21335Re: [EMHL] Re: Happy New Year and Nik's locus for 2013
- Jan 2, 2013Dear Paul,
A little generalization:
If A'B'C' is the cevian triangle of a point P
and A"BC, B"CA, C"AB are similar isosceles triangles
outside of ABC with base agnle w then the triangles
ABC, A*B*C* where A* is the mid point of A'A" . . .
are perspective if and only if the locus of P
is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
The point S lies on the Kiepert hyperbola
and corresponds to A", B", C".
The point S'= ((1/(S_A+3S_w):..:..)) lies also
on the Kiepert hyperbola and corresponds to the
Centroids of triangles A"BC, B"CA, C"AB.
The point Q is the barycentric product of S
and the isotomic of S'.
> I think we can go a little bit further and get a Euclidean
> constructible example. The point $Q$ can be regarded as the
> isogonal conjugate of the barycentric product of $X(15)$
> (isodynamic point) and $X(17)$ (Napoleon point):
> Q = (X(15)*X(17))^*.
> It is an interior point of triangle $ABC$ if all angles are
> less than $120^\circ$.
> In this case, the barycentric square root $\sqrt Q$ is a
> point on the locus. With $P = \sqrt Q$, the perspector is
> the barycentric square root of
> (1/((S_A+S \sqrt 3)(S_A+S/\sqrt 3)) : ... : ...)
> = (1/(S_A+S \sqrt 3)^2 : ... : ...)*
> ((S_A+S Sqrt)/(S_A+S/Sqrt) : ... : ...)
> = (1/(S_A+S \sqrt 3)^2 : ... : ...)*Q,
> and is the barycentric product $X(17)*P$.
> Best regards
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