## Re: more ex-extra perspectors

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• Re: more ex-extra perspectors ... [snip] ... Based on some recent observations by Steve, I went back to that last result, to try to find out just when the
Message 1 of 40 , Jul 28, 2000
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Re: more ex-extra perspectors

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.
[snip]
> 1.4 The twelve points of a desmic structure lie on a cubic curve.
> However, the ex-mate of a desmic quad does not generally lie on this
> same cubic curve.

Based on some recent observations by Steve, I went back to that last
result, to try to find out just when the ex-mate of a desmic quad
actually does lie on that cubic. Apparently I didn't simplify the
result of the old calculations, because my answer is: it ALWAYS does.

The general desmic structure contains 12 points and 16 lines, and
contains three related quadrangles. However we are only looking at
the special case where one of these three quadrangles is of the form
(D,A,B,C), where A,B,C is the triangle of reference, and D is the
desmon. In this case each of the other two quadrangles is called a
desmic structure, which then determines a cubic curve that passes
through all twelve of its points. That curve will be called the
cubic determined by the desmic quadrangle.

Theorem 1.4 revised: The ex-mate of a desmic quadrangle always lies
on the cubic determined by that quadrangle.

Proof: Use the formulas I gave in message 1072 to calculate the ex-mate
of a desmic quadrangle, and plug them into the equation of the cubic.

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 writing all the above, I just saw Steve's latest messages. He wrote:
> Now the new results: the ex-extra operation, if defined for a row, is an
> operation on this cubic, taking a particular row to another row.
[snip]
> I have done this for 4 cubics; the ex-extra operation gives a new point on
> the cubic. I noticed this as I was computing group tables for most cubics. I
> will be working on a proof while I am hiking.

Steve, I can prove it. First, if four points on a cubic form one of
the three quadrangles of a desmic structure, and if all 12 points of
that structure lie on the cubic, then those first four points are all
in the same row of the group table of the cubic.

Partial proof: Use k for the constant of the cubic. Then
Io + IIo + IIIo = k
Io + IIa + IIIa = k
Ia + IIo + IIIa = k
Ia + IIa + IIIo = k
Adding and cancelling, Io + Io = Ia + Ia. This means
that Io and Ia are in the same row of the group table.

Combine this with my calculations for arbitrary desmic quads and
their ex-mates. This calculation shows that the ex-mate is desmic
and (after I redid that calc correctly) also lies on the cubic.

Also, almost every row of the group table is such a desmic quad.
Just use the D A B C row as the second quad, and find the third quad.

I'll comment on your last message (number 1149) later.
--
Barry Wolk <wolkb AT cc.umanitoba.ca>
• ... 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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