## 17-Point Cubic

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• The 17-Point cubic (aka Thomson cubic) passes through the (3) vertices of the reference triangle, the (3) vertices of the medial triangle, the (3) midpoints of
Message 1 of 16 , Sep 23, 2000
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The 17-Point cubic (aka Thomson cubic) passes through the (3) vertices of
the reference triangle, the (3) vertices of the medial triangle, the (3)
midpoints of the altitudes, the (4) in/excenters, and the two dyads (2x2)
of isogonals (H,O), (G,K), in total: 3+3+3+4+4 = 17 points.

Kimberling, with his program CUBIC, found four more:
X_9, and its isog. X_57; X_223, and its isog. X_282 (TCCT, p. 240).

Are there other known notable points lying on this curve?

Antreas
• ... Antreas, this is one of the canonical cubics that we have discussed here, the isogonal cubic with pivot G. Additional point that I know of are The four
Message 2 of 16 , Sep 23, 2000
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on 9/23/00 11:24 AM, xpolakis@... wrote:

> The 17-Point cubic (aka Thomson cubic) passes through the (3) vertices of
> the reference triangle, the (3) vertices of the medial triangle, the (3)
> midpoints of the altitudes, the (4) in/excenters, and the two dyads (2x2)
> of isogonals (H,O), (G,K), in total: 3+3+3+4+4 = 17 points.
>
> Kimberling, with his program CUBIC, found four more:
> X_9, and its isog. X_57; X_223, and its isog. X_282 (TCCT, p. 240).
>
> Are there other known notable points lying on this curve?

Antreas, this is one of the canonical cubics that we have discussed here,
the isogonal cubic with pivot G.

Additional point that I know of are

The four Mittenpunkts and their isogonic conjugates and the ex-extras of the
isogonic conjugates of the Mittenpunkts (and their conjugates).

That's 16 more.

Where does the name Thompson cubic come from?

Steve
• ... I checked Kimberling, and his points are the central parts of the quadrangles I quoted above. Remember that points on a cubic come in fours, so in this day
Message 3 of 16 , Sep 23, 2000
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on 9/23/00 6:00 PM, Steve Sigur wrote:

> on 9/23/00 11:24 AM, xpolakis@... wrote:
>
>> The 17-Point cubic (aka Thomson cubic) passes through the (3) vertices of
>> the reference triangle, the (3) vertices of the medial triangle, the (3)
>> midpoints of the altitudes, the (4) in/excenters, and the two dyads (2x2)
>> of isogonals (H,O), (G,K), in total: 3+3+3+4+4 = 17 points.
>>
>> Kimberling, with his program CUBIC, found four more:
>> X_9, and its isog. X_57; X_223, and its isog. X_282 (TCCT, p. 240).
>>
>> Are there other known notable points lying on this curve?
>
> Antreas, this is one of the canonical cubics that we have discussed here,
> the isogonal cubic with pivot G.
>
> Additional point that I know of are
>
> The four Mittenpunkts and their isogonic conjugates and the ex-extras of the
> isogonic conjugates of the Mittenpunkts (and their conjugates).
>
> That's 16 more.

I checked Kimberling, and his points are the central parts of the

Remember that points on a cubic come in fours, so in this day and time it is
not possible to know "17 points" on a cubic.

I reproduce the group table below. It is a much simpler group table than the
Neuberg cubic. The work of Barry and me over the summer guarantees that we
can _construct_ an infinite number more. About two weeks ago I summarized
our work with a posting to Hyacinthos about constructing points on cubics.

Many of the cubics we have discussed extensively can be thought of as having
two operations that generate points. For this cubic, it is the isogonic
conjugate and the ex-extra operation (aka, Cevian quotient with respect to
G). Using these we can generate as many points (with their corresponding
colinearities) as we wish.

For the Darboux cubic the two operations are the isogonic conjugate and
reflection in the circumcenter.

Another relevant comment is that the odd/even-ness inherent in the group
table guarantees that, if the desmon is a strong point, an infinite number
of weak, quartile points (such as the Mittenpunkts) will be on the cubic,
occupying every other row of the group table. The other rows will be strong
points.

Here is the group table, listing 36 points. By the way we would love to know
a geometric interpretation of the third intersection of AO, BO, and CO with
the cubic.

Thompson centroidal isogonic cubic
precevian of R = pedal of P
isogonal

o a b c
-------------
-V
-IV
-III gM'M'o :gM'M'b:
-II H ***
-I M'o M'a M'b M'c
0 K A_G B_G C_G
I Io Ia Ib Ic
II G A B C
III Mo Ma Mb Mc
IV O H+A H+B H+C
V M'M'o :M'M'b:

pivot G
center K
constant O = row IV

Notation: Mo = Mittenpunkt = X(9), M'o = mate of Mo = isogonal conjugate of
Mo = X(37), A_G = a-trace of G , H+A = midpt of HA, M'M'o= ex-extra pf M'o
= M'o/G = X(223).

*** - the points in this row are the third meets of lines AO, BO, CO with
the cubic.

The I row are the incenters, which are self conjugate. Corresponding rows
above and below row I are isogonally conjugate.
• Dear Steve and friends, [Steve] ... they are the isogonals of the midpoints of the altitudes. I just want to add a well-known fact that might give other points
Message 4 of 16 , Sep 24, 2000
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Dear Steve and friends,

[Steve]
>By the way we would love to know
> a geometric interpretation of the third intersection of AO, BO, and CO with
> the cubic.

they are the isogonals of the midpoints of the altitudes.

I just want to add a well-known fact that might give other points on this
cubic :
the homothecy h, center G, ratio -1/2, maps the Lucas cubic (L) to the
Thomson cubic (T) therefore (T) is isotomic wrt the medial triangle.

For example, X253 = isotomic of X20 (de Longchamps) being on (L), the point
h(X253) is on (T) but maybe it's one of the points already mentionned.

BTW, dear Steve, it's called Thomson and not Thompson but I don't know who
Thomson was... [Antreas, any idea ?]

The name 17-point cubic comes from the very extraordinary fact that this
cubic is (probably) the only one meeting 22 usual lines in only 17 points :
3 sidelines, 3 altitudes, 3 medians, 3 symmedians, 3 perp. bisectors, 6
bisectors and the Euler line.

Denote that the tangents in A,B,C are the symmedians, the tangents in the
midpoints of the sides are the perp. bisectors, the tangents in Ia,Ib,Ic are
the internal bisectors.
• ... Dear Bernard, There were some geometers named Thomson but I don t know which one is of the Thomson cubic. Most likely one the following: James Thomson
Message 5 of 16 , Sep 24, 2000
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Bernard Gibert wrote:

>BTW, dear Steve, it's called Thomson and not Thompson but I don't know who
>Thomson was... [Antreas, any idea ?]

Dear Bernard,

There were some geometers named Thomson but I don't know which one
is of the Thomson cubic.

Most likely one the following:

James Thomson (1786-1849), the author of:
Thomson, James: Elements of plane and spherical trigonometry, with the first
principles of analytic geometry . 2nd ed.
Belfast : Simms & McIntyre, 1830. 108 p.

Thomson, James: First six and the eleventh and twelfth books of Euclid's
elements : with notes and illustrations and an appendix in five books.
Edinburgh : A. & C. Black, 1834. v+386 p.

Herman Ivah Thomson author of:
Thomson, Herman Ivah: The Osculants of Plane Rational Quartic Curves.
Reprint from _American Journal of Mathematics 32:3 (1910) 207-234_
Baltimore, Johns Hopkins Univ., Diss., 1909.

I searched Zbl for "Thomson cubic" and here are two related articles
(which probably contain references to Thomson, but haven't them handy):

___________________________________________

Cundy, Henry Martyn - Parry, Cyril Frederick: Some cubic curves associated
with a triangle.
J. Geom. 53, No.1-2, 41-66 (1995).

Keywords: trilinear coordinates; isogonal conjugates; auto-isogonal cubics;
cubic curves; polar conic; circular cubics; Euler pencil; Darboux cubic;
McCay cubic; Thomson cubic; Neuberg cubic; Neuberg group; Moebius
involution

-------------------------------------------

Pinkernell, Guido M.: Cubic curves in the triangle plane.
J. Geom. 55, No.1-2, 141-161 (1996).

The term ``triangle plane'' in the title means a plane in which a triangle
ABC is singled out. The author studies two pencils of cubic curves that are
the result of certain constructions in the triangle plane. The two pencils
contain nearly all the important special cubics, for example the Darboux
cubic, the 17-point cubic (also known as Thomson's cubic), the Neuberg
cubic, and the Lucas cubic.
[ E.J.F.Primrose (Leicester) ]

___________________________________________

BTW, which is McCay cubic ?

PS: I will post a query on Thomson to Historia Matematica list.

