# Finding a curve that intersects any line on the plane

Question Is there a curve on plane such that any line on the plane meets it (a non zero ) finite times ?

What are the bounds on the number of such intersections.

My question was itself inspired by this "Can you draw circles on the plane so that every line intersects at least one of them but no more than 100 of them?"

• Clearly if your curve is the graph of an odd polynomial, the number of intersections is at most the degree of the polynomial. – Cameron Williams Sep 18 '13 at 15:08
• One possibility is a pair of parabolas, say one concave up and one concave down, situated so that they touch tangentially (or intersect) and therefore cannot be separated by a hyperplane. – hardmath Sep 18 '13 at 15:08
• Please try to make titles of questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Lord_Farin Sep 18 '13 at 15:08
• @CameronWilliams Correct, but just being an odd degree polynomial; does not guarantee that every line in the plane will intersect it – ARi Sep 18 '13 at 15:10
• @ARi: Actually it does imply that, except degree 1. – hardmath Sep 18 '13 at 15:13

Cubic parabola $$y=x^3$$ has this property. The max number of such intersections is given by the Fundamental theorem of algebra:

$$x^3=ax+b$$ can have at most 3 solutions.

• Yes but will all of the solutions be real for all real (a,b) pairs – ARi Sep 18 '13 at 15:14
• For any $a$, there will be only one real solution for all sufficiently large $|b|$. – hardmath Sep 18 '13 at 15:17
• I think the important fact here (i.e., relevant to the question), is that $\,x^3-ax-b=0\;$ has always at least one real solution and, of course, at most three different ones. Nice answer. +1 – DonAntonio Sep 18 '13 at 15:22
• So the upper bound for intersections will be three and the lower one 0. – ARi Sep 18 '13 at 15:29
• Well, a tighter and always attainable lower bound is one, in fact...and this fulfills the question's conditions. – DonAntonio Sep 18 '13 at 15:31

As already shown, any cubic polynomial (and indeed, any odd-degree polynomial) has the requisite property by the fundamental theorem of algebra.

What's more, a simple perturbation argument should be enough to show that any (sufficiently) smooth curve that meets every line in at least one point will meet some lines in at least three points. Consider a point tangent to the curve where the second derivative 'with respect to the tangent line' is non-zero; that is, a non-reflex tangent point, or locally extremal point. (Such points must exist if the curve is non-trivial). Now, consider pencils of lines 'near' this intersection point; displaced infinitesimally one way from the tangent, they must have another point of intersection with the curve, and this point can be made 'generic' so that it doesn't vanish under small perturbations. Then displace infinitesimally the other direction; the 'generic' point of intersection is still a point of intersection, but the tangent turns into two points of intersection.