## Re: [EMHL] Looking for an explanation

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• Dear Jean-Pierre and Steve ... These 3 points P lie on the Thomson cubic. We ve already seen this hyperbola but where ? ... the Darboux cubic ... that s right.
Message 1 of 6 , Mar 31, 2006
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Dear Jean-Pierre 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.

We've already seen this hyperbola but where ?

>> 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?
>
>
> [SS] So the pedal triangle of M is a Cevian triangle of an infinite
> point.
> This puts it on one of the famous cubics (I cannot remember which),

the Darboux cubic

> and the infinite point must be an infinite point on that cubic. If so
> there must be 2 other points that serve as M.

that's right. they are 3 common points of the Darboux cubic and C(O,3R).

there's a close analogy with the bottom of my web page on the Darboux
cubic.

Best regards

Bernard

[Non-text portions of this message have been removed]
• 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 2 of 6 , Apr 1, 2006
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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 3 of 6 , Apr 1, 2006
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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 4 of 6 , Apr 1, 2006
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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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