Antreas
• ... Here is Julio s informative response: FWD MESSAGE ---------------------------------------------------------- Date: Thu, 28 Sep 2000 22:05:18 -0300 To:
Message 6 of 16 , Sep 29, 2000
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[Bernard Gibert]:

>>BTW, dear Steve, it's called Thomson and not Thompson but I don't know who
>>Thomson was... [Antreas, any idea ?]

[APH]:

>PS: I will post a query on Thomson to Historia Matematica list.

Here is Julio's informative response:

FWD MESSAGE ----------------------------------------------------------

Date: Thu, 28 Sep 2000 22:05:18 -0300
To: "Antreas P. Hatzipolakis" <xpolakis@...>
From: Julio Gonzalez Cabillon <jgc@...>
Subject: Re: [HM] Thomson
Cc: historia-matematica@..., guido.pinkernell@...

On 24 Sep 2000, Antreas P. Hatzipolakis asked:

"Who was Thomson of the well-known triangle cubic
named after him (Thomson cubic aka 17point cubic)?"

-------

Dear Antreas,

I am sorry to say that, as far as I can tell, virtually nothing
has been recorded of this Thomson, other than the issue that he
posed his geometrical locus as a problem in the enjoyable British
_Educational Times_ journal (August 1864).

F. D. Thomson ( as he used to sign most of his short notices [*])
was a habitue of that journal, having submitted many solutions of
questions posed there.

[*] Thomson, F. D.
"Notes on the Geometry of a Cubic Curve", _Messenger_ volume V
(1869) pp 27-30.
"The Equation to the Axes of a Conic",_Messenger_ volume V (1869)
pp 65-66.

In 1865, Thomson's problem appeared also in the well-known French
journal _Nouvelles annales de mathematiques_:

_Question 735_

"Le lieu des centres des coniques tangentes aux cotes d'un
triangle, et telles que les normales menees par les points
de contact, se rencontrent en un meme point est une courbe
du troisieme degre qui passe par les sommets du triangle, le
point de rencontre des hauteurs, le centre de gravite, les
centres incrit et ex-inscrits, les milieux des cotes, les
milieux des hauteurs." (2a. serie, t. IV, p 144)

Interestingly enough, this "Question 735" was solved by a junior
named Arthur Poussart, a student of 'Lycee de Douai' (class of Mr
Painvin), and his demonstration appeared on pages 469-473 of the
same journal (2a. serie, t. IV).

Joseph Neuberg (1840--1926) is responsible for the popularization
of term THOMSON CUBIC [ "La cubique de THOMSON" ], in his article
"Bibliographie du triangle et du tetraedre" (Mathesis 1923).

There are at least two interesting and brand new articles on the
subject of cubic curves on the plane in which the given triangle
is considered. For instance, Guido Pinkernell studies two pencils
of cubic curves which contain basically the main special cubics;
namely, Darboux's, Lucas's, Neuberg's & Thomson's.

Cf. Pinkernell, Guido M:
"Cubic Curves in the Triangle Plane",_Journal of Geometry_ vol 55
(1996), no 1-2, pp 141-161.

Cundy, Henry Martyn; Parry, Cyril Frederick:
"Some Cubic Curves Assoc. with a Triangle", _Journal of Geometry_
vol 53 (1995), no 1-2, pp 41-66.

Best wishes from Montevideo, Julio

PS Antreas... I apologise for having forgotten an important event
in mid August. Better late than never... they say.

END ------------------------------------------------------------------

Very many Thanks, dear Julio!

Greetings from Athens

Antreas
• Dear Antreas, Julio and other Hyacinthists, ... Very interesting, indeed. So if we look at the locus of the centers of the conics tangent to the sides and such
Message 7 of 16 , Sep 29, 2000
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Dear Antreas, Julio and other Hyacinthists,
Antreas quoted :

> "Le lieu des centres des coniques tangentes aux cotes d'un
> triangle, et telles que les normales menees par les points
> de contact, se rencontrent en un meme point est une courbe
> du troisieme degre qui passe par les sommets du triangle, le
> point de rencontre des hauteurs, le centre de gravite, les
> centres incrit et ex-inscrits, les milieux des cotes, les
> milieux des hauteurs." (2a. serie, t. IV, p 144)
>

Very interesting, indeed.
So if we look at the locus of the centers of the conics tangent to
the sides and such as the three lines going through the contact
points and making a given angle Phi with the corresponding side
concur, we get a cubic C(Phi) member of the pencil generated by the
Thomson cubic and the union of the three sides : all those cubics go
through the midpoints of the sides and are tangent at A, B, C to the
corresponding symedian.
The isogonal conjugate of C(phi) is C(-phi) and, in the particular
case of the Thomson cubic, the common point of the three normals
should lie on the Darboux cubic.
A lot of work for Bernard !!

