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Re: more ex-extra perspectors

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  • Barry Wolk
    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.
      > 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 quadrangle. Each desmic quadrangle determines an entire
      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.
      > 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>
      Winnipeg Manitoba Canada
    • Jean-Pierre.EHRMANN
      ... 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
        > 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)
        > that the three asymptotes of (C) concur at P0.
        > Hence the polar conic of P (which goes through the points of contact of
        > 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,
        > 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
        > 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
        > 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
        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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