TITLE: Real intersections of plane cubic curves 
QUESTION [5 upvotes]: Let $C$ and $D$ be plane cubic curves over the real numbers. Suppose that the real loci in the projective plane of $C$ and $D$ consist of two connected components. Denote $C_0$ and $D_0$ the bounded component (which is often referred as "oval") of $C$ and $D$, and denote $C_1$ and $D_1$ the unbounded component of $C$ and $D$ respectively. 
I have following question: 
What is the maximal number of real intersections of $C_1$ and $D_1$ in the projective plane?
Of course a trivial bound is 9 given by Bezout's theorem, but I conjecture that the bound is  rather 5. I think this should be well known, but I couldn't find any reference.

REPLY [3 votes]: Another approach, building on Noam Elkies's answer:
$$(y - 5) (x - 2 y + 10.5) (x + 2 y - 10.5) = -0.001$$
$$(x - 5) (y - 2 x + 10.5) (y + 2 x - 10.5) = -0.001$$
The images below show the same graph at different resolutions. Unfortunately, getting far enough out to see all 9 points makes it hard to see the ovals cleanly. But the basic idea is to choose two triples of lines which form triangles (not which pass through a common point as in Noam's answer) and deform a little to make the topology work.
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