## Re: [EMHL] Re: Schiffler (reply to part 3)

Expand Messages
• Dear friends, ... the locus is a tricircular nonic (9th degree) having the perp. bisectors of ABC as real asymptotes. A, B, C are double. The tangents at A are
Message 1 of 11 , Sep 3, 2003
Dear friends,

--- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
<darij_grinberg@w...> wrote:
> I am quite interested in the locus of all points P
> so that the Euler lines of triangles BPC, CPA, APB
> intercept the vertices of a cevian triangle. This
> should pass through the incenters, the orthocenter,
> the circumcenter, but any other well-known points??
>

the locus is a tricircular nonic (9th degree) having the perp.
bisectors of ABC as real asymptotes.

A, B, C are double. The tangents at A are the perp. at A to AB, AC.

it contains
- the antipodes of A, B, C on C(O,R),
- the reflections of A, B, C in the sidelines of ABC
- in/excenters
- O, H

but certainly not the Fermats !

I expect there are some other centers on the curve.

Best regards

Bernard

[Non-text portions of this message have been removed]
• ... The locus of these points is Neuberg Cubic ++ There was paper by a tetrad of hyacinthos listmembers on concurrent Euler lines (and other lines through O)
Message 2 of 11 , Sep 3, 2003
On Tuesday, September 2, 2003, at 07:53 AM, andreyantonov wrote:

>> I joined the group in the beginning of August.
>> I immediately checked that points Ia, Ib, Ic and I are well
>> replacements of the above point P, so the Euler lines of triangles
>> APC, ABP, BCP and ABC concur at one point. This was also mentioned by
>> Steve Sigur and Darij Grinberg (and maybe others).
>> I also found that points X(13) TORRICELLI POINT makes the statement
>> true.

The locus of these points is Neuberg Cubic ++

There was paper by a tetrad of hyacinthos listmembers
on concurrent Euler lines (and other lines through O)
of the triangles PBC,PCA,PAB,
which can be found here:

http://forumgeom.fau.edu/FG2001volume1/FG200109index.html

If I remember correctly JPE mentioned also cases with tripolars
of points. I think that if the Lemoine axes of PBC,PCA,PAB
are concurrent, then all four Lemoine axes of ABC, PBC,PCA,PAB
are concurrent.
I don't remember if there was a discussion on the locus
of P with this property. (JP may correct my memory!)

Antreas

--
• Dear Antreas ... You ve proved in your paper that the locus of P such as the Brocard axis of PBC, PCA, PAB concur is again the Neuberg cubic and that, if they
Message 3 of 11 , Sep 3, 2003
Dear Antreas

> There was paper by a tetrad of hyacinthos listmembers
> on concurrent Euler lines (and other lines through O)
> of the triangles PBC,PCA,PAB,
> which can be found here:
>
> http://forumgeom.fau.edu/FG2001volume1/FG200109index.html
>
> If I remember correctly JPE mentioned also cases with tripolars
> of points. I think that if the Lemoine axes of PBC,PCA,PAB
> are concurrent, then all four Lemoine axes of ABC, PBC,PCA,PAB
> are concurrent.
> I don't remember if there was a discussion on the locus
> of P with this property. (JP may correct my memory!)

You've proved in your paper that the locus of P such as the Brocard
axis of PBC, PCA, PAB concur is again the Neuberg cubic and that, if
they concur, they concur on the Brocard axis of ABC.
If the Lemoine axis of PBC, PCA, PAB concur, they concur on the
Lemoine axis of ABC but the locus of such P is a tricircular
circumsextic and I'm unable to find any interesting point on this
sextic.
Friendly. Jean-Pierre
Your message has been successfully submitted and would be delivered to recipients shortly.