## Re: [EMHL] Re: more ex-extra perspectors

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• ... [then a mistake, corrected in:] ... ... after which I snipped some stuff that suggests both Steve and Barry are reinventing some wheels. The rows of these
Message 1 of 40 , Aug 1 2:10 PM
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On Fri, 28 Jul 2000, Barry Wolk wrote:
> On June 20, in message 1049 I said:
>
> > Theorem 1: If the quadrangle PoPaPbPc is desmic, then it has
> > an ex-mate, and this ex-mate is also desmic

[then a mistake, corrected in:]

> Theorem 1.4 revised: The ex-mate of a desmic quadrangle always lies
> on the cubic determined by that quadrangle....
>
> This is getting more and more interesting. It proves that, in the
> group tables for cubic curves that Steve has been posting, the
> ex-mate of each row will be another row of the same table. I now
> suspect that, for the special case of a desmic quadrangle, this
> ex-mate construction is a generalization of some other construction
> that is already known for cubic curves. It would be nice to remove
> the requirement that the cubic passes through A,B and C.

... after which I snipped some stuff that suggests both Steve
and Barry are reinventing some wheels.

The rows of these group tables are all things of the form

P, P+x, P+y, P+z

where x,y,z are the three elements of order two in the group.
Also, three points P,Q,R of the cubic are in line just if

P+Q+R is the particular element I like to call "line",

and it then follows that their three rows form a desmic system
(provided they're distinct). In particular, the "desmic mate" of
row P is always another row, namely that of

line - P.

I hope that sets the record straight!

I'm very intrigued indeed by Barry's result that the ex-mate of
a row is always another row. What is the relation between the
corresponding group elements? The fact that the ex-mate relation is
symmetric suggests that it has the form

P + Q might be a certain constant, which I'll call "wolk".

If this is the case, then P + Q + (line - wolk) = line, so
that the line PQ passes through a fixed point

R = line - wolk

on the cubic. Is this in fact the case, and if so, what is this
fixed point R? Is it isogonally symmetric to D, the fourth
point of the row A B C D ?

John Conway

PS : could you remind me of the definition of ex-mate - I'm here
without any notes. JHC
• ... coordinates ... such ... the ... tangent at ... equation of ... y0, ... first and ... equation for ... Exactly. The Bernard cubic Kp is the locus of P such
Message 40 of 40 , Nov 12, 2000
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Dear Fred and other Hyacinthists, Fred wrote :

> I want to explain you my computations and deductions for the problem of
> Jean-Pierre:
>
> "Is the Darboux cubic the only isogonal cubic with
> pivot for which the asymptots are real and concur on the cubic ?"
>
> Step 1)
> The general isogonic cubic (C) with pivot (f,g,h) has in trilinear
coordinates
> the following equation:
>
> (C) h*(x^2 - y^2)*z + f*x*(y^2 - z^2) + g*y*(z^2 - x^2) = 0
>
> Step 2) We are looking for the existence on (C) of a point P0(x0,y0,z0)
such
> that the three asymptotes of (C) concur at P0.
> Hence the polar conic of P (which goes through the points of contact of
the
> tangents from P0) must degenerate in a line (necessary an inflexional
tangent at
> P0) and the line of infinity.
>
> Hence the equation of the polar conic of P0 must be divisible by the
equation of
> the line at infinity. (I do the division on the variable x).
> It gives an homogen system of 3 equations with three unknowns, namely x0,
y0,
> z0.
>
> (S):
>
> -f*x0*a^2 + g*y0*a^2 - 2*c*h*x0*a + 2*c*f*z0*a - c^2*g*y0 + c^2*h*z0 = 0
>
> -2*h*y0*a^2 + 2*g*z0*a^2 + 2*c*g*x0*a - 2*b*h*x0*a -
> 2*c*f*y0*a + 2*b*f*z0*a - 2*b*c*g*y0 + 2*b*c*h*z0 = 0
>
> f*x0*a^2 - h*z0*a^2 + 2*b*g*x0*a - 2*b*f*y0*a - b^2*g*y0 + b^2*h*z0 = 0
>
> Step 3)
> For a nontrivial solution, the determinant must be 0.
> This gives the first condition on f, g h.
> Here is this condition:
>
> (I)
> f*g^2*a^3 - f*h^2*a^3 + 2*b*g^3*a^2 - 2*c*h^3*a^2 -
> b*g*h^2*a^2 - b*f^2*g*a^2 + c*f^2*h*a^2 +
> c*g^2*h*a^2 - 2*b^2*f^3*a + 2*c^2*f^3*a +
> b^2*f*g^2*a - c^2*f*g^2*a + b^2*f*h^2*a -
> c^2*f*h^2*a - 2*b*c^2*g^3 + 2*b^2*c*h^3 + b^3*g*h^2 +
> b*c^2*g*h^2 - b^3*f^2*g + b*c^2*f^2*g + c^3*f^2*h -
> b^2*c*f^2*h - c^3*g^2*h - b^2*c*g^2*h = 0
>...

