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## Centers

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• And to think research in euclidean geometry was declared dead years ago! Is the following known? It relates to a problem from Antreas. Let P be a point in the
Message 1 of 19 , Feb 3, 2000
And to think research in euclidean geometry was declared dead years ago!

Is the following known? It relates to a problem from Antreas.

Let P be a point in the plane of triangle ABC with centroid G,, let Ga,
Gb, and Gc be the centroids of triangles PBD, PAC, and PAB respectively,
then AGa, BGb, and CGc are concurrent at a point that is three-fourths
the way from P to G.

Antreas had looked at using incenters, but the problem is not universal
as for the centroid.

Conjecture: let X be a triangle center, then there exists a point P so
that if Xa, Sb, Xc are the centers of the triangles PBC, PAC, and PAB
then AXa, BXb, and CXc are concurrent.

It actually appears that for a given center, there are infintely many
such points P. But only for the centroid is it true for every point.

The theorem I have proved.

Michael Keyton
St. Mark's School of Texas
• on 2/3/00 10:01 PM Michael Keyton wrote ... Yes, this is at least in part known. ... Clark Kimberling proved most of this assertion in one of his papers on
Message 2 of 19 , Feb 4, 2000
on 2/3/00 10:01 PM Michael Keyton wrote

>Is the following known? It relates to a problem from Antreas.

Yes, this is at least in part known.

>
>Let P be a point in the plane of triangle ABC with centroid G,, let Ga,
>Gb, and Gc be the centroids of triangles PBD, PAC, and PAB respectively,
>then AGa, BGb, and CGc are concurrent at a point that is three-fourths
>the way from P to G.
>
>Antreas had looked at using incenters, but the problem is not universal
>as for the centroid.
>
>Conjecture: let X be a triangle center, then there exists a point P so
>that if Xa, Sb, Xc are the centers of the triangles PBC, PAC, and PAB
>then AXa, BXb, and CXc are concurrent.
>
>It actually appears that for a given center, there are infintely many
>such points P. But only for the centroid is it true for every point.

Clark Kimberling proved most of this assertion in one of his papers on
triangle centers about 5 or 6 years ago. He also gave a computational
apparatus in trilinears to compute the new center. The part of this
theorem that may be new is the last statement.

Steve
• Dear all, ... Has anyone investigated the problem of finding the locus of such points P. That must be a big big challenge although several special cases are
Message 3 of 19 , Feb 4, 2000
Dear all,

> on 2/3/00 10:01 PM Michael Keyton wrote
>
>>
>> Conjecture: let X be a triangle center, then there exists a point P so
>> that if Xa, Sb, Xc are the centers of the triangles PBC, PAC, and PAB
>> then AXa, BXb, and CXc are concurrent.
>>
>> It actually appears that for a given center, there are infintely many
>> such points P. But only for the centroid is it true for every point.
>
Has anyone investigated the problem of finding the locus of such points P.

That must be a big big challenge although several special cases are well
known.

regards

bernard
• on 2/4/00 10:26 AM Bernard Gibert wrote ... We have discussed some cases in the swarthmpore college groups. The article by Clark has some such discussion. He
Message 4 of 19 , Feb 4, 2000
on 2/4/00 10:26 AM Bernard Gibert wrote

>Has anyone investigated the problem of finding the locus of such points P.
>
>That must be a big big challenge although several special cases are well
>known.

We have discussed some cases in the swarthmpore college groups.

The article by Clark has some such discussion. He gives systematic ways
to compute the centers if P is the incenter.

This subject is s big challange but I also agree that it is worth
exploring.

Steve
• ... De : Bernard Gibert À : Envoyé : vendredi 4 février 2000 16:26 Objet : Re: [EMHL] Centers ... Bernard
Message 5 of 19 , Feb 4, 2000
----- Message d'origine -----
De : "Bernard Gibert" <b.gibert@...>
� : <Hyacinthos@onelist.com>
Envoy� : vendredi 4 f�vrier 2000 16:26
Objet : Re: [EMHL] Centers

