## Re: [EMHL] Pedal triangles

Expand Messages
• 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
Message 1 of 37 , Nov 24, 2008
• 0 Attachment
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]
• 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
Message 37 of 37 , Dec 30, 2008
• 0 Attachment
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

> 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
> >
> > > Dear Nikos
> > > Thank you very much for your help.
> 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
> > >
> > > > 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
> > > >
>
>
>
> ------------------------------------
>