Friendly from France. Jean-Pierre.
• ... May I ask where this quotation s from? I gather that lieu in such contexts means locus - is the Latin word not used in French? ... I agree that this
Message 8 of 16 , Sep 29, 2000
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> > "Le lieu des centres des coniques tangentes aux cotes d'un
> > triangle, et telles que les normales menees par les points
> > de contact, se rencontrent en un meme point est une courbe
> > du troisieme degre qui passe par les sommets du triangle, le
> > point de rencontre des hauteurs, le centre de gravite, les
> > centres incrit et ex-inscrits, les milieux des cotes, les
> > milieux des hauteurs." (2a. serie, t. IV, p 144)

May I ask where this quotation's from? I gather that "lieu"
in such contexts means "locus" - is the Latin word not used in
French?

> So if we look at the locus of the centers of the conics tangent to
> the sides and such as the three lines going through the contact
> points and making a given angle Phi with the corresponding side
> concur, we get a cubic C(Phi) member of the pencil generated by the
> Thomson cubic and the union of the three sides : all those cubics go
> through the midpoints of the sides and are tangent at A, B, C to the
> corresponding symedian.

I agree that this is very interesting. It's a pity that for
general Phi these cubics involve a term XYZ and so aren't
either isotomic or isogonal ones (unless I misunderstand something?).

However, the fact that they go through the midpoints mA,mB,mC
of the sides means that their "dilated" forms have a chance of being
isogonal or isotomic. Are they? I'd check myself, but of course
can't remember which one the Thomson cubic is - can somebody please
remind me (preferably by giving its equation)?

Regards, John Conway
• ... I m not terribly surprised, since of course all these words mean the same thing, but this does make me wonder why English (alone?) continues to use the
Message 9 of 16 , Sep 29, 2000
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On Fri, 29 Sep 2000, Antreas P. Hatzipolakis wrote:

> Yes, in the French math. literature "lieu" = locus.
>
> In the Greek, by the way, topos = locus.

I'm not terribly surprised, since of course all these words
mean the same thing, but this does make me wonder why English
(alone?) continues to use the Latin word for this specialised
sense.

> > However, the fact that they go through the midpoints mA,mB,mC
> >of the sides means that their "dilated" forms have a chance of being
> >isogonal or isotomic. Are they? I'd check myself, but of course
> >can't remember which one the Thomson cubic is - can somebody please
> >remind me (preferably by giving its equation)?
>
> bcx(y^2 - z^2> +....(cyclically) = 0

Thanks. This looks like the form in (orthogonal) trilinears,
so I'll rescale it to abbccx(yy-zz) + &c = 0, and replace
ax,by,cz by X,Y,Z to get X(ccYY-bbZZ) + &c = 0, or equivalently

| X Y Z |
|aaYZ bbZX ccXY| = 0,
| 1 1 1 |

showing that this is the isogonal cubic with pivot/perspector G.

Oh - I've just realised that I didn't need to ask, since the
fact that it passes through the midpoints of the sides forces G
to be the pivot. But thanks anyway!

OK So the general cubic of the pencil is

X(ccYY-bbZZ) + ... + kXYZ = 0

which I dilate by putting Y+Z,Z+X,X+Y for X,Y,Z:

(Y+Z)[cc(XX+2ZX+ZZ) - bb(XX+2XY+YY)] + ... + k(Y+Z)(Z+X)(X+Y) = 0.

I sort terms, using term' and term" for the 2nd and 3rd copies
of the first bracket here.

Coeff(ZZZ) = cc - (bb)' = 0.

Coeff(XYZ) = (2cc-2bb)+ that' + that" + 2k = 2k

Aha - so this kills my hope that perhaps all these (dilated) cubics
would be isotomic/isogonal. But let's continue with the Thomson one
(k=0), looking for terms in things like XXY

(cc-bb)XXY 2ccZZX -bbYYZ ccYZZ (cc-bb)ZXX -2bbXYY
': (aa-cc)YYZ 2aaXXY -ccZZX aaZXX (aa-cc)XYY -2ccYZZ
": (bb-aa)ZZX 2bbYYZ -aaXXY bbXYY (bb-aa)YZZ -2aaZXX
--------------------------------------------------------------
(cc-bb+aa)XXY + &c (-bb+aa-cc)XYY &c

giving SA.ZZX + SB.XXY + SC.YYZ = SA.XYY + SB.YZZ + SC.ZXX

or SA.X(YY-ZZ) + &c = 0, or

| SA SB SC |
| YZ ZX XY | = 0
| X Y Z |
for the dilated form, ie., the isotomic cubic with perspector
dK = (SA:SB:SC).

