- Dear Bernard and Steve

> >> [JP] consider the rectangular hyperbola through G, O, K

through

> >> (symedian), X[110).

> >> His center is the midpoint of GX[110] and the hyperbola goes

> >> the infinite points of the Jerabek hyperbola, X[154], X[354], X

I don't think so

> >> [392], X

> >> [1201],...

> >> Consider a common point P (not X[110]) of the hyperbola with the

> >> circumcircle;

>

> These 3 points P lie on the Thomson cubic.

> >> let M be the homothetic of P in (O, -3) and A'B'C' the

Darboux

> >> pedal triangle of M.

> >> Then AA', BB', CC' are parallel.

> >> Any explanation?

> there's a close analogy with the bottom of my web page on the

> cubic.

Yes, indeed. These three points are UVW in your Darboux page. many

>

> http://perso.wanadoo.fr/bernard.gibert/Exemples/k004.html

thanks.

It is quite interesting to see that these points are the only points

P in the plane such as, if A'B'C' is the pedal triangle of P, the

lines AA', BB', CC' are parallel (of course they are parallel to the

asymptots of Lucas - or Thomson -)

Friendly. Jean-Pierre - Dear Jean-Pierre,

>>>> [JP] consider the rectangular hyperbola through G, O, K

THEY DO, dear Jean-Pierre !

>>>> (symedian), X[110).

>>>> His center is the midpoint of GX[110] and the hyperbola goes

> through

>>>> the infinite points of the Jerabek hyperbola, X[154], X[354], X

>>>> [392], X

>>>> [1201],...

>>>> Consider a common point P (not X[110]) of the hyperbola with the

>>>> circumcircle;

>>

>> These 3 points P lie on the Thomson cubic.

>

> I don't think so

Best regards

Bernard

[Non-text portions of this message have been removed] - Dear Bernard
> THEY DO, dear Jean-Pierre !

Of course, you're right - in fact I was thinking to your points UVW -

Looking at your nice page about Darboux, I have an obvious explanation

of the nice property of the pedal triangle of U,V,W.

Suppose that A'B'C' is the pedal triangle of M; then BB' and CC' are

parallel iff M lies on your hyperbola Ha.

As U,V,W are the common points of Ha,Hb,Hc, for these points, AA',

BB', CC' are parallel.

Thus, there is nothing new in what I said.

Friendly.Jean-Pierre