Dear Nikos, Paul and friends,

> [ND] 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'.

If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is pK(Ω, S) with

pole Ω = (SA k + (2-k)Sw) / (SA + Sw) : : ,

pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,

isopivot : SA k + (2-k)Sw : : , on the line X2-X69.

When k varies, the locus of Ω is a line through X2,

When w varies, the locus of Ω is a nodal circum-cubic with node X2, the nodal tangents being X2-X6 and Euler line.

This cubic is a member of the pencil generated by :

the three medians (k = 1)

Kiepert + Euler (k = 0)

line X2-X6 + isotomic of Euler (k = 2).

----------------------------------------------------------

The locus of X is another pK(Ω', S') with

pole Ω' = 1 / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.

Best regards and Happy New Year to you all.

Bernard

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