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Can one find 4 distinct points on the above curve in $\mathbb R^3$, such that a line going through the first and third point intersects with the one passing through the other two?

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  • $\begingroup$ (Eventually, but not necessarily more advanced.) $\endgroup$ Commented Jun 15, 2016 at 5:18
  • $\begingroup$ Somewhat related: math.stackexchange.com/questions/1101424/… (Notice that the question contains argument why no plane can intersect this curve at four different points. Although the argument is closely related to the argument given in the answer already posted.) $\endgroup$ Commented Jun 15, 2016 at 8:39

1 Answer 1

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Suppose the $4$ points lie on an affine plane $W$ (exercise: why is this equivalent to your problem?)

Then, $W=\{(x,y,z)\,:\,ax+by+cz+d=0\}$

But this means that $at^3+bt^2+ct+d=0$ has four distinct solutions.

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    $\begingroup$ Another way to show they can't lie in an affine plane is to construct the Cayley-Menger determinant for four points. In the present case, it has the form of a Vandermonde determinant which is nonvanishing if the corresponding $t$'s are distinct; hence the four points cannot lie in the same plane. (Though it's definitely simpler to appeal to properties of cubics than of Vandermonde determinants.) $\endgroup$ Commented Jun 9, 2016 at 4:36
  • $\begingroup$ Thanks, @Semiclassical, for the cool information. $\endgroup$
    – user228113
    Commented Jun 9, 2016 at 4:58

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