## Re: [EMHL] inSteiner foci

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• Dear Bernard, Let A be the midpoint between B and C. The locus of a point M such that BC is a bisector of A M and A M (M isogonal conjugate of M) is a
Message 1 of 16 , Sep 24, 2001
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Dear Bernard,

Let A' be the midpoint between B and C. The locus of a point M such
that BC is a bisector of A'M and A'M' (M' isogonal conjugate of M)
is a isogonal circular focal cubic, without pivot, with root at infinity on
BC.

The foci of the Steiner inellipse are on this cubic K and on the cubic
reflected
of K through G.

Similarly with B' and C'.

We obtain 6 cubics which are 4 common points : the 4 foci of the
Steiner inellipse. The 4 foci are on the "second" Brocard cubic.

Gilles

Bernard Gibert a *crit :

> Dear Fred, Jean-Pierre and friends,
> >
> > I was just trying a joke : as I knew that Edward, Paul and Bernard
> > were looking for a nice form of the coordinates of the focii of the
> > Steiner inellipse, I've sent this little problem because
> > your "CircumcevianCentroids" are those focii - it's quite funny to
> > prove that -
> > I'm sorry for for the time you've lost because of my stupid joke.
>
> I don't think Fred lost its time since he confirms the coordinates I had
> found, although mine are symmetrical but absolutely horrendous.
> The trilinears are 4.97002 & 0.288722
>
> Thanks a lot Fred !
>
> (Edward : I've checked those foci are on the "second" Brocard cubic)
>
> Bernard
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• Dear Gilles and Edward, [GB] ... [cubics denoted by K ,K ] ... That s very good indeed ! Hence the three focals K,K ,K and the second Brocard are in the same
Message 2 of 16 , Sep 24, 2001
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Dear Gilles and Edward,

[GB]
> Let A' be the midpoint between B and C. The locus of a point M such
> that BC is a bisector of A'M and A'M' (M' isogonal conjugate of M)
> is a isogonal circular focal cubic, without pivot, with root at infinity on
> BC.
>
> The foci of the Steiner inellipse are on this cubic K and on the cubic
> reflected
> of K through G.
>
> Similarly with B' and C'.

[cubics denoted by K',K"]
>
> We obtain 6 cubics which are 4 common points : the 4 foci of the
> Steiner inellipse. The 4 foci are on the "second" Brocard cubic.
>
That's very good indeed !

Hence the three focals K,K',K" and the second Brocard are in the same pencil
of focals through A,B,C,the two circular points at infinity and the four
foci.

In this pencil there is also a strophoid with node at I, root at X514, focus
at X106, infinity at X519. It is the locus of foci of inscribed conics whose
center is on GI.
The tangents at I are parallel to the asymptotes of the Feuerbach hyperbola
but I don't think the cubic is its inversive image wrt some circle centered
at I. It's too late to go on...

Equation for Edward and §4.3.2 :

(b-c) x (c^2y^2+b^2z^2) + cyclic + 2(b-c)(c-a)(a-b) xyz = 0

Similarly, we can define three other strophoids with excenters.

Best regards

Bernard
• Dear Hyacinthists, the circumcircle of PBC intersects again the line AB at Ab and the line AC at Ac. A = BAc inter CAb Similarly, define B , C . Find the
Message 3 of 16 , Sep 26, 2001
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Dear Hyacinthists,
the circumcircle of PBC intersects again the line AB at Ab and the
line AC at Ac.
A' = BAc inter CAb
Similarly, define B', C'.
Find the locus of P such as ABC and A'B'C' are perspective.
I guess that Bernard will be happy to meet one of his best friends.
Friendly. Jean-Pierre
• Dear Jean-Pierre and Hyacinthists, [JP] ... Good old B2 ! (second Brocard cubic) (b^2 - c^2) x (c^2y^2 + b^2z^2) + cyclic = 0 (barys) It is very nice to denote
Message 4 of 16 , Sep 27, 2001
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Dear Jean-Pierre and Hyacinthists,

[JP]
> the circumcircle of PBC intersects again the line AB at Ab and the
> line AC at Ac.
> A' = BAc inter CAb
> Similarly, define B', C'.
> Find the locus of P such as ABC and A'B'C' are perspective.
> I guess that Bernard will be happy to meet one of his best friends.