> From: Bernard Gibert <b.gibert@...>
>
> Dear Bernard and all,
Bernard wrote
>
> > on 2/3/00 10:01 PM Michael Keyton wrote
> >
> >>
> >> Conjecture: let X be a triangle center, then there exists a point P so
> >> that if Xa, Sb, Xc are the centers of the triangles PBC, PAC, and PAB
> >> then AXa, BXb, and CXc are concurrent.
> >>
> >> It actually appears that for a given center, there are infintely many
> >> such points P. But only for the centroid is it true for every point.
> >
> Has anyone investigated the problem of finding the locus of such points P.
>
> That must be a big big challenge although several special cases are well
> known.
>
>
> regards
>
> bernard
>
> If the center is the Lemoine point , - and if I didn't mistake - your
locus is a nice bicircular curve of degree 5, with an asymptot parallel to
GK - G centroid -
This curve goes through
A, B, C, H, K, the reflections of A, B, C, H w.r.t the sides, the two
isodynamic points and a lot of other ones I don't know.
If the center is O (circumcenter), I saw the answer on Hyacinthos : the
union of the circumcircle and the Neuberg cubic.
Friendly from France. JPE
• Dear Bernard and all, if Pa, Pb, Pc are the Lemoine points of PBC, PCA, PAB, the locus of P in order to have Apa, BPb, CPc concurrent is, as I told, a
Message 6 of 19 , Feb 4, 2000
Dear Bernard and all,
if Pa, Pb, Pc are the Lemoine points of PBC, PCA, PAB, the locus of P in
order to have Apa, BPb, CPc concurrent is, as I told, a bicircular curve of
degree 5.
We need 16 conditions to determine a bicircular curve of degree 5.
This one
- goes through H; K; the two isodynamic points; the reflections of A, B, C
w.r.t BC, CA, AB;
- is tangent to AK at A, to BK at B, to CK at C
- the three other points on the circumcircle lie on the lines AK, BK, CK
Total = 16
In fact, we know the six points on the circumcircle; hence the curve cannot
go the through the reflections of H w.r.t the sides, as I said a bit too
quickly.
If A1 is the point of AK lying on the circumcircle, I think the curve goes
through A2 where
Vect(AA2) = -1/2 Vect(AA1).
The curve goes through the infinite point of GK too.
Friendly from France.
Jean - Pierre
• ... De : Steve Sigur À : Envoyé : vendredi 4 février 2000 18:47 Objet : Re: [EMHL] Centers ... Dear Steve,
Message 7 of 19 , Feb 6, 2000
----- Message d'origine -----
De : "Steve Sigur" <ssigur@...>
� : <Hyacinthos@onelist.com>
Envoy� : vendredi 4 f�vrier 2000 18:47
Objet : Re: [EMHL] Centers

> From: Steve Sigur <ssigur@...>

Dear Steve, Bernard and all,
Steve wrote :
>
> on 2/4/00 10:26 AM Bernard Gibert wrote
>
> >Has anyone investigated the problem of finding the locus of such points
P.
> >
> >That must be a big big challenge although several special cases are well
> >known.
>
> We have discussed some cases in the swarthmpore college groups.
>
> The article by Clark has some such discussion. He gives systematic ways
> to compute the centers if P is the incenter.
>
> This subject is s big challange but I also agree that it is worth
> exploring.
>
> Steve
>
I found the following result - may be, it is well known, but I don't think
so - :
Let X(ABC) a triangle center such as
Vect(OX) = k Vect(OH) - O circumcircle, H orthocenter, k constant - and X'
such as Vect(OX') = 1/k Vect(OH).
Then the locus of P such as the lines A_X(PBC),
B_X(PCA), C_X(PAB) are concurrent is the uninon of the circumcircle and of
the cubic F going through A, B, C, H, X, X' and the reflections A', B', C'
of the vertices w.r.t. the sides.
For instance :
X = H (k = 1) : F = the union of the three altitudes
X = O (k = 0) : F = The Neuberg cubic
X = Longchamps ( k = -1) : F = The cubic discovered by Bernard - this curves
goes through A,B,C,H,A',B',C', the eight centers of Soddy circle and is
tangent to the Euler_line at Longchamps
X = nine point center (k = 1/2) ...and so on.

I think it would be interesting to try to generalize this property
1) when k is a symmetric polynomial in a, b, c
2) when X(A, B, C) is barycentric f(a, b,c) | f(b,c,a) | f(c, a, b) where f
is an homogeneous polynomial
I conjecture that the locus of P is allways an algebraic curve through A, B,
C, H, A', B', C', X.

