- Dear 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.

Bernard

[Non-text portions of this message have been removed] - Dear Nik, Bernard, and Francisco Javier,

Thank you for these generalizations. FJ, I did not realize that the locus contains G!

Best regards

Sincerely

Paul

________________________________________

From: Hyacinthos@yahoogroups.com [Hyacinthos@yahoogroups.com] on behalf of Bernard Gibert [bg42@...]

Sent: Wednesday, January 02, 2013 8:00 AM

To: Hyacinthos@yahoogroups.com

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

Dear 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.

Bernard

[Non-text portions of this message have been removed]

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

Yahoo! Groups Links - Happy New year 2013 .

From Sevilla

Ricardo

http://personal.us.es/rbarroso/trianguloscabri/

[Non-text portions of this message have been removed] - Dear all,

If A'B'C' is the pedal triangle of P (instead of cevian triangle of P), and A", B", C" as before,

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 a central cubic passing through X3 with asymptotes parallel to the altitudes of ABC and concurring on the Brocard axis, very similar to the Darboux cubic which is obtained when k = 0.

The locus of the perspector X is pK(X2, T) where

T = (-a^4+b^4+c^4)k-4 SA Sw : : , on the line X4-X69.

If k = 0, T = X69 and this is the Lucas cubic.

Best regards

Bernard

[Non-text portions of this message have been removed]