## Re: [EMHL] Looking for an explanation

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• Dear Bernard and Steve ... through ... I don t think so ... Darboux ... Yes, indeed. These three points are UVW in your Darboux page. many thanks. It is quite
Message 1 of 6 , Apr 1, 2006
Dear Bernard and Steve

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

> >> let M be the homothetic of P in (O, -3) and A'B'C' the
> >> 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
Darboux
Yes, indeed. These three points are UVW in your Darboux page. many
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, ... THEY DO, dear Jean-Pierre ! Best regards Bernard [Non-text portions of this message have been removed]
Message 2 of 6 , Apr 1, 2006
Dear Jean-Pierre,

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

THEY DO, dear Jean-Pierre !

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard ... 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
Message 3 of 6 , Apr 1, 2006
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
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