## 21335Re: [EMHL] Re: Happy New Year and Nik's locus for 2013

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• Jan 2, 2013
Dear 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'.

Best regards
Nik

> 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
> $$> (1/((S_A+S \sqrt 3)(S_A+S/\sqrt 3)) : ... : ...) > = (1/(S_A+S \sqrt 3)^2 : ... : ...)* > ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...) > = (1/(S_A+S \sqrt 3)^2 : ... : ...)*Q, >$$
> and is the barycentric product $X(17)*P$.
>
> Best regards
> Sincerely
> Paul
>
>
>
>
> ------------------------------------
>