- Dear friends,
>

after rereading that, I feel it is not very clear :

> I conjecture that the three conics form a mesh iff M is a point of the

> Neuberg cubic. If so, they must have two other (real or not) points N1 & N2

> in common also on the Neuberg cubic.

I just wanted to say that the 3 conics have 3 points in common (one of them

being M) iff M is a point of the Neuberg cubic and then the two other points

N1 & N2 also lie on it.

No mesh in it !!!

Best regards

Bernard - Dear Bernard

> GaGbGc is the antimedial triangle,

the

> A',B',C' are the reflections of A / BC, etc.

>

> for any point M,

> Ca is the conic through B, C, A', Ga, M,

> Cb, Cc defined likewise.

>

> I conjecture that the three conics form a mesh iff M is a point of

> Neuberg cubic. Can somebody help ?

As you wrote in another message, your conjecture is that the three

conics have two other common points (real or not) apart M iff M lies

on the Neuberg cubic.

This is true but I didn't find an easy proof.

Here is a summary :

Suppose that M is barycentric (p : q : r); consider the triangle

UaUbUc inscribed in the antimedial triangle such as (algebrically)

GaUc/GbUc = p (2.SA.r + c^2.q)/q/(2.SB.r + c^2.p); Ua,Ub cyclic

Now, using combination of their equations, we get that the union of

the lines CM and UaUb is a degenerated conic of the pencil generated

by Ca and Cb.

Hence, the three conics will have two other common points (apart M)

iff Ua, Ub, Uc lie on a same line, ie (Menelaus)

((2.SA.r + c^2.q)/(2.SB.r + c^2.p))*cyclic*cyclic = 1, which means

exactly that M lies on the Neuberg cubic.

Friendly. Jean-Pierre - --- jpehrmfr <jean-pierre.ehrmann@...> wrote:
> Dear Bernard

three

>

> > GaGbGc is the antimedial triangle,

> > A',B',C' are the reflections of A / BC, etc.

> >

> > for any point M,

> > Ca is the conic through B, C, A', Ga, M,

> > Cb, Cc defined likewise.

> >

> > I conjecture that the three conics form a mesh iff M is a

> > point of the Neuberg cubic. Can somebody help ?

>

> As you wrote in another message, your conjecture is that the

> conics have two other common points (real or not) apart M iff

M lies

> on the Neuberg cubic.

triangle

> This is true but I didn't find an easy proof.

> Here is a summary :

> Suppose that M is barycentric (p : q : r); consider the

> UaUbUc inscribed in the antimedial triangle such as

(algebrically)

> GaUc/GbUc = p (2.SA.r + c^2.q)/q/(2.SB.r + c^2.p); Ua,Ub

cyclic

> Now, using combination of their equations, we get that the

union of

> the lines CM and UaUb is a degenerated conic of the pencil

generated

> by Ca and Cb.

(apart M)

> Hence, the three conics will have two other common points

> iff Ua, Ub, Uc lie on a same line, ie (Menelaus)

means

> ((2.SA.r + c^2.q)/(2.SB.r + c^2.p))*cyclic*cyclic = 1, which

> exactly that M lies on the Neuberg cubic.

I verified the conjecture by a different method. With M as

> Friendly. Jean-Pierre

above, and P=(x:y:z), the equation for Ca can be written as

z/x = (k1 + k2 t)/(k3 + k4 t)

where t=y/x, for some ki which are polynomials in a,b,c,p,q,r.

Plugging this into the equation for Cb, we get a cubic

polynomial Pb(t), with a known root t=q/p. Treating Cc similarly

gives Pc(t), and we want the two poly's Pb and Pc to have the

same roots. Comparing their coefficients leads quickly to the

Neuberg cubic.

The two additional points lying on the Neuberg cubic will come

from the other two roots of Pb (or of Pc). They have irrational

coordinates. but they lie on a line with rational coordinates.

--

Barry Wolk

__________________________________________________

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http://hotjobs.yahoo.com/ - Dear Jean-Pierre, Nikos, Paul and all Hyacinthos!

In the October of 2002 we discussed the problem in subject.

In message 5999 Jean-Pierre Ehrmann asked:

> the former message from Sergei suggests me the following problem :

In message 6000 Paul Yiu answered:

> for reasons of continuity, there exist two perpendicular lines D, D'

> dividing ABC in 4 parts of equal area. How can we construct these

> lines? (I don't know the answer) I presume that this problem is very

> old one.

> Euler has written 2 long papers on this problem:

I have found this Euler's 200 years old paper in Russian State Library,

>

> [729] Dilucidationes super problemate geometrico de quadrisectione,

> M\'emoires de l'acad\'emie des sciences de St. P\'etersbourg,

> {1803/6}{1809}{26--48} I26, 286--303.

>

> [730] Solutio completa problematis de quadrisectione trianguli

> per duas rectas inter se normales, ibid. {1803/6}{1809}{49--87}

> I26, 304--333.

>

> These papers are in volume 26 of his opera omnia.

and scanned it so that those of you who are not happy to have this book

in local library will still be able to read Euler's original text.

Article is in Latin (not in Russian).

http://www.mccme.ru/~markelov/29142-3.pdf

Size is 4698445 bytes (48 pages). Sorry for slow downloading from

Russia.

Interested in this problem should also have a look at the message

number 6009 by Nikolaos Dergiades:

> this problem is Leibniz's problem and leads to an equation of eighth

Another related posting was my message number 5997 with solution to the

> degree. F.G-M 1624a. If a point D is on the side AB then we can

> construct the point E on AC such that 2(ADE) = (ABC) To construct

> now a line MP that divides the triangle ADE and the quadrilateral

> DBCE in two equal parts is Huygens' problem (1650) F.G.-M 1674c.

> The above construction needs the common tangent MP to two

> hyperbolae but since they have DE as common asymptote this

> construction leads to a second degree equation and is ruler and

> compass possible. Now my ancestors would move the point D on AB until

> the lines DE, MP be perpendicular.

easier problem (constructing line through given point bisecting the

area of the given triangle).

Best regards,

Sergei Markelov