Friendly from France.
Jean-Pierre
• ... I wrote ... curves ... In fact all those cubics are members of the pencil of cubics going through A, B, C, A , B , C and tangent to HM at H, where M is
Message 8 of 19 , Feb 6, 2000
>Dear Steve, Bernard and all,
I wrote
> Let X(ABC) a triangle center such as
> Vect(OX) = k Vect(OH) - O circumcircle, H orthocenter, k constant - and X'
> such as Vect(OX') = 1/k Vect(OH).
> Then the locus of P such as the lines A_X(PBC),
> B_X(PCA), C_X(PAB) are concurrent is the uninon of the circumcircle and of
> the cubic F going through A, B, C, H, X, X' and the reflections A', B', C'
> of the vertices w.r.t. the sides.
> For instance :
> X = H (k = 1) : F = the union of the three altitudes
> X = O (k = 0) : F = The Neuberg cubic
> X = Longchamps ( k = -1) : F = The cubic discovered by Bernard - this
curves
> goes through A,B,C,H,A',B',C', the eight centers of Soddy circle and is
> tangent to the Euler_line at Longchamps
> X = nine point center (k = 1/2) ...and so on.
>

In fact all those cubics are members of the pencil of cubics going through
A, B, C, A', B', C' and tangent to HM at H, where M is the isogonal
conjugate of the nine-point center.
More over, the center of curvature at H is the same one for all those
cubics, but I didn't succeed in recognizing this point among the well known
triangle centers.
Friendly from France. Jean-Pierre
• Dear Jean-Pierre and all, Nice piece of work, Jean-Pierre ! Bravo. I just want to add several tiny notes as I was working on the same subject but Jean-Pierre
Message 9 of 19 , Feb 6, 2000
Dear Jean-Pierre and all,

Nice piece of work, Jean-Pierre ! Bravo.
I just want to add several tiny notes as I was working on the same subject
but Jean-Pierre has been much faster than me...
> Jean-Pierre wrote
>> Let X(ABC) a triangle center such as
>> Vect(OX) = k Vect(OH) - O circumcircle, H orthocenter, k constant - and X'
>> such as Vect(OX') = 1/k Vect(OH).
>> Then the locus of P such as the lines A_X(PBC),
>> B_X(PCA), C_X(PAB) are concurrent is the uninon of the circumcircle and of
>> the cubic F going through A, B, C, H, X, X' and the reflections A', B', C'
>> of the vertices w.r.t. the sides.
>> For instance :
>> X = H (k = 1) : F = the union of the three altitudes
>> X = O (k = 0) : F = The Neuberg cubic
>> X = Longchamps ( k = -1) : F = The cubic discovered by Bernard - this
>> curves
>> goes through A,B,C,H,A',B',C', the eight centers of Soddy circle and is
>> tangent to the Euler_line at Longchamps
>> X = nine point center (k = 1/2) ...and so on.
>>
>
> In fact all those cubics are members of the pencil of cubics going through
> A, B, C, A', B', C' and tangent to HM at H, where M is the isogonal
> conjugate of the nine-point center.

You need a nineth condition to be sure you have a pencil of cubics : you
only need to notice all the cubics have a triple contact in H because the
polar conic of H is a rectangular hyperbola going through H, L and the 3
harmonic conjugates of H wrt A & A', B & B', C & C' which means, as
Jean-Pierre said, they have the same center of curvature in H.

I checked M in Clark's ETC : it's called X(54) = Kosnita point. Never heard
of it before...

As I said to Jean-Pierre in a private message, I am not very happy with the
fact this locus is a quintic [5th degree]. I would prefer a sextic [6th
degree] decomposed in a quintic and the line at infinity. It would be more
coherent when we use complex projective notions like cyclic points, circular
curves, etc.

It seems this quintic-sextic degenerates when the center is on the Euler
line. It doesn't when it is K for example.

QUESTION :
when does the locus degenerates ?

I was looking for a geometric description of the cubic I found recently.
It came out from something apparently not related at a first sight showing
this cubic is in fact a close cousin of the Neuberg cubic.