Good old B2 ! (second Brocard cubic)
(b^2 - c^2) x (c^2y^2 + b^2z^2) + cyclic = 0 (barys)

It is very nice to denote that the perspector Q is on B2 as well and is what
I call Psi(P) : if F1 & F2 are the foci of the inSteiner ellipse, Psi is the
product of the symmetry about the line F1F2 and the inversion wrt the circle
with diameter F1F2.
This is a really nice involution with many properties.

JP : I'm happy to find a geometrical description of B2 and to see that our
work on orthopivotal cubics reappears.

Thank you

Bernard
• Dear Hyacinthists, P* is the isogonal conjugate of P The lines PB and PC intersects again the Kiepert hyperbola respectively at B and C . Find the locus of P
Message 5 of 16 , Sep 27, 2001
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Dear Hyacinthists,

P* is the isogonal conjugate of P
The lines PB and PC intersects again the Kiepert hyperbola
respectively at B' and C'.
Find the locus of P for which the lines AP*, B'C' intersect on the
Brocard axis OK.
Friendly. Jean-Pierre

PS : I apologize if my message is duplicated, but i never received
the first one
• Dear Jean-Pierre and Hyacinthists, ... [JP] ... It s the day of B2 ! Thanks JP. in the style of the same nobody : let A = B C / AP*, B & C likewise what
Message 6 of 16 , Sep 27, 2001
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Dear Jean-Pierre and Hyacinthists,
>
[JP]
> P* is the isogonal conjugate of P
> The lines PB and PC intersects again the Kiepert hyperbola
> respectively at B' and C'.
> Find the locus of P for which the lines AP*, B'C' intersect on the
> Brocard axis OK.

It's the day of B2 ! Thanks JP.

in the style of the same nobody :

let A" = B'C' /\ AP*, B" & C" likewise

what is the locus of P such that A",B",C" are collinear ?
---------------------------------------------------------
generalization in the style of Antreas :

replace OK with any line through O and Kiepert with the isogonal conjugate
of the line.

what are the 2 loci above ?
---------------------------------------------------------

generalization of the generalization in the style of Antreas :

replace OK with any line d and Kiepert with the isoconjugate of the line.

what are the 2 loci above ?

Best regards

Bernard
• Dear Bernard The 2 loci are the same : an isogonal cubic without pivot, root at the isogonal of the trilinear pole of the lline d. Friendly Gilles
Message 7 of 16 , Sep 27, 2001
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Dear Bernard

The 2 loci are the same : an isogonal cubic without pivot, root at the
isogonal
of the trilinear pole of the lline d.

Friendly

Gilles

Bernard Gibert a *crit :

> Dear Jean-Pierre and Hyacinthists,
> >
> [JP]
> > P* is the isogonal conjugate of P
> > The lines PB and PC intersects again the Kiepert hyperbola
> > respectively at B' and C'.
> > Find the locus of P for which the lines AP*, B'C' intersect on the
> > Brocard axis OK.
>
> It's the day of B2 ! Thanks JP.
>
> in the style of the same nobody :
>
> let A" = B'C' /\ AP*, B" & C" likewise
>
> what is the locus of P such that A",B",C" are collinear ?
> ---------------------------------------------------------
> generalization in the style of Antreas :
>
> replace OK with any line through O and Kiepert with the isogonal conjugate
> of the line.
>
> what are the 2 loci above ?
> ---------------------------------------------------------
>
> generalization of the generalization in the style of Antreas :
>
> replace OK with any line d and Kiepert with the isoconjugate of the line.
>
> what are the 2 loci above ?
>
> Best regards
>
> Bernard
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• Dear Gilles, ... Good ! I was just attempting to symmetrize JP s idea. Best regards Bernard
Message 8 of 16 , Sep 27, 2001
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Dear Gilles,
>
> The 2 loci are the same : an isogonal cubic without pivot, root at the
> isogonal
> of the trilinear pole of the lline d.
>
Good ! I was just attempting to "symmetrize" JP's idea.

