Dear Bernard

Thank you for your nice remarks and your references.

I guessed that A'B'C' had some name but I am losing my memory.

Quelle tristesse!

There is a vast literature on inner Steiner ellipse foci and I don't

remember if this construction has already be given. No doubt yes but what is

interesting is the long choo-choo way I have followed to find it. Alas, I

have no room here to tell you the story.

In fact, the same choo-choo theorem gives as a special case the construction

of 2 isogonal conjugate points with given middle. I remember Paul Yiu gave

such a construction and maybe you or Jean-Pierre had sent me another one.

As for me, I have already found one also based on circular geometry but less

beautiful than this last one derived from choo-choo theory.

I can't resist to give you the general construction I found and you will

tell me if it is the same as Paul's one or yours.

So we want to construct 2 isogonal conjugate points F and F' wrt triangle

ABC, with given middle P.

They are the foci of the ABC-inner conic with center P.

It is easy to get the perspector J of the conic as the isotomic conjugate of

the anticomplement of P.

I hope I don't mix up here complement and anticomplement, funny names for

homotheties!

The cevian triangle U V W of P is the contact triangle of the conic with the

sides of ABC.

Now comes the construction strictly speaking:

Let T be the isogonal conjugate of P wrt ABC.

Line AT cuts again the (A V W) - circumcircle in A'.

Line BT cuts again the (B W U) - circumcircle in B'

Line CT cuts again the (C U V) - circumcircle in C'.

Here you can check that the 4 points A', B', C', T are on a same circle!

Let f be the direct circular map such that: f(A) = A', f(B) = B', f(C) = C'

Then f is of order 2, that is to say involutive.

The central point of f is P, f(P) = Infty and f(Infty) = P

The fixed points of f are the foci F and F' of the ABC-inner conic of center

P.

Hence we have 3 harmonic quadrangles (A, A', F, F'), (B, B', F, F'), (C, C',

F, F'), so giving an easy construction of these foci.

Friendly

Francois

PS

I like this construction for it is very symmetric though maybe a little

long!

Is it new ? I doubt it

On Nov 12, 2007 7:23 AM, Bernard Gibert <bg42@...> wrote:

> Dear Francois ,

>

>

> > So we start with our usual beloved triangle ABC, its circumcenter O

> > and its

> > Lemoine point K.

> > By the way, why this damned Lemoine point is always labelled K ?

>

> L is the de Longchamps point...

>

> > Let A', B', C' respectively the orthogonal projections of O onto the

> > symedians AK, BK, CK. ( They are on the Brocard circle.)

>

> A'B'C' is the second Brocard triangle.

>

> > Let f be the direct circular map such that: f(A) = A', f(B) = B', f

> > (C) =

> > C'.

> > Then f is of order 2, that is to say involutive.

>

> I suspect that your map is the one described in "orthocorrespondence

> and orthopivotal cubics" §5.

>

> f(M) is the product of an inversion and a reflection. It is also :

>

> - the center of the polar conic of M wrt the McCay or Kjp cubic,

>

> - the isoconjugate of M in the triangle formed by G and the circular

> points at infinity. In other words, the pole of M in the pencil of

> rectangular hyperbolas passing through the 4 foci of the Steiner

> inellipse.

>

> Best regards

>

> Bernard

>

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

>

>

>

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