Isn't fascinating ?

regards and congratulations to Jean-Pierre again.

bernard
• Dear Jean-Pierre, ... I ve just realized (and proved) the cubic we get for k is the same we get for 1/k. So, let s investigate now for k between -1 and 1.
Message 10 of 19 , Feb 6, 2000
Dear Jean-Pierre,
>
>> Dear Steve, Bernard and all,
> I wrote
>> Let X(ABC) a triangle center such as
>> Vect(OX) = k Vect(OH) - O circumcircle, H orthocenter, k constant - and X'
>> such as Vect(OX') = 1/k Vect(OH).
>> Then the locus of P such as the lines A_X(PBC),
>> B_X(PCA), C_X(PAB) are concurrent is the uninon of the circumcircle and of
>> the cubic F going through A, B, C, H, X, X' and the reflections A', B', C'
>> of the vertices w.r.t. the sides.
>> For instance :
>> X = H (k = 1) : F = the union of the three altitudes
>> X = O (k = 0) : F = The Neuberg cubic
>> X = Longchamps ( k = -1) : F = The cubic discovered by Bernard - this
> curves
>> goes through A,B,C,H,A',B',C', the eight centers of Soddy circle and is
>> tangent to the Euler_line at Longchamps
>> X = nine point center (k = 1/2) ...and so on.
>>
>
> In fact all those cubics are members of the pencil of cubics going through
> A, B, C, A', B', C' and tangent to HM at H, where M is the isogonal
> conjugate of the nine-point center.
> More over, the center of curvature at H is the same one for all those
> cubics, but I didn't succeed in recognizing this point among the well known
> triangle centers.

I've just realized (and proved) the "cubic" we get for k is the same we get
for 1/k.
So, let's investigate now for k between -1 and 1.

regards

bernard
• ... I have long used the name Ep (in which that p is a subscript) for the point O + p(H-O), the letter E , of course, standing for Eulerian since
Message 11 of 19 , Feb 7, 2000
On Sun, 6 Feb 2000, Bernard Gibert wrote:

> From: Bernard Gibert <b.gibert@...>
>
> Dear Jean-Pierre,
> >
> >> Dear Steve, Bernard and all,
> > I wrote
> >> Let X(ABC) a triangle center such as
> >> Vect(OX) = k Vect(OH) - O circumcircle, H orthocenter, k constant - and X'
> >> such as Vect(OX') = 1/k Vect(OH).

I have long used the name Ep (in which that "p" is a subscript) for the
point O + p(H-O), the letter "E", of course, standing for "Eulerian"
since the standard centers of this type are the Eulerian ones

E0 = O, E1 = H, E(1/3) = G, E(1/2) = N

and we can add the de Longchamps point E(-1) = L, and the point at infinity
on the Euler line, E(infinity).

So, if I'm reading it correctly, the result is that the locus of P for
which the Ek-centers of PBC,PCA,PAB are in perspective with ABC is the
union of the circumcircle and a certain cubic, let's call it Ck, that
generalizes the Neuberg cubic (k=0), and degenerates to the union of the
altitudes when k = 1.

Do you have the barycentric equation of this cubic?

John Conway
• ... De : John Conway À : Cc : Envoyé : lundi 7 février 2000 18:14 Objet :
Message 12 of 19 , Feb 7, 2000
----- Message d'origine -----
De : "John Conway" <conway@...>
� : <Hyacinthos@onelist.com>
Cc : <Hyacinthos@onelist.com>
Envoy� : lundi 7 f�vrier 2000 18:14
Objet : Re: [EMHL] Centers

> From: John Conway <conway@...>
>
>
>
> On Sun, 6 Feb 2000, Bernard Gibert wrote:
>
> > > Dear John, Bernard and all,
John wrote :
> I have long used the name Ep (in which that "p" is a subscript) for
the
> point O + p(H-O), the letter "E", of course, standing for "Eulerian"
> since the standard centers of this type are the Eulerian ones
>
> E0 = O, E1 = H, E(1/3) = G, E(1/2) = N
>
> and we can add the de Longchamps point E(-1) = L, and the point at
infinity
> on the Euler line, E(infinity).
>
> So, if I'm reading it correctly, the result is that the locus of P
for
> which the Ek-centers of PBC,PCA,PAB are in perspective with ABC is
the
> union of the circumcircle and a certain cubic, let's call it Ck, that
> generalizes the Neuberg cubic (k=0), and degenerates to the union of the
> altitudes when k = 1.
>
> Do you have the barycentric equation of this cubic?
>
> John Conway
>
>The Neuberg cubic should be
N = a^2(c^2 SC -2 SA SB) Z Y^2 - a^2 (b^2 SB -2 SA SC ) Z^2 Y + b^2(a^2
SA -2 SB SC) X Z^2 - b^2 (c^2 SC -2 SB SA ) X^2 Z + c^2(b^2 SB -2 SC SA) Y
X^2 - c^2 (a^2 SA -2 SC SB ) Y^2 X = 0
The three altitudes
H = (SA X - SB Y) (SB Y - SC Z) (SC Z - SA X) = 0
and the locus
(k - 1)^2 N + 4 k H = 0
Friendly from France.
Jean-Pierre
• ... De : John Conway À : Cc : Envoyé : lundi 7 février 2000 18:14 Objet :
Message 13 of 19 , Feb 9, 2000
----- Message d'origine -----
De : "John Conway" <conway@...>
� : <Hyacinthos@onelist.com>
Cc : <Hyacinthos@onelist.com>
Envoy� : lundi 7 f�vrier 2000 18:14
Objet : Re: [EMHL] Centers

