Hyacinthos is a Restricted Group with 3 members.
 Hyacinthos

 Restricted Group,
 3 members
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 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 threefourths
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
>Is the following known? It relates to a problem from Antreas.
Yes, this is at least in part known.
>
Clark Kimberling proved most of this assertion in one of his papers on
>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 threefourths
>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.
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,
> on 2/3/00 10:01 PM Michael Keyton wrote
Has anyone investigated the problem of finding the locus of such points P.
>
>>
>> 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.
>
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
>Has anyone investigated the problem of finding the locus of such points P.
We have discussed some cases in the swarthmpore college groups.
>
>That must be a big big challenge although several special cases are well
>known.
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   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@...>
Bernard wrote
>
> Dear Bernard and all,
>
locus is a nice bicircular curve of degree 5, with an asymptot parallel to
> > 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
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 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   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 :>
P.
> on 2/4/00 10:26 AM Bernard Gibert wrote
>
> >Has anyone investigated the problem of finding the locus of such points
> >
I found the following result  may be, it is well known, but I don't think
> >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
>
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.
JeanPierre >Dear Steve, Bernard and all,
I wrote
> Let X(ABC) a triangle center such as
curves
> 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
> goes through A,B,C,H,A',B',C', the eight centers of Soddy circle and is
In fact all those cubics are members of the pencil of cubics going through
> tangent to the Euler_line at Longchamps
> X = nine point center (k = 1/2) ...and so on.
>
A, B, C, A', B', C' and tangent to HM at H, where M is the isogonal
conjugate of the ninepoint 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. JeanPierre Dear JeanPierre and all,
Nice piece of work, JeanPierre ! Bravo.
I just want to add several tiny notes as I was working on the same subject
but JeanPierre has been much faster than me...> JeanPierre wrote
You need a nineth condition to be sure you have a pencil of cubics : you
>> 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 ninepoint center.
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
JeanPierre 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 JeanPierre 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 quinticsextic 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 JeanPierre again.
bernard  Dear JeanPierre,
>
I've just realized (and proved) the "cubic" we get for k is the same we get
>> 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 ninepoint 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.
for 1/k.
So, let's investigate now for k between 1 and 1.
regards
bernard  On Sun, 6 Feb 2000, Bernard Gibert wrote:
> From: Bernard Gibert <b.gibert@...>
I have long used the name Ep (in which that "p" is a subscript) for the
>
> Dear JeanPierre,
> >
> >> 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).
point O + p(HO), 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 Ekcenters 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   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@...>
John wrote :
>
>
>
> On Sun, 6 Feb 2000, Bernard Gibert wrote:
>
> > > Dear John, Bernard and all,
> I have long used the name Ep (in which that "p" is a subscript) for
the
> point O + p(HO), the letter "E", of course, standing for "Eulerian"
infinity
> 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
> on the Euler line, E(infinity).
for
>
> So, if I'm reading it correctly, the result is that the locus of P
> which the Ekcenters 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
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
> 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
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.
JeanPierre   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@...>
the
>
>
>
> > > >
> > >> Dear John, Bernard and all,
> > > John wrote
> I have long used the name Ep (in which that "p" is a subscript) for
> point O + p(HO), the letter "E", of course, standing for "Eulerian"
infinity
> 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
> on the Euler line, E(infinity).
for
>
> So, if I'm reading it correctly, the result is that the locus of P
> which the Ekcenters 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
generated by the Neuberg cubic and the three altitudes going through Ek.
> 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
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
JeanPierre  Hello Hyacinthers,
 In Hyacinthos@egroups.com, "JeanPierre.EHRMANN"
<JeanPierre.EHRMANN@w...> wrote:> Dear Bernard and all,
P in
> if Pa, Pb, Pc are the Lemoine points of PBC, PCA, PAB, the locus of
> 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.
A, B, C
> This one
>  goes through H; K; the two isodynamic points; the reflections of
> w.r.t BC, CA, AB;
BK, CK
>  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,
> Total = 16
curve
> In fact, we know the six points on the circumcircle; hence the
cannot> go the through the reflections of H w.r.t the sides, as I said a
bit
too> quickly.
curve goes
> If A1 is the point of AK lying on the circumcircle, I think the
> through A2 where
I have found the following results:
> Vect(AA2) = 1/2 Vect(AA1).
> The curve goes through the infinite point of GK too.
> Friendly from France.
> Jean  Pierre
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.
Excuseme for my bad english.
Regards
Fred Lang   In Hyacinthos@egroups.com, "JeanPierre.EHRMANN"
<JeanPierre.EHRMANN@w...> wrote:>
subscript) for
>  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
> the
"Eulerian"
> > point O + p(HO), the letter "E", of course, standing for
> > since the standard centers of this type are the Eulerian ones
at
> >
> > 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
> infinity
of P
> > on the Euler line, E(infinity).
> >
> > So, if I'm reading it correctly, the result is that the locus
> for
ABC is
> > which the Ekcenters of PBC,PCA,PAB are in perspective with
> the
that
> > union of the circumcircle and a certain cubic, let's call it Ck,
> > generalizes the Neuberg cubic (k=0), and degenerates to the union
of the
> > altitudes when k = 1.
b^2(a^2
> >
> > 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 +
> 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
Dear All,
> 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.
> JeanPierre
I find for the locus (in trilinear)
(k  1)^2 N + k H = 0
Regards.
Fred.   In Hyacinthos@egroups.com, "JeanPierre.EHRMANN"
<JeanPierre.EHRMANN@w...> wrote:
....> Let X(ABC) a triangle center such as
and X'
> Vect(OX) = k Vect(OH)  O circumcircle, H orthocenter, k constant 
> such as Vect(OX') = 1/k Vect(OH).
and of
> 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
> 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.
this curves
> 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 
> goes through A,B,C,H,A',B',C', the eight centers of Soddy circle and
is
> tangent to the Euler_line at Longchamps
where f
> 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)
> is an homogeneous polynomial
through A, B,
> I conjecture that the locus of P is allways an algebraic curve
> C, H, A', B', C', X.
Hello!
>
> Friendly from France.
> JeanPierre
The pencil of cubics F is very interesting indeed.
In this pencil, the Neuberg cubic is the only cubic isoinvariant.(
For example, there'is no isotomicinvariant) and the only circular
too.
The conjecture of JeanPierre seams quite difficult to proove. (I try
with centers on line OK and OI without success).
Bye.
Fred.  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> > >
Thanks. So it's an isogonal cubic, as I hoped. I meant
> > >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.
> > JeanPierre
>
> Dear All,
> I find for the locus (in trilinear)
> (k  1)^2 N + k H = 0
to say more, but some folks have just arrived to take me out to lunch!
John Conway  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:
lunch!
> > [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
>
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. JeanPierre  On Wed, 24 Jan 2001, jp ehrmann wrote:
> Dear John, Fred and other Hyacinthists
As I said, some folks arrived to take me out to lunch! I was about to
> 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. JeanPierre
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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