- Dear Francois
> Dear Bernard and dear friends

P recur

> It is known that the shapes of repeated triangles wrt a fixed point

> 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.

isometric

> Do you know the locus of P for which this third pedal triangle is

> with the reference triangle or more generally similar with a given

ratio k >

> 0?

If I don't mistake,if P is inside ABC, the third pedal triangle

> I think this is some sextic?

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

> 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

> > > >

>

>

>

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

>

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