Best regards

Bernard

>
> Bernard Gibert a *crit :
>
>> Dear Jean-Pierre and Hyacinthists,
>>>
>> [JP]
>>> P* is the isogonal conjugate of P
>>> The lines PB and PC intersects again the Kiepert hyperbola
>>> respectively at B' and C'.
>>> Find the locus of P for which the lines AP*, B'C' intersect on the
>>> Brocard axis OK.
>>
>> It's the day of B2 ! Thanks JP.
>>
>> in the style of the same nobody :
>>
>> let A" = B'C' /\ AP*, B" & C" likewise
>>
>> what is the locus of P such that A",B",C" are collinear ?
>> ---------------------------------------------------------
>> generalization in the style of Antreas :
>>
>> replace OK with any line through O and Kiepert with the isogonal conjugate
>> of the line.
>>
>> what are the 2 loci above ?
>> ---------------------------------------------------------
>>
>> generalization of the generalization in the style of Antreas :
>>
>> replace OK with any line d and Kiepert with the isoconjugate of the line.
>>
>> what are the 2 loci above ?
>>
>> Best regards
>>
>> Bernard
>>
>>
>>
>>
>> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
• Dear Bernard, Gilles and other Hyacinthists, As you ve noticed, if we consider a line D, an isoconjugation i and the second common points Pa, Pb, Pc of AP, BP,
Message 9 of 16 , Sep 27, 2001
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Dear Bernard, Gilles and other Hyacinthists,

As you've noticed, if we consider a line D, an isoconjugation i and
the second common points Pa, Pb, Pc of AP, BP, CP with the
circumconic i(D), the following properties are equivalent, for P not
on D and not on i(D) :
1) i(P)A, PbPc, D concur
2) i(P)B, PcPa, D concur
3) i(P)C, PaPb, D concur
4) UPa, VPb, WPc concur(U,V,W = common points of D with the sides of
ABC)
5) P lies on the i(D)-invariant circumcubic with root the tripole of
D and parameter 0 (which means without xyz term or such as the
tangent at U, V, W concur)
6).... (using duality or anything else)

It becomes quite nice with Kjp ( D = line at infinity, i(D) =
circumcircle ) : P lies on Kjp if AP* and B'C' are parallel, ...
A'B'C' = circumcevian triangle of P.(May be, Bernard, you can write
that in the isocubics paper)

Starting from this remark, I've immediately generalized by
projectivity and then I've applied that to B2 to propose you this
locus in nobody's style.

In fact, it appears that, among the non-pivotal isocubics,the ones
such as the tangents at U,V,W concur have a lot of special
properties, which are similar or dual of the properties of pivotal
isocubics.

So, I propose to use a special term - if it doesn't still exist - for
this kind of isocubics : something like "rooted" or root-pivotal
isocubics.

What do you think of that ?
Friendly. Jean-Pierre

PS : As my mail is long enough, I didn't repeat the previous ones.
• Dear Hyacinthists, I wrote ... not ... of ... of ... The root is not the tripole of D but his isoconjugate (the perspector of the conic) and the concurrent
Message 10 of 16 , Sep 27, 2001
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Dear Hyacinthists,
I wrote

> As you've noticed, if we consider a line D, an isoconjugation i and
> the second common points Pa, Pb, Pc of AP, BP, CP with the
> circumconic i(D), the following properties are equivalent, for P
not
> on D and not on i(D) :
> 1) i(P)A, PbPc, D concur
> 2) i(P)B, PcPa, D concur
> 3) i(P)C, PaPb, D concur
> 4) UPa, VPb, WPc concur(U,V,W = common points of D with the sides
of
> ABC)
> 5) P lies on the i(D)-invariant circumcubic with root the tripole
of
> D and parameter 0 (which means without xyz term or such as the
> tangent at U, V, W concur)

The root is not the tripole of D but his isoconjugate (the perspector
of the conic) and the concurrent tangents are the tangent at the
common points of the trilinear polar of the root with the sidelines
of ABC.

