## Re: Pedal triangles

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• Dear Francois ... P recur ... the third ... isometric ... ratio k ... If I don t mistake,if P is inside ABC, the third pedal triangle cannot be isometric
Message 1 of 37 , Nov 11, 2007
Dear Francois
> Dear Bernard and dear friends
> It is known that the shapes of repeated triangles wrt a fixed point
P recur
> with period 3 and the Mac Cay cubic is the locus for P for which
the third
> pedal triangle is homothetic to the reference triangle.
> Do you know the locus of P for which this third pedal triangle is
isometric
> with the reference triangle or more generally similar with a given
ratio k >
> 0?
> I think this is some sextic?

If I don't mistake,if P is inside ABC, the third pedal triangle
cannot be isometric with the reference triangle because the ratio of
similarity k is <=1/8.
In fact k = (PPa.PPb.PPc)/(PA.PB.PC) ( with PaPbPc = pedal triangle
of P)and the Oppenheim inequality says that PA.PB.PC >=(Pb+Pc)(Pc+Pa)
(Pa+Pb) and
(PPb+PPc)(PPc+PPa)(PPa+PPb)>= 8.PPa.PPb.PPc
Friendly. Jean-Pierre
• 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
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
> > > >
>
>
>
> ------------------------------------
>