## Re: Happy New Year and Nik's locus for 2013

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• Dear Nik and Francisco Javier, ND: If A B C is the cevian triangle of P and A BA1, CA A2 are equilateral triangles and A* is the mid point of A1, A2 in the
Message 1 of 14 , Jan 1, 2013
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Dear Nik and Francisco Javier,

ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2
are equilateral triangles and A* is the mid point of A1, A2
in the opposite side of BC ralative to A, and similarly define B*,
C*, what is the locus of P such that ABC, A*B*C* are perspective?

PY: If A_0BC, B_0CA, C_0AB are the equilateral triangles
constructed externally on the sides BC, CA, AB respectively,
... A*B*C* is perspective with ABC if and only if P [lies on]
the isocubic K(Q,X(13)), where
Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
has (6-9-13) search number 0.245203113787â¦

***

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
• Dear Nikos, Paul Happy New Year!!! 1. We can take A B C as the pedal triangle of P and ask for the locus. 2. We can take A*,B*,C* as triangle centers
Message 2 of 14 , Jan 1, 2013
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Dear Nikos, Paul

Happy New Year!!!

1. We can take A'B'C' as the pedal triangle of P and ask for the locus.

2. We can take A*,B*,C* as triangle centers (Circumcenters, Orthocenters
etc) of
the triangles A'A1A2, B'B1B2, C'C1C2 and ask of the loci of P such taht
ABC,A*B*C* be perspective.

I think that interesting is the case of the circumcenters.

Season's Greetings

Antreas

> **
>
>
> Dear Antreas
> Happy New Year to you
> and all friends of Hyacinthos.
>
> A locus for the begining of 2013.
> If A'B'C' is the cevian triangle of P
> and A'BA1, CA'A2 are equilateral triangles
> and A* is the mid point of A1, A2 in the
> opposite side of BC ralative to A,
> and similarly define B*, C* what is the locus
> of P such that ABC, A*B*C* are perspective.
> Best regards
>
>

[Non-text portions of this message have been removed]
• 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
Message 3 of 14 , Jan 2, 2013
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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
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
• Dear Antreas, if A B C is the pedal triangle and not the cevian then the locus is a cubic but I think complicated. The same holds if we take circumenters. I
Message 4 of 14 , Jan 2, 2013
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Dear Antreas,
if A'B'C' is the pedal triangle and not
the cevian then the locus is a cubic
but I think complicated. The same holds
if we take circumenters.
I think that the most interesting case is:

If A'B'C' is the cevian triangle of a point P,
A1A'B, A2CA'are similar isosceles triangles
outside of ABC with base angle w,
(the bases are opposite of A1 and A2)
A* is the centroid of triangle A'A1A2
and similarly define B*, C*
then the triangles ABC, A*B*C* 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+5S_w)/(S_A+S_w):..:..).

Best regards
Nikos

> Dear Nikos, Paul
>
> Happy New Year!!!
>
> 1. We can take A'B'C' as the pedal triangle of P and ask for
> the locus.
>
> 2. We can take A*,B*,C* as triangle centers (Circumcenters,
> Orthocenters
> etc) of
> the triangles A'A1A2, B'B1B2, C'C1C2 and ask of the loci of
> P such taht
> ABC,A*B*C* be perspective.
>
> I think that interesting is the case of the circumcenters.
>
> Season's Greetings
>
> Antreas
>
> On Tue, Jan 1, 2013 at 2:15 AM, Nikolaos Dergiades <ndergiades@...>wrote:
>
> > **
> >
> >
> > Dear Antreas
> > Happy New Year to you
> > and all friends of Hyacinthos.
> >
> > A locus for the begining of 2013.
> > If A'B'C' is the cevian triangle of P
> > and A'BA1, CA'A2 are equilateral triangles
> > and A* is the mid point of A1, A2 in the
> > opposite side of BC ralative to A,
> > and similarly define B*, C* what is the locus
> > of P such that ABC, A*B*C* are perspective.
> > Best regards
> >
> >
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
• Dear Nikos, Paul and friends, ... 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
Message 5 of 14 , Jan 2, 2013
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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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• 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
Message 6 of 14 , Jan 2, 2013
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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'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

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

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

• Happy New year 2013 . From Sevilla  Ricardo http://personal.us.es/rbarroso/trianguloscabri/ [Non-text portions of this message have been removed]
Message 7 of 14 , Jan 2, 2013
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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*
Message 8 of 14 , Jan 2, 2013
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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]
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