Sorry if the terminology's wrong, I don't know differential topology. Also, this is more of a brain-teaser than a bona fide research question, but it's hopefully a "real mathematician"-level brain-teaser.


So the crossing number inequality gives a lower bound for how many intersections you have to have if you draw a graph with enough edges in the plane. The thing is, the crossing number inequality counts intersections "with multiplicity." It's pretty easy to see that if you don't count intersections with multiplicity, you can draw any finite graph in the plane with only one "crossing point."


However, while there's an easy explicit description of such a drawing, the edges aren't smooth (or even first-differentiable!) So my question is in two parts:


1. Is there a drawing of K\_n in the plane such that there is exactly one point where edges can intersect, and all edges are smooth embeddings of the (open) unit interval in R^2?
2. Is there an explicit description of such a drawing? (E.g., can you write the edges as real algebraic curves?)


(Note to moderators: you might want to tag this as "recreational" or "brain-teaser" or something of that sort; I don't have 250 reputation and so can't. :-/)



Smooth immersion(?) of graphs into the plane
If a graph can be embedded into the plane with smooth edges and one point of crossing, then it can be embedded smoothly into the projective plane without crossings, by blowing up the crossing point (and conversely). But not every graph can be embedded into the projective plane, so not every graph has a smooth embedding of the type you describe.


