          > > ABC, A#B#C# are perspective ==> ABC, A*B*C* are perspective
          > > for all Ps
          > >
          > >
          > > Synthetic proof??
          > >
          >
          > Hey Antreas!
          >
          > Have you given any more thought to this theorem?
          >
          > Sincerely, Jeff
          >
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