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1818Re: [EMHL] Re: more ex-extra perspectors

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  • Jean-Pierre.EHRMANN
    Nov 12, 2000
      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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