> Step 4) Now we have to solve two of the equations of (S) (I choose the
first and
> the last), this gives a "potential" point P0.
> This point must be on the general cubic (C), this gives the second
equation for
> (f,g,h),
> the septic (II):
>
> (II): .... = 0
>
> f*h^6*a^9 - f^3*h^4*a^9 - f*g^2*h^4*a^9 +... = 0
>

> Step 5)
> (f,g,h) must be on the intersection of (I) anf (II), 21 potential points.
>
> The tangent at I to the cubic (I) is the line KI.
> Is this cubic the cubic from Bernard?

Exactly. The Bernard cubic Kp is the locus of P such as the asymptots of
C(P) concur - C(P) = self-isogonal cubic with pivot P -
> It's easy to prove that I Ia Ib Ic and L are common points of the curves
(I) and
> (II).
> And that I Ia Ib Ic are double points of the curve (II) and that L is a
simple
> point of L.
> The curves (I) and (II) are not tangent at I Ia Ib Ic L.
>
> This gives 2 x 4 + 1 = 9 common points.
> What are the other 21 - 9 = 12 points?
>
> With a numerical case, the (quite complicated) graphics tell me that there
are
> another six common points, three of them are triple points of (II).
> What are these six points?
> I don't know.
> Are my deductions correct?
> What is the excentral triangle? Ia Ib Ic perhaps.
> I'll try to calculate the cusps of the Steiner hypocycloid of this
triangle to
> see if they are other common points

I'm sure that everything is quite correct.
Because of a particular choice of a variable in Step 2, the septic (II) is
not invariant under circular permutation - for instance, we see a^3 in the
equation (II) but not b^3 and c^3
That's the reason why the three
triple points cannot be solutions of the problem : with another choice of a
variable, we get another septic with different triple points - in fact, I
think that the circle going through the triple points of each of those
septics goes through a vertice of ABC and only one -
Thus, it remains three simple points; those points are necesseraly the cusps
of the Steiner hypocycloid of the excentral triangle - Ia, Ib, Ic - for the
following reason :
Take any point M on the line at infinity; the homothecy (I, 2) maps the
isogonal conjugate of M to a point M' on the excentral circumcircle. An easy
computation shows that the line at infinity touches C(P) at M iff P is the
characteristic point of the Simson line of M' w.r.t. the excentral triangle.
More over, C(P) has a triple contact at M with the line at infinity iff the
Simson line of M' goes through O and that means that P is a cusp of the
hypocycloid.
Of course those cusps are not constructible and we could get their
coordinates with the angles A/3, B/3, C/3 but it is useless for our problem.
Finally, we have 8 solutions :
- L gives the Darboux cubic
- I, Ia, Ib, Ic give three degenerated cubics
- the three cusps give three "tridents"; clearly those cubics cannot have
three asymptots
So, we can conclude that P = L is the only solution of the following problem
:
Find P such as C(P) is non degenerated and has three asymptots that concur
at a point of C(P).

My approach of the problem was slightly different : I used the fact that if
a line intersects a cubic c at p, q, r and if the tangents to c at p,q
intersect at a point m lying on c, then the tangent to c at r goes through m
iff m is a flex of c.
As Fred did, I came to 11 solutions, three of them being wrong.
I think that it will be interesting to extend this problem to self-isogonal
cubics without pivot.

Something else :
Steve wrote :
'' Knowing that many of you like old math books, here is a source of high
quality reporductions by the mathematics department at Cornell University,
both for purchase and online.

http://www.math.cornell.edu/~library/reformat.html#Index ''

You can use too the Bibliotheque Nationale de France - it is long enough -
and they often change the available books

http://gallica.bnf.fr/ + Catalogues

Friendly from France. Jean-Pierre
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