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Consider a set of nontrivial algebraic curves on the plane groovy if that set is closed under rotation, dilation, and translation, and has the property that no two members of the set intersect more than twice.

The set containing all circles and all straight lines is groovy. Furthermore, it is a maximal groovy set, since no proper superset can be groovy.

Is there a way to characterize all maximal groovy sets?

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@Anton: To be honest, I'm having trouble constructing a single other maximal groovy set! For instance, among closed curves, I'm pretty sure only the circle can generate a groovy set, since for any other closed curve you should be able to, by a rotation juxtapose the "wide" direction of the curve with the "narrow" direction, and create more than 2 intersections. –  Darren Ong Jan 26 '12 at 12:23
sorry I did not read "closed under rotation, dilation, and translation". In this case only circles and lines. –  Anton Petrunin Jan 26 '12 at 17:28

1 Answer 1

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I hope, by "nontrivial" curves you mean the curves of infinitely many points.

For the convenience, when speaking on the similarity transformation, we always assume that they preserve the orientation.

Take an irreducible algebraic curve $C$ (we assume that it contains infinitely many points) and consider a family $F$ of all its images under the similarity transformations. We claim that if $F$ is groovy then $C$ is either a circle or a line.

Consider all the triples of distinct points on $C$. There are infinitely many pairwise similar triples among them (since the triples up to similarity form a 2-dimensional projective manifold). Let $\phi$ be the similarity transformation passing one such triple to another one; then $C$ and $\phi(C)$ share three common points, hence they should coincide. Thus we have found nan infinite group of transformation preserving $C$.

Now, if among these transformations there is one with a nontrivial scale factor, then, applying it repeatedly, you obtain infinitely many collinear points on $C$, hence $C$ is a line. The same is valid for the translations.

In the remaining case, all the transformations are rotations, and you obtain infinitely many points on one circle, unless all the angles are rational multiples of $\pi$, and there are different centers of rotation. In the latter case, though, combining these rotations you can obtain a nontrivial translation, which is impossible.

Hence all the irreducible components of the curves in a groovy family are either circles or lines. It is easy to see then that the curves themselves are circles or lines. Thus your family is the unique maximal one.

On the other hand, it seems that a family generated by a (short enough) segment of a spiral $r=a^\varphi$ is groovy.

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