Friendly. JP
• Dear Jean-Pierre and other Hyacinthists, this is very interesting indeed ! I will only add several minor remarks... ... *** The point they concur at is the
Message 11 of 16 , Sep 28, 2001
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Dear Jean-Pierre and other Hyacinthists,

this is very interesting indeed !
I will only add several minor remarks...
>
> As you've noticed, if we consider a line D, an isoconjugation i and
> the second common points Pa, Pb, Pc of AP, BP, CP with the
> circumconic i(D), the following properties are equivalent, for P not
> on D and not on i(D) :
> 1) i(P)A, PbPc, D concur
> 2) i(P)B, PcPa, D concur
> 3) i(P)C, PaPb, D concur
> 4) UPa, VPb, WPc concur(U,V,W = common points of D with the sides of
> ABC)

*** The point they concur at is the pole of the line PP* in the circumconic
i(D).

> 5) P lies on the i(D)-invariant circumcubic with root the tripole of
> D and parameter 0 (which means without xyz term or such as the
> tangent at U, V, W concur)
> 6).... (using duality or anything else)
>
> It becomes quite nice with Kjp ( D = line at infinity, i(D) =
> circumcircle ) : P lies on Kjp if AP* and B'C' are parallel, ...
> A'B'C' = circumcevian triangle of P.(May be, Bernard, you can write
> that in the isocubics paper)

*** I will, dear JP ! That's another characterization of your favourite
cubic.
>
> Starting from this remark, I've immediately generalized by
> projectivity and then I've applied that to B2 to propose you this
> locus in nobody's style.

*** I've tried with other known lines/conics/cubics :
isogonal examples,
D=Euler line, i(D)=Jerabek, the cubic is B3 (third Brocard cubic)
D=OI, i(D)=Feuerbach, the cubic is the "Pelletier" cubic.

In fact, any line through O corresponds to a rectangular hyperbola and to a
circular focal cubic.

I'll try to investigate and try to relate our "new" cubics to lines and/or
conics.

> In fact, it appears that, among the non-pivotal isocubics,the ones
> such as the tangents at U,V,W concur have a lot of special
> properties, which are similar or dual of the properties of pivotal
> isocubics.
>
> So, I propose to use a special term - if it doesn't still exist - for
> this kind of isocubics : something like "rooted" or root-pivotal
> isocubics.

*** I must admit I'm not thrilled neither with "rooted" (since any
nK=non-pivotal isocubic has a root) nor with "root-pivotal" (since it might
create confusion with pK= pivotal isocubic). For the paper I propose to
denote them by nK_0 and let's try to find something better for the name.
>
> What do you think of that ?

THAT'S A GREAT IDEA ! BRAVO, JP !

best regards

Bernard
• Dear Bernard and other Hyacinthists, we ve seen that if i is an isoconjugation D a line intersecting the sides of ABC at U, V, W i(D) the circumconic
Message 12 of 16 , Sep 28, 2001
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Dear Bernard and other Hyacinthists,

we've seen that if
i is an isoconjugation
D a line intersecting the sides of ABC at U, V, W
i(D) the circumconic isoconjugate of D
if the line AP intersects again i(D) at A',..
the following properties are equivalent
1)Ai(P), B'C', D concur
2)Bi(P), C'A', D concur
3)Ci(P), A'B', D concur
4)UA', VB', WC' concur
5)P lies on the nK_0 i-invriant cubic with root the perspector of i
(D) (= isoconjugate of the tripole of D)

Now if I, J is the pair of isoconjugate points lying on D, if we use
the involution on D with fixed points I, J, we can associate to the
nK_0 a pivotal isocubic.

In fact the following properties are equivalent :
1)(Ai(P) inter D) and (B'C' inter D) are harmonic conjugate wrt I,J
2)(Bi(P) ...)
3)(Ci(P)...)
4)U'A', V'B', W'B' concur where U' is the harmonic conjugate of U wrt
I,J
5)P lies on the pivotal isocubic invariant by i with pivot the pole
of D wrt i(D)

I propose to call this pivotal cubic the pivotal sister of the nK_0
If the nK_0 is ux(ry^2+qz^2)+.. = 0, then the pivotal sister is
(-u^2/p+v^2/q+w^2/r)ux(ry^2-qz^2) + ... = 0
For instance, the pivotal sister of Kjp is the Mac Cay cubic, the
pivotal sister of B2 is the Napoleon cubic ...
Of course, it is a strange family, because a pivotal isocubic is the
sister of 3 nK_0.
I'm sure that Fred will find a very nice isomorphism between those
cubics.