> From: John Conway <conway@...>
>
>
>
> > > >
> > >> Dear John, Bernard and all,
> > > John wrote
> I have long used the name Ep (in which that "p" is a subscript) for
the
> point O + p(H-O), the letter "E", of course, standing for "Eulerian"
> since the standard centers of this type are the Eulerian ones
>
> E0 = O, E1 = H, E(1/3) = G, E(1/2) = N
>
> and we can add the de Longchamps point E(-1) = L, and the point at
infinity
> on the Euler line, E(infinity).
>
> So, if I'm reading it correctly, the result is that the locus of P
for
> which the Ek-centers of PBC,PCA,PAB are in perspective with ABC is
the
> union of the circumcircle and a certain cubic, let's call it Ck, that
> generalizes the Neuberg cubic (k=0), and degenerates to the union of the
> altitudes when k = 1.
>
>
> John Conway
>
> As Bernard - or myself - wrote, C(k) = C(1/k) is the member of the pencil
generated by the Neuberg cubic and the three altitudes going through Ek.
All those cubics go through A,B,C, their reflections w.r.t. the sides, H and
have the same center of curvature at H - they are tangent at H to the line
H - isogonal conjugate of nine points center.

I would like to add something : ( I incenter)
If the line I Ek intersects
- the Feuerbach hyperbola at I, G(k)
- the cubic C(k) at Ek, S1, S2
Then (I, G(k), S1, S2) is harmonic.
Of course, we have similar properties for the excenters.
Hence, as G(-1) = Gergonne, it becomes easiest to understand why the
"Bernard cubic " - C(-1) - goes through the eight Soddy centers.
I think there are a lot of other nice relations between the three following
pencils :
- the self isogonal cubics with pivot on the Euker line (they go through
vertices, incenter, excenters, orthocenter, circumcenter)
- the C(k) pencil
- the rectangular hyperbolas through A,B,C
We can notice too that
G(0) = incenter; G(-1) = Gergonne;
G(1) = orthocenter; G(1/3) = Nagel;
G(inf) = X_79; G(1/2) = X_80;
G(something) = Mittenpunkt ;
G(sthing else) = Schiffler ...

Friendly from France
Jean-Pierre
• Hello Hyacinthers, ... P in ... curve of ... Let s call (P) the quintic. Let s call Q the point of concurrence. The transformation f:P- Q is quadratic, so the
Message 14 of 19 , Sep 25, 2000
Hello Hyacinthers,

--- In Hyacinthos@egroups.com, "Jean-Pierre.EHRMANN"
<Jean-Pierre.EHRMANN@w...> wrote:
> Dear Bernard and all,
> if Pa, Pb, Pc are the Lemoine points of PBC, PCA, PAB, the locus of
P in
> order to have Apa, BPb, CPc concurrent is, as I told, a bicircular
curve of
> degree 5.

Let's call (P) the quintic.
Let's call Q the point of concurrence.
The transformation f:P->Q is quadratic, so the locus of Q should be a
10th degree curve (Q).

> We need 16 conditions to determine a bicircular curve of degree 5.
> This one
> - goes through H; K; the two isodynamic points; the reflections of
A, B, C
> w.r.t BC, CA, AB;
> - is tangent to AK at A, to BK at B, to CK at C
> - the three other points on the circumcircle lie on the lines AK,
BK, CK
> Total = 16
> In fact, we know the six points on the circumcircle; hence the
curve
cannot
> go the through the reflections of H w.r.t the sides, as I said a
bit
too
> quickly.
> If A1 is the point of AK lying on the circumcircle, I think the
curve goes
> through A2 where
> Vect(AA2) = -1/2 Vect(AA1).
> The curve goes through the infinite point of GK too.
> Friendly from France.
> Jean - Pierre

