21337Re: [EMHL] Re: Happy New Year and Nik's locus for 2013
- Jan 2, 2013Dear Nikos, Paul and friends,
> [ND] A little generalization: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
> 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'.
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.
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