- Dear Francois,

you are right.

A simpler construction is the following:

Let O be the circumcenter of ABC.

The perpendicular from O to BC meets the

rotation of line CO about C by the signed angle B'A'C'

at the point A1. Similarly

the perpendicular from O to CA meets the

rotation of line AO about A by the signed angle C'B'A'

at the point B1.

The reflection of C in A1B1 is the required point P.

Best regards

Nikos Dergiades

> Dear Nikos

___________________________________________________________

> Excuse me for answering you so late.

> You are right of course provided that you are using

> oriented, (or signed

> ?), angles of lines.

> I give you two constructions of this point but as I think

> they are too

> complex, that's why I ask for a more direct one.

> Triangles ABC and A'B'C' are given and are

> supposed to be not inversely

> similar for in this case the following constructions lead

> to nowhere.

> 1° First construct a ABC-inscribed triangle <abc>

> homothetic with A'B'C' ,

> (very known construction) and then the point P is the

> common point of

> circles (bcA), (caB), (abC), (the Miquel point).

> 2° First construct a ABC-circumscribed triangle

> <abc> homothetic with

> A'B'C', (straightforward!), then the common

> point P* of circles (aBC),

> (bCA), (cAB).

> P is the isogonal conjugate of P* wrt ABC.

>

> Friendly

> Francois

> PS

> Another way still more elaborate to recover P is the

> following:

> You look at the direct circular map <f> sending ABC

> onto A'B'C' and you look

> at the pôle P' of <f> such that f(P') =

> Infty, the point at infinity.

> Then P is the point inverse of P' wrt circle ABC.

> Or you can look at the indirect circular map <g>

> sending ABC onto A'B'C',

> then P is the pôle of <g> such that g(P) = Infty.

> So any simple construction of a pôle of a circular map

> must lead to a simple

> construction of P.

>

>

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

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

>

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

Yes.

And the points A1, B1, C1

are found in this case as previously

but by negative rotations.

Have a Happy New Year.

Best regards

Nikos Dergiades

> Dear Nikos and Francois,

___________________________________________________________

>

> So, if instead of the triangle A'B'C', we use a

> triangle inversely

> similar to A'B'C', then we find the pedal

> triangle of the point P*

> inverse of P in the ABC circumcircle?

>

> Sincerely, Jeff

>

>

> [ND]

> > Dear Francois,

> > I think you mean if we construct the point C1

> > as we constructed the points A1, B1,

> > if the reflections

> > of C in A1B1,

> > of A in B1C1

> > of B in C1A1

> > coincide?

> > Yes.

> >

> > Best regards

> > Nikos Dergiades

> >

> > > Dear Nikos

> > > Thank you very much for your help.

> > > Now, I will think about your construction and

> compare it

> > > with mines.

> > > Have you already look at the configuration you

> get when you

> > > perform your

> > > construction with other vertices?

> > > Friendly

> > > Francois

> > >

> > > 2008/11/24 Nikolaos Dergiades

> <ndergiades@...>

> > >

> > > > Dear Francois,

> > > > you are right.

> > > > A simpler construction is the following:

> > > >

> > > > Let O be the circumcenter of ABC.

> > > > The perpendicular from O to BC meets the

> > > > rotation of line CO about C by the signed

> angle

> > > > B'A'C' at the point A1.

> Similarly

> > > > the perpendicular from O to CA meets the

> > > > rotation of line AO about A by the signed

> angle

> > > > C'B'A' at the point B1.

> > > > The reflection of C in A1B1 is the required

> point P.

> > > > Best regards

> > > > Nikos Dergiades

> > > >

>

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

>

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