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

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• [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
[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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