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

>

>

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

> Hyacinthos-fullfeatured@yahoogroups.com

>

>