Dear Fred and other Hyacinthists, Fred wrote :
> I want to explain you my computations and deductions for the problem of
> "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
> P0) and the line of infinity.
> Hence the equation of the polar conic of P0 must be divisible by the
> 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,
> -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:
> 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
> the last), this gives a "potential" point P0.
> This point must be on the general cubic (C), this gives the second
> 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
> 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
> 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
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.
You can use too the Bibliotheque Nationale de France - it is long enough -
and they often change the available books
Friendly from France. Jean-Pierre