I recognize this as what I used to call "the antipodal cubic",
since it's equal to its reflection in O. So (good) the Thomson
cubic is one of the three great rationally equivalent cubics I
wrote about some time ago. (I doubt if any two of the other
well-known cubics are rationally equivalent - does anyone know
of any such pairs?)

John Conway
• Dear John, ... A cubic with a term in xyz can be isogonal (or isotomic) but, in this case, there s no pivot. These cubics are very interesting but difficult to
Message 10 of 16 , Sep 29, 2000
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Dear John,

> I agree that this is very interesting. It's a pity that for
> general Phi these cubics involve a term XYZ and so aren't
> either isotomic or isogonal ones (unless I misunderstand something?).
>
A cubic with a term in xyz can be isogonal (or isotomic) but, in this case,
there's no pivot.

These cubics are very interesting but difficult to deal with.
I know a (small ) number of them and I still have a lot of work to do to
understand them a little bit better.

Regards

Bernard
• Dear Jean-Pierre and friends, ... It does, JP, and the perspector is on the isogonal cubic with pivot X69 = isotomic of H. What about the locus of the foci ?
Message 11 of 16 , Sep 29, 2000
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Dear Jean-Pierre and friends,

> The isogonal conjugate of C(phi) is C(-phi) and, in the particular
> case of the Thomson cubic, the common point of the three normals
> should lie on the Darboux cubic.

It does, JP, and the perspector is on the isogonal cubic with pivot X69 =
isotomic of H.
What about the locus of the foci ? It seems it's a much higher degree curve
(12, I think) which decomposes into the Lucas cubic and ???...)

Regards

Bernard
• ... [...] ... This is a matter of how we use language. The term isogonal cubic is usually used to mean a cubic that s not only self-isogonal but also passes
Message 12 of 16 , Sep 29, 2000
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On Fri, 29 Sep 2000, Bernard Gibert wrote:

> Dear John,

[...]

> A cubic with a term in xyz can be isogonal (or isotomic) but, in this case,
> there's no pivot.

This is a matter of how we use language. The term "isogonal cubic"
is usually used to mean a cubic that's not only self-isogonal but also
passes through the incenters, or equivalently "has a pivot" P (meaning
that it's the locus of Q for which Q,gQ.P are collinear). Similarly
"isotomic cubic" means one that's not only self-isotomic but also
passes through G and its associates, or "has a pivot" in the sense
that Q,tQ,P are collinear.

I agree this convention isn't universal, but we've discussed it
before, and the convention is a good one because there's no point
in making "isogonal" and "self-isogonal" mean exactly the same thing!

> These cubics are very interesting but difficult to deal with.
> I know a (small ) number of them and I still have a lot of work to do to
> understand them a little bit better.

My feeling as far as the Triangle Book is concerned is that the
isotomic and isogonal cubics in the above sense will be as far as
we go, because their (P,Q,tQ) and (P,Q,gQ) definitions are quite
geometrical, whereas other high-degree curves really involve more
algebra than geometry.

Despite that, I've done some investigations of particular ones,
most notably the locus of P whose pedal and prepedal triangles
are at a given angle theta. This generalized Thomson cubic sounds
very similar.

Regards,
John Conway
• ... Bernard. By my way of thinking the cubics are not bad at all, rather it is the conics that are confusing. The reason I think they are not confusing is that
Message 13 of 16 , Sep 29, 2000
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on 9/29/00 3:36 PM, Bernard Gibert wrote:

> These cubics are very interesting but difficult to deal with.
> I know a (small ) number of them and I still have a lot of work to do to
> understand them a little bit better.

Bernard.

By my way of thinking the cubics are not bad at all, rather it is the conics
that are confusing. The reason I think they are not confusing is that to me
they are more fundamental. I take points and lines to be the most
fundamental objects and the cubics organize them both.

