## The vanLamoen's circle

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• Dear Floor and Hyacinthists, a new purely proof has being put on my website: volume (2008) Le cercle de van Lamoen http://perso.orange.fr/jl.ayme Sincerely
Message 1 of 8 , Mar 30, 2008
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Dear Floor and Hyacinthists,
a new purely proof has being put on my website: volume (2008) Le cercle de van Lamoen
http://perso.orange.fr/jl.ayme
Sincerely
Jean-Louis

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• Dear Jean-Louis, dear Floor and dear friends Very nice proof, Jean-Louis, written in your inimitable style! As I cannot sleep, I ask myself if something can
Message 2 of 8 , Apr 1, 2008
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Dear Jean-Louis, dear Floor and dear friends
Very nice proof, Jean-Louis, written in your inimitable style!
As I cannot sleep, I ask myself if something can be generalized in this
Floor drawing.
So I start with our beloved triangle ABC and draw the cevian triangle A'B'C'
of some point P in the plane and look at the centers of the circles
circumscribed to PBA', PCA', PCB', PAB', PAC', PBC', maybe there was
something to look at and in fact Cabri tells me that these 6 points are on a
same conic <Gamma>.
Of course I was too tired to prove anything, so I let you folks to do the
job and please find for me:
1° the locus of the point P for which <Gamma> is a circle.
For the moment, I can only prove that the centroid G is on this locus,
that's better than nothing!
2° the locus of the point P for which <Gamma> is a rectangular hyperbola.
3° the locus of the point P for which <Gamma> degenerates in 2 lines.
Friendly
Francois

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• Dear Francois, dear Jean-Louis, dear Floor and dear friends In a joined file (cercle de Van Lamoen), I have calculate the barycentrics coordinates of the 6
Message 3 of 8 , Apr 4, 2008
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Dear Francois, dear Jean-Louis, dear Floor and dear friends
In a joined file (cercle de Van Lamoen), I have calculate the
barycentrics coordinates of the 6 centers of the 6 circles through
the centroide of triangle (ABC)
as descrived in Jean-Louis text
It is very difficult to calculate the equation of the circle through
the 6 centers without a automate calculus
In Cabri, I try with a point P and its traces
If P is inside the triangle, the 6 centers are on a ellips
If P is outside, it's an hyperbola
The questions are always opened

Friendly
Michel

> [Non-text portions of this message have been removed]
>
• Dear Michel ... I don t think that the nature of your conic depends only of the position of P wrt the triangle ABC but of the position of P wrt the sidelines
Message 4 of 8 , Apr 7, 2008
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Dear Michel
> In a joined file (cercle de Van Lamoen), I have calculate the
> barycentrics coordinates of the 6 centers of the 6 circles through
> the centroide of triangle (ABC)
> as descrived in Jean-Louis text
> It is very difficult to calculate the equation of the circle through
> the 6 centers without a automate calculus
> In Cabri, I try with a point P and its traces
> If P is inside the triangle, the 6 centers are on a ellips
> If P is outside, it's an hyperbola
> The questions are always opened

I don't think that the nature of your conic depends only of the
position of P wrt the triangle ABC but of the position of P wrt the
sidelines of ABC and some curve of degree 7
A curious case : if P is X(110), focus of the Kiepert parabola, the
conic degenerates and the 6 centers lie 3 by 3 on two lines
intersecting at X(1576). Any explanation? It seems that the locus of
the points P with a degenerated associated conic is some bicircular
circumquartic.
Friendly. Jean-Pierre
• Dear Jean-Pierre Thank you answering my third question. What about the first and the second? I think there is only the centroid with an associated circle but I
Message 5 of 8 , Apr 7, 2008
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Dear Jean-Pierre
Thank you answering my third question.
What about the first and the second?
I think there is only the centroid with an associated circle but I have no
proof!
In such a problem where calculations are tremendous, only a software like
Mathematica or Maple can do the job and I have only my chinese abacus and my
slide rule!
But maybe there is also a synthetic proof for the existence of the conic
based on Carnot theorem?
Friendly
Francois

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• Dear Francois, About your first question, please see: http://forumgeom.fau.edu/FG2005volume5/FG200516index.html. is a circle if and only if P is the
Message 6 of 8 , Apr 7, 2008
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Dear Francois,

http://forumgeom.fau.edu/FG2005volume5/FG200516index.html.
<Gamma> is a circle if and only if P is the centroid or the
orthocenter of ABC.

Best regards,
Sincerely,
Cosmin

--- In Hyacinthos@yahoogroups.com, "Francois Rideau"
<francois.rideau@...> wrote:
>
> Dear Jean-Pierre
> Thank you answering my third question.
> What about the first and the second?
> I think there is only the centroid with an associated circle but I
have no
> proof!
> In such a problem where calculations are tremendous, only a software
like
> Mathematica or Maple can do the job and I have only my chinese
abacus and my
> slide rule!
> But maybe there is also a synthetic proof for the existence of the conic
> based on Carnot theorem?
> Friendly
> Francois
>
>
> [Non-text portions of this message have been removed]
>
• Dear Francois ... conic ... Pascal is better : U,V,W,U ,V ,W are respectively the circumcenters of PBA , PCB , PAC , PBC , PCA , PAB . The lines UV and WW
Message 7 of 8 , Apr 8, 2008
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Dear Francois
> But maybe there is also a synthetic proof for the existence of the
conic
> based on Carnot theorem?

Pascal is better :
U,V,W,U',V',W' are respectively the circumcenters of PBA', PCB', PAC',
PBC', PCA', PAB'.
The lines UV' and WW' are parallel (they are the perpendicular
bisectors of PA' and PA)
same way, the lines UU' and VW' are parallel and the lines VV' and U'W
are parallel.
The result follows.
Friendly. Jean-Pierre
• Dear Jean-Pierre Thank you for your proof. I was too short-sighted to see a Pascal line far away at infinity! Friendly Francois [Non-text portions of this
Message 8 of 8 , Apr 8, 2008
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Dear Jean-Pierre