I have found the following results:

f(A1) = point of intersection of AK and BC.
f(A) = A
f(H) = X(53)
f(K) = X(251)
f(point of infinity of GK) = G
f(isodynamics) = {X(61),X(62)}
A2 is on the quintic and f(A2) = A
points KPQ are always colinears (may be it is true for all triangle
center?)
The quintic (P) intersects the line AK at {A,A,A1,A2,K}.
The quintic (P) intersects the line at infinity at the point of
infinity of GK and twice the cyclic points.
The quintic (P) intersects the line circumcircle (O) at
{A,B,C,A1,B1,C1, twice cyclic points

We can do very nice pictures of (P), but for (Q), I have no equation
for the moment.

Excuse-me for my bad english.

Regards

Fred Lang
• ... subscript) for ... Eulerian ... at ... of P ... ABC is ... that ... of the ... b^2(a^2 ... SC SA) Y ... Dear All, I find for the locus (in trilinear)
Message 15 of 19 , Jan 23, 2001
--- In Hyacinthos@egroups.com, "Jean-Pierre.EHRMANN"
<Jean-Pierre.EHRMANN@w...> wrote:
>
> ----- Message d'origine -----
> De : "John Conway" <conway@m...>
> À : <Hyacinthos@onelist.com>
> Cc : <Hyacinthos@onelist.com>
> Envoyé : lundi 7 février 2000 18:14
> Objet : Re: [EMHL] Centers
>
>
> > From: John Conway <conway@m...>
> >
> >
> >
> > On Sun, 6 Feb 2000, Bernard Gibert wrote:
> >
> > > > Dear John, Bernard and all,
> John wrote :
> > I have long used the name Ep (in which that "p" is a
subscript) for
> the
> > point O + p(H-O), the letter "E", of course, standing for
"Eulerian"
> > since the standard centers of this type are the Eulerian ones
> >
> > E0 = O, E1 = H, E(1/3) = G, E(1/2) = N
> >
> > and we can add the de Longchamps point E(-1) = L, and the point
at
> infinity
> > on the Euler line, E(infinity).
> >
> > So, if I'm reading it correctly, the result is that the locus
of P
> for
> > which the Ek-centers of PBC,PCA,PAB are in perspective with
ABC is
> the
> > union of the circumcircle and a certain cubic, let's call it Ck,
that
> > generalizes the Neuberg cubic (k=0), and degenerates to the union
of the
> > altitudes when k = 1.
> >
> > Do you have the barycentric equation of this cubic?
> >
> > John Conway
> >
> >The Neuberg cubic should be
> N = a^2(c^2 SC -2 SA SB) Z Y^2 - a^2 (b^2 SB -2 SA SC ) Z^2 Y +
b^2(a^2
> SA -2 SB SC) X Z^2 - b^2 (c^2 SC -2 SB SA ) X^2 Z + c^2(b^2 SB -2
SC SA) Y
> X^2 - c^2 (a^2 SA -2 SC SB ) Y^2 X = 0
> The three altitudes
> H = (SA X - SB Y) (SB Y - SC Z) (SC Z - SA X) = 0
> and the locus
> (k - 1)^2 N + 4 k H = 0
> Friendly from France.
> Jean-Pierre

Dear All,
I find for the locus (in trilinear)
(k - 1)^2 N + k H = 0

Regards.
Fred.
• ... wrote: ... and X ... and of ... B , C ... this curves ... is ... where f ... through A, B, ... Hello! The pencil of cubics F
Message 16 of 19 , Jan 23, 2001
--- In Hyacinthos@egroups.com, "Jean-Pierre.EHRMANN"
<Jean-Pierre.EHRMANN@w...> wrote:

....
> Let X(ABC) a triangle center such as
> Vect(OX) = k Vect(OH) - O circumcircle, H orthocenter, k constant -
and X'
> such as Vect(OX') = 1/k Vect(OH).
> Then the locus of P such as the lines A_X(PBC),
> B_X(PCA), C_X(PAB) are concurrent is the uninon of the circumcircle
and of
> the cubic F going through A, B, C, H, X, X' and the reflections A',
B', C'
> of the vertices w.r.t. the sides.
> For instance :
> X = H (k = 1) : F = the union of the three altitudes
> X = O (k = 0) : F = The Neuberg cubic
> X = Longchamps ( k = -1) : F = The cubic discovered by Bernard -
this curves
> goes through A,B,C,H,A',B',C', the eight centers of Soddy circle and
is
> tangent to the Euler_line at Longchamps
> X = nine point center (k = 1/2) ...and so on.
>
> I think it would be interesting to try to generalize this property
> 1) when k is a symmetric polynomial in a, b, c
> 2) when X(A, B, C) is barycentric f(a, b,c) | f(b,c,a) | f(c, a, b)
where f
> is an homogeneous polynomial
> I conjecture that the locus of P is allways an algebraic curve
through A, B,
> C, H, A', B', C', X.
>
> Friendly from France.
> Jean-Pierre