Geometrically they are more simple; by themselves, they only use projective
properties of concurrence and colinearity. Mathematically this is expressed
by a group law between these points. Unfamiliar and abstract, yes, but not
difficult. We have picked up a trick or two in the last century and perhaps
the best way to understand the ancient mathematical creation of the Greeks
is to use the less ancient mathematical creations of the French and Germans.

What we get from this is that points are arranged in quadrangles (whence the
triangle has 4 "sides," the in-side, the a-side, the b-side, and the
c-side--this comment is supposed to be as much a pun as geometry). These
weak and strong points. If the cubic has a pivot that is strong, then there
is a powerful organization and inter-relation between the strong and weak,
quartile points.

So cubics, by themselves, are actually easy and profound to understand.

But of course their interactions with other geometric structures may be
harder to understand, which is what you were referring to.

Friendly from Princeton,

STeve
• Dear Jean-Pierre and friends, ... please correct a typo : it s not isogonal but isotomic cubic (Lucas cubic = locus of points P such that a cevian triangle of
Message 14 of 16 , Sep 29, 2000
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Dear Jean-Pierre and friends,
>
>> The isogonal conjugate of C(phi) is C(-phi) and, in the particular
>> case of the Thomson cubic, the common point of the three normals
>> should lie on the Darboux cubic.

I wrote :

> It does, JP, and the perspector is on the isogonal cubic with pivot X69 =
> isotomic of H.

please correct a typo : it's not isogonal but isotomic cubic (Lucas cubic =
locus of points P such that a cevian triangle of P is a pedal triangle of Q.
That's why Q is on Darboux)

It's nice to observe [I didn't prove it yet but it shouldn't be very
difficult] that the line PQ goes through L=X20 and the line through Q and
the center of the inconic [which is on Thomson] goes through H.

Another little thing : the in-Steiner ellipse has its foci isogonal and on a
line through G therefore those foci are on Thomson and on the locus below.

> What about the locus of the foci ? It seems it's a much higher degree curve
> (12, I think) which decomposes into the Lucas cubic and ???...)

I don't think now this is correct... I have to redo this part if I can...
The vertices A,B,C are at least triple, the in/excenters at least double,
and the feet of the altitudes are on it : that makes at least 7 points on
each side !!??

Any idea, JP ?

Sorry for so much carelessness.

Regards

Bernard
• Steve Sigur wrote: the cubics ... ^^^^^^ ^^^^^^^ ...rather of francophones (= French, Belgians etc) and germanophones (Germans, Austrians, etc) And of the
Message 15 of 16 , Sep 30, 2000
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Steve Sigur wrote:

the cubics
>Geometrically they are more simple; by themselves, they only use projective
>properties of concurrence and colinearity. Mathematically this is expressed
>by a group law between these points. Unfamiliar and abstract, yes, but not
>difficult. We have picked up a trick or two in the last century and perhaps
>the best way to understand the ancient mathematical creation of the Greeks
>is to use the less ancient mathematical creations of the French and Germans.
^^^^^^ ^^^^^^^
...rather of francophones (= French, Belgians etc) and germanophones (Germans,
Austrians, etc)

And of the British/Irish as well!
(It was Newton who first classified the cubics)
There was also a great Dutch geometry tradition (which is still alive!)

Antreas
• ... and the Greekophone tradition is still alive, if all those books you quote are an indication! In the above paragraph I was thinking of the development of
Message 16 of 16 , Sep 30, 2000
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on 9/30/00 7:30 PM, xpolakis@... wrote:

> Steve Sigur wrote:
>
> the cubics
>> Geometrically they are more simple; by themselves, they only use projective
>> properties of concurrence and colinearity. Mathematically this is expressed
>> by a group law between these points. Unfamiliar and abstract, yes, but not
>> difficult. We have picked up a trick or two in the last century and perhaps
>> the best way to understand the ancient mathematical creation of the Greeks
>> is to use the less ancient mathematical creations of the French and Germans.
> ^^^^^^ ^^^^^^^
> ...rather of francophones (= French, Belgians etc) and germanophones (Germans,
> Austrians, etc)
>
> And of the British/Irish as well!
> (It was Newton who first classified the cubics)
> There was also a great Dutch geometry tradition (which is still alive!)
>

and the Greekophone tradition is still alive, if all those books you quote
are an indication!

In the above paragraph I was thinking of the development of abstract algebra
and group theory, rather than of cubics.

While we are listing strong geometric traditions, there is an influx of
Eastern European mathematicians into the US, and their geometry is very
strong. While geometry fades into obscurity here in the US, their tradition
is still taught to their young and is vibrant.

Steve
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