Friendly. Jean-Pierre
• Dear Jean-Pierre and other Hyacinthists, The tripole of D is the common point of the tangents at U,V,W. Regards Gilles
Message 13 of 16 , Sep 28, 2001
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Dear Jean-Pierre and other Hyacinthists,

The tripole of D is the common point of the tangents at U,V,W.

Regards

Gilles

jean-pierre.ehrmann@... a *crit :

> Dear Hyacinthists,
> I wrote
>
> > As you've noticed, if we consider a line D, an isoconjugation i and
> > the second common points Pa, Pb, Pc of AP, BP, CP with the
> > circumconic i(D), the following properties are equivalent, for P
> not
> > on D and not on i(D) :
> > 1) i(P)A, PbPc, D concur
> > 2) i(P)B, PcPa, D concur
> > 3) i(P)C, PaPb, D concur
> > 4) UPa, VPb, WPc concur(U,V,W = common points of D with the sides
> of
> > ABC)
> > 5) P lies on the i(D)-invariant circumcubic with root the tripole
> of
> > D and parameter 0 (which means without xyz term or such as the
> > tangent at U, V, W concur)
>
> The root is not the tripole of D but his isoconjugate (the perspector
> of the conic) and the concurrent tangents are the tangent at the
> common points of the trilinear polar of the root with the sidelines
> of ABC.
>
> Friendly. JP
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• Dear Jean-Pierre, Bernard, Gilles and all Hyacinthists, The subject of JP is a special case of jacobian: General: If C(1), C(2) and C(3) are three independant
Message 14 of 16 , Sep 29, 2001
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Dear Jean-Pierre, Bernard, Gilles and all Hyacinthists,

The subject of JP is a special case of jacobian:

General:
If C(1), C(2) and C(3) are three independant conics (i.e. C3 not in
the pencil generated by C1 and C2), they form a net of conics.
Let p(i) be the polar of P relative to C(i).
The locus of P such that the p(i) concur is a cubic J, known as the
jacobian of the net of conics.

J is the locus of the double points of the degenerated conics of the
net.

Special:
Take for C(1), C(2) two generators of the pencil of conics which
defines the isoconjugation i and put C(3) = i(D), then J is the nK_0
i-invariant cubic of JP.

Example:
C(1) = (I Ia)(Ib Ic), C(2) = (I Ib)(Ia Ic), C(3) = (O), i = isogonal
transformation and J = KjP.

In the article, I sended to Paul for FG, you'll see that:
1) Every non-pivotal isocubic is the jacobian of a net.
2) If C(3) is circumscribed to the autopolar triangle of C(1), C(2),
then you receive a nK_0 cubic, and in this case the circumcircle of
ABC is in the net.
Later I plan:
3)properties of the special case nK_0
4) group structure (difficult, but I have the elliptic invariant of
the cubics)

Indeed, in Durège you have 100 pages of properties relative to the
jacobian...

I think it could be a good idea to write something together for FG
(JPE + BG + FL +...)

Regards.

Fred.

--- In Hyacinthos@y..., jean-pierre.ehrmann@w... wrote:

> we've seen that if
> i is an isoconjugation
> D a line intersecting the sides of ABC at U, V, W
> i(D) the circumconic isoconjugate of D
> if the line AP intersects again i(D) at A',..
> the following properties are equivalent
> 1)Ai(P), B'C', D concur
> 2)Bi(P), C'A', D concur
> 3)Ci(P), A'B', D concur
> 4)UA', VB', WC' concur
> 5)P lies on the nK_0 i-invriant cubic with root the perspector of i
> (D) (= isoconjugate of the tripole of D)
>
> Now if I, J is the pair of isoconjugate points lying on D, if we use
> the involution on D with fixed points I, J, we can associate to the
> nK_0 a pivotal isocubic.
>
> In fact the following properties are equivalent :
> 1)(Ai(P) inter D) and (B'C' inter D) are harmonic conjugate wrt I,J
> 2)(Bi(P) ...)
> 3)(Ci(P)...)
> 4)U'A', V'B', W'B' concur where U' is the harmonic conjugate of U
wrt
> I,J
> 5)P lies on the pivotal isocubic invariant by i with pivot the pole
> of D wrt i(D)
>
> I propose to call this pivotal cubic the pivotal sister of the nK_0
> If the nK_0 is ux(ry^2+qz^2)+.. = 0, then the pivotal sister is
> (-u^2/p+v^2/q+w^2/r)ux(ry^2-qz^2) + ... = 0
> For instance, the pivotal sister of Kjp is the Mac Cay cubic, the
> pivotal sister of B2 is the Napoleon cubic ...
> Of course, it is a strange family, because a pivotal isocubic is the
> sister of 3 nK_0.
> I'm sure that Fred will find a very nice isomorphism between those
> cubics.
>
> Friendly. Jean-Pierre
• Dear Fred, thank you for your very interesting remarks. They should lead to alot ofnice results. ... the ... nK_0 ... The tangent to i(D) intersect BC, CA, AB
Message 15 of 16 , Sep 29, 2001
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Dear Fred,
thank you for your very interesting remarks.
They should lead to alot ofnice results.

> The subject of JP is a special case of jacobian:
>
> General:
> If C(1), C(2) and C(3) are three independant conics (i.e. C3 not in
> the pencil generated by C1 and C2), they form a net of conics.
> Let p(i) be the polar of P relative to C(i).
> The locus of P such that the p(i) concur is a cubic J, known as the
> jacobian of the net of conics.
>
> J is the locus of the double points of the degenerated conics of
the
> net.
>
> Special:
> Take for C(1), C(2) two generators of the pencil of conics which
> defines the isoconjugation i and put C(3) = i(D), then J is the
nK_0
> i-invariant cubic of JP.

The tangent to i(D) intersect BC, CA, AB respectively at U, V, W
lying on a same line
Thus a conic is member of your net iff the polar of A, B, C wrt the
conic go respectively through U,V,W.
>
> Example:
> C(1) = (I Ia)(Ib Ic), C(2) = (I Ib)(Ia Ic), C(3) = (O), i =
isogonal
> transformation and J = KjP.
>
> In the article, I sended to Paul for FG, you'll see that:
> 1) Every non-pivotal isocubic is the jacobian of a net.
> 2) If C(3) is circumscribed to the autopolar triangle of C(1), C
(2),
> then you receive a nK_0 cubic, and in this case the circumcircle of
> ABC is in the net.

I don't understand that; I think that the only circumcubic in the net
is the circumconic with perspector the root of the cubic - but, may
be, you were talking of the special case of Kjp -

> Later I plan:
> 3)properties of the special case nK_0
> 4) group structure (difficult, but I have the elliptic invariant of
> the cubics)
>
> Indeed, in Durège you have 100 pages of properties relative to the
> jacobian...
>
> I think it could be a good idea to write something together for FG
> (JPE + BG + FL +...)

Why not!!
>
> Regards.
>
> Fred.
• Dear JP, ... Yes. ... of ... net ... That s right, It s circumconic and not circumcircle. Sorry. Bye Fred
Message 16 of 16 , Oct 1, 2001
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Dear JP,
--- In Hyacinthos@y..., jean-pierre.ehrmann@w... wrote:

> The tangent to i(D) intersect BC, CA, AB respectively at U, V, W
> lying on a same line
> Thus a conic is member of your net iff the polar of A, B, C wrt the
> conic go respectively through U,V,W.

Yes.

> > 2) If C(3) is circumscribed to the autopolar triangle of C(1), C
> (2),
> > then you receive a nK_0 cubic, and in this case the circumcircle
of
> > ABC is in the net.
>
> I don't understand that; I think that the only circumconic in the
net
> is the circumconic with perspector the root of the cubic - but, may
> be, you were talking of the special case of Kjp -

That's right, It's circumconic and not circumcircle.
Sorry.
Bye
Fred
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