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Isogonal Conjugate of X2164 (was: Re: Some Points)

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  • Antreas
    [APH] ... We have the answer, thanks to Francisco: They are. The point of concurrence is the isogonal conjugate of X2164 The above points are the points X278,
    Message 1 of 3 , Jan 4, 2011
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      [APH]
      > > Let ABC be a triangle and AA', BB', CC' its altitudes.
      > > The circle (B,BA') intersects AB at B1, B2 [B2 on the
      > > extension of AB]
      > > and the circle (C,CA') intersects the AC at C1, C2
      > > [C2 on the extension of AC]
      > > Let 1 be the intersection of BC1 and CB1,
      > > 1' B2C1 and C2B1
      > > 1" BC2 and CB2
      > >
      > > In a similar way we define the points 2, 2', 2" and 3, 3', 3".
      > >
      > > Then the triangles 123, 1'2'3', 1"2"3" are in perspective
      > > with ABC.
      > >
      > > Perspectors (according to my quick computations):
      > >
      > > 123 and ABC :
      > >
      > > (tan(A/2)/cosA ::) in trilinears
      > >
      > >
      > > 1'2'3' and ABC
      > >
      > > (tanA ::) in trilinears
      >
      > This is CLAWSON POINT
      >
      > >
      > > 1"2"3" and ABC :
      > >
      > > (1+cosA /cosA ::) = (secA + 1 ::) in 3linears
      >
      > This is wrong. The correct is: in Barycentrics.
      >
      > In trilinears the coordinates are: ((cot(A/2)/cosA ::)
      >
      > >
      > >
      > > Problem:
      > > Are the points 1,1',1" collinear? (similarly the points 2,2',2"
      > > and 3,3',3")
      >
      > Obviously 1,1',1" are collinear (and also (2,2',2") - (3,3',3")
      > by Pappous Hexagon Theorem
      >
      > > If yes: Are the lines 11'1", 22'2", 33'3" concurrent?
      >
      > ARE THEY??????

      We have the answer, thanks to Francisco:

      They are. The point of concurrence is the isogonal conjugate
      of X2164

      The above points are the points X278, X19 [=Clawson Point]
      and X281, resp., in ETC.

      Antreas
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