Ed, If you work projectively (i.e. introduce a new variable z and work with the set of monomials in x,y,z which have degree 3), you find that there are precisely 10 cubic monomials in 3 variables. If you have 9 points, then you'll find that if the determinant of the linear system that you get from the 9 points is non-zero this gives you a unique solution (this is essentially what Stirling was probably thinking of). So, for there to be more than one curve passing through the 9 points it is necessary and sufficient that the determinant be 0. This conditions will be some large polynomial in all of the coordinates of all the points which must vanish. Once you have this you'll have a whole pencil of solutions. Victor On Fri, Nov 13, 2009 at 12:40 AM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
Stirling (in 1717) prove that 9 points uniquely define a cubic.
McLaurin (in 1720) proved that two cubics intersect in at most nine points. Bezout gets credit for it, due to his incorrect proof which came years later.
Around 1750, Euler and Cramer noticed that these seem to contradict, since different cubics are passing through the same nine points.
With that set-up, here's my question. Are there some really nice sets of 9 points that have multiple interesting cubics passing through all nine points? In a Mathematica wrapper, here's a set of four cubics that pass through the same 9 points.
ContourPlot[{ 41x - 5x^3 == 12y, -41y + 5y^3 == 12x, 365x - 41x^3 == 12y^3, -365y + 41y^3 == 12x^3 }, {x, -4, 4}, {y, -4, 4}, Frame -> False]
Cubics in the Triangle Plane http://pagesperso-orange.fr/bernard.gibert/index.html likely has a set of nine triangle centers that leads to a variety of odd cubics, but I haven't figured out a good way to isolate a good case.
--Ed _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun