## Re: [EMHL] Re: Involutive transformation

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• Dear Jeff Of course, I don t know the answer! But in case of a circle, inversion have infinitely many fixed points when isoconjugation have only 4 fixed points
Message 1 of 10 , Jul 25, 2007
Dear Jeff
Of course, I don't know the answer!
But in case of a circle, inversion have infinitely many fixed points when
isoconjugation have only 4 fixed points real or not, inversion is defined on
the extended conformal plane when isoconjugation is defined on the extended
projective plane minus 3 points, so the situation is very tricky but maybe
this problem is not formulated well enough?

On 7/25/07, Francois Rideau <francois.rideau@...> wrote:
>
> Dear Jeff
> When you look at some isoconjugation (m <--> m') with 4 real fixed points,
> for example isotomic conjugation or isogonal conjugation, then m and m' are
> conjugate wrt every conic on these 4 fixed points, that is to say conics in
> a linear pencil;
> That's why I ask the converse problem: given a conic <Gamma>, what are the
> (algebraic) involutive maps (m <--> m') such that m and m' are conjugate wrt
> <Gamma>?
> If <Gamma> is a circle, inversion wrt <Gamma> is such a map but what are
> the others?
> Friendly
> Francois
>
>
> On 7/25/07, Jeff <cu1101@...> wrote:
> >
> > Dear Francois,
> >
> > I am attempting to tackle this problem with a direct proof. But, my
> > methods are long and laborious. It seems that the tangent line at
> > <p> works for all lines (even those not tangent to the incircle.)
> >
> > Am I wrong?
> >
> > Sincerely, Jeff
> >
> >
> > > Dear friends
> > > I give you an example of such an involutive map that I have found
> > > while doing some chowchows.
> > > I start with an equilateral triangle ABC with <Gamma> as
> > > circumcircle and <gamma> as incircle.
> > > I choose a point <p> on the incircle <gamma> of which the tangent
> > > at <gamma> cuts side BC in <a>, side CA in <b> and side AB in <c>.
> > >
> > > Now given any point M in the plane, I give the construction of the
> > > point M'= f(M), depending on the choice of <p> on the incircle
> > > <gamma>.
> > >
> > > Line AM cuts again circumcircle <Gamma> in U.
> > > Line aU cuts again circumcircle <Gamma> in U'.
> > >
> > > In the same way, we get the points V' and W'.
> > >
> > > Line BM cuts again circumcircle <Gamma> in V.
> > > Line bV cuts again circumcircle <Gamma> in V'.
> > >
> > > Line CM cuts again circumcircle <Gamma> in W.
> > > Line cW cuts again circumcircle <Gamma> in W'.
> > >
> > > Then lines AU', BV' and CW' are on point M'.
> > >
> > > Now it is easy to see that this construction is involutive.
> > > Moreover M and M' are conjugate wrt the equalizer conic of ABC,
> > > with barycentric equation: x² + y² + z² = 0.
> > > That is to say M' is on the dual line of M wrt ABC and of course M
> > > is also on the dual line of M' wrt ABC.
> >
> > > Friendly
> > > Francois
> > >
> > > PS
> > > 1° Equilateral triangles are not needed. Of course, all this stuff
> > > is affine (as the chowchow theory), replacing ABC by any triangle
> > > and circles by both Steiner ellipses.
> > > But construction is easier with circles than with ellipses.
> > > 2° I guess M and M' are isoconjugate in some isoconjugation.
> > >
> >
> >
> >
>
>

[Non-text portions of this message have been removed]
• ... I don t think I understood your problem, so I ask the following: The reflection about a line that passes through the center of the given circle is
Message 2 of 10 , Jul 26, 2007
Dear Francois, you wrote:
> > given a
> conic <Gamma>, what are the
> > (algebraic) involutive maps (m <--> m') such that
> m and m' are conjugate wrt
> > <Gamma>?
> > If <Gamma> is a circle, inversion wrt <Gamma> is
> such a map but what are
> > the others?

I don't think I understood your problem, so I
The reflection about a line that passes through
the center of the given circle <Gamma> is such a map?
Best regards

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• Dear Francois, ... I like the setup but I don t know what is the equalizer conic of ABC, with barycentric equation x^2 + y^2 + z^2 = 0. Why is this conic
Message 3 of 10 , Jul 27, 2007
Dear Francois,

>> the situation is very tricky but maybe
>> this problem is not formulated well enough?

I like the setup but I don't know what is the "equalizer conic of ABC,
with barycentric equation x^2 + y^2 + z^2 = 0." Why is this conic
called 'equalizer'?

Sincerely, Jeff
• ... Dear Francois, Sorry, I just now saw the reference in msg 14109. Sincerely, Jeff
Message 4 of 10 , Jul 30, 2007
--- In Hyacinthos@yahoogroups.com, "Jeff" <cu1101@...> wrote:
>
> Dear Francois,
>
> >> the situation is very tricky but maybe
> >> this problem is not formulated well enough?
>
> I like the setup but I don't know what is the "equalizer conic of
> ABC, with barycentric equation x^2 + y^2 + z^2 = 0." Why is this
> conic called 'equalizer'?
>
> Sincerely, Jeff
>

Dear Francois,
Sorry, I just now saw the reference in msg 14109.
Sincerely, Jeff
• Dear Jeff Yes, I think this conic is the equalizer, following Steve notations. Of course this conic has no real points but it does not matter! The point m with
Message 5 of 10 , Jul 30, 2007
Dear Jeff
Yes, I think this conic is the equalizer, following Steve notations.
Of course this conic has no real points but it does not matter!
The point m with barycentrics (a:b:c) has for polar line wrt this conic the
line with barycentric equation:
a.x + b.y + c.z = 0
But if you don't like this point of view, you can also use duality wrt the
(x,y,z) -->x^2 + y^2 + z^2
Friendly
Francois

On 7/28/07, Jeff <cu1101@...> wrote:
>
> Dear Francois,
>
> >> the situation is very tricky but maybe
> >> this problem is not formulated well enough?
>
> I like the setup but I don't know what is the "equalizer conic of ABC,
> with barycentric equation x^2 + y^2 + z^2 = 0." Why is this conic
> called 'equalizer'?
>
> Sincerely, Jeff
>
>
>

[Non-text portions of this message have been removed]
• Dear Nikos Nope! Look at the circle x^2 + y^2 = 1 and at the symmetry: (x,y) -- (x, -y) wrt the x-axis The points (a,b) and (a, -b) are in general never
Message 6 of 10 , Jul 30, 2007
Dear Nikos
Nope!
Look at the circle x^2 + y^2 = 1 and at the symmetry: (x,y) --> (x, -y) wrt
the x-axis
The points (a,b) and (a, -b) are in general never conjugate wrt this this
circle.
I want to find all involutive transformations (x,y) --> f(x,y) defined on
some open set of the projective (resp conformal) plane such that points
(x,y) and f(x,y) are always conjugate wrt some conic (resp circle)
Friendly
Francois

>
> Dear Francois, you wrote:
> > > given a
> > conic <Gamma>, what are the
> > > (algebraic) involutive maps (m <--> m') such that
> > m and m' are conjugate wrt
> > > <Gamma>?
> > > If <Gamma> is a circle, inversion wrt <Gamma> is
> > such a map but what are
> > > the others?
>
> I don't think I understood your problem, so I
> The reflection about a line that passes through
> the center of the given circle <Gamma> is such a map?
> Best regards
>
>
>
> ___________________________________________________________
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