Hello!
The pencil of cubics F is very interesting indeed.
In this pencil, the Neuberg cubic is the only cubic iso-invariant.(
For example, there'is no isotomic-invariant) and the only circular
too.
The conjecture of Jean-Pierre seams quite difficult to proove. (I try
with centers on line OK and OI without success).

Bye.
Fred.
• ... [The locus of P for which the E(k)-centers of PBC,APC,ABP are in perspective.] John Conway ... Thanks. So it s an isogonal cubic, as I hoped.
Message 17 of 19 , Jan 24, 2001
On Tue, 23 Jan 2001 fred.lang@urbanet. chquoted and wrote:

> > > Do you have the barycentric equation of this cubic?

[The locus of P for which the E(k)-centers of PBC,APC,ABP are
in perspective.] > > > John Conway
> > >
> > >The Neuberg cubic should be
> > N = a^2(c^2 SC -2 SA SB) Z Y^2 - a^2 (b^2 SB -2 SA SC ) Z^2 Y +
> b^2(a^2
> > SA -2 SB SC) X Z^2 - b^2 (c^2 SC -2 SB SA ) X^2 Z + c^2(b^2 SB -2
> SC SA) Y
> > X^2 - c^2 (a^2 SA -2 SC SB ) Y^2 X = 0
> > The three altitudes
> > H = (SA X - SB Y) (SB Y - SC Z) (SC Z - SA X) = 0
> > and the locus
> > (k - 1)^2 N + 4 k H = 0
> > Friendly from France.
> > Jean-Pierre
>
> Dear All,
> I find for the locus (in trilinear)
> (k - 1)^2 N + k H = 0

Thanks. So it's an isogonal cubic, as I hoped. I meant
to say more, but some folks have just arrived to take me out to lunch!

John Conway
• Dear John, Fred and other Hyacinthists ... lunch! ... I m sorry, but I don t understand that. As the cubic goes through the orthocenter, the cubic, if it was
Message 18 of 19 , Jan 24, 2001
Dear John, Fred and other Hyacinthists
In Hyacinthos@egroups.com, John Conway <conway@m...> wrote:
> On Tue, 23 Jan 2001 fred.lang@urbanet. chquoted and wrote:
> > [Fred] Dear All,
> > I find for the locus (in trilinear)
> > (k - 1)^2 N + k H = 0
>
> Thanks. So it's an isogonal cubic, as I hoped. I meant
> to say more, but some folks have just arrived to take me out to
lunch!
>
I'm sorry, but I don't understand that.
As the cubic goes through the orthocenter, the cubic, if it was an
isogonal one, should go through the circumcenter, which occurs only
for k = 0.
Best regards. Jean-Pierre
• ... As I said, some folks arrived to take me out to lunch! I was about to think my way through this, but when they came I had to finish my message quickly,
Message 19 of 19 , Jan 24, 2001
On Wed, 24 Jan 2001, jp ehrmann wrote:

> Dear John, Fred and other Hyacinthists
> In Hyacinthos@egroups.com, John Conway <conway@m...> wrote:
> > On Tue, 23 Jan 2001 fred.lang@urbanet. chquoted and wrote:
> > > [Fred] Dear All,
> > > I find for the locus (in trilinear)
> > > (k - 1)^2 N + k H = 0
> >
> > Thanks. So it's an isogonal cubic, as I hoped. I meant
> > to say more, but some folks have just arrived to take me out to
> lunch!
> >
> I'm sorry, but I don't understand that.
> As the cubic goes through the orthocenter, the cubic, if it was an
> isogonal one, should go through the circumcenter, which occurs only
> for k = 0.
> Best regards. Jean-Pierre

As I said, some folks arrived to take me out to lunch! I was about to
think my way through this, but when they came I had to finish my message
quickly, and just mentally miscalculatedthat this cubic was isogonal.
Sorry!

JHC
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