- Dear Francois, you wrote:
> > given a

I don't think I understood your problem, so I

> 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?

ask the following:

The reflection about a line that passes through

the center of the given circle <Gamma> is such a map?

Best regards

Nikos Dergiades

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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr - Dear Francois,

>> the situation is very tricky but maybe

I like the setup but I don't know what is the "equalizer conic of ABC,

>> this problem is not formulated well enough?

with barycentric equation x^2 + y^2 + z^2 = 0." Why is this conic

called 'equalizer'?

Sincerely, Jeff - --- In Hyacinthos@yahoogroups.com, "Jeff" <cu1101@...> wrote:
>

Dear Francois,

> 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

>

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 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

quadratic form:

(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 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

On 7/26/07, Nikolaos Dergiades <ndergiades@...> wrote:

>

> 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

> ask the following:

> The reflection about a line that passes through

> the center of the given circle <Gamma> is such a map?

> Best regards

> Nikos Dergiades

>

>

>

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