We have a Maple script which computes the universal eliminant, as well as the discriminant. The universal eliminant is a polynomial of degree 6 in the variable x25 and the parameters s, t, and u. The discriminant has 388 terms and has degree 30 in the parameters s, t, and u. The problem is to show the discriminant is a sum of squares. By the first column of the table of Section 3.vi, this would then prove the conjecture of Shapiro and Shapiro in this case.
Here is another challenge. There are 6 2-planes in C6 satisfying (J2)2(J1)4, that is meeting 2 3-planes and 4 4-planes. We have a Maple script which computes eliminant, and the discriminant. The universal eliminant is a polynomial of degree 6 in the variable x25 and the parameters s, t, u, and v. This discriminant factors as the square of a cubic form and one form of degree 24 in the parameters s, t, u, and v with 1289 terms. The problem is to show the discriminant is a sum of squares. Since all 6 2-planes are real when s, t, u, v = 1, 2, 3, 4, this would prove the conjecture in this case.
Another challenge is the following. In the case of 14 2-planes which meet 8 given 4-planes in C6, we are unable to compute anything if we allow all 8 points of osculation to vary, or even when we fix two, but let the other 6 vary. We have a specialization which does allow us to compute an eliminant: Fix one point of osculation to be at infinity, another to be at zero, and the remaining 6 to be in pairs s, -s, t, -t, and u, -u. We have a Maple V.5 script which computes a universal eliminant (a polynomial of degree 14 in the variable x12) and the discriminant. The discriminant has several factors, the most interesting being its third factor, which is a square of a degree 72 polynomial in s, t, and u with 595 terms.
The universal eliminant has only even powers of the variable x12, so we obtain a simpler eliminant of degree 7 by substituting the square root of x12. This also give a simpler discriminant which has three factors, the first is inconsequential, the second is a square, but the third has degree 48 in s, t, and u with 217 terms.
Both discriminants are polynomials in s2, t2, and u2, and so we may substitute the square roots of the variables, to obtain even simpler `discriminants'. Here is the simplification of the first discriminant, and here is the simplification of the second. The challenge here is to show that these are sums of squares, and also to properly interpret this geometrically.