Real intersections of plane cubic curves  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. 
 A: [Revised because I read too quickly and thought the problem was to find
two connected cubics meeting in nine points]
Here's a version of the construction with two sets of three lines
that may be easier to parse visually:
   (image source)
Each triplet of lines forms an equilateral triangle;
these are plots of $C=40c$ and $C=240c$ where
$$
C = (y-3) (y+6-3^{1/2}x) (y+6+3^{1/2}x),
$$
$$
c = (x+1) (x-2-3^{1/2}y) (x-2+3^{1/2}y).
$$
A: 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.
       (source)
       (source)
       (source)
A: The answer is $9$. Choose a curve $C$ with the required topology. Choose $8$ points $x_1$, $x_2$, ..., $x_8$ on $C_1$. The conditions of passing through the $x_i$ impose $8$ linear conditions on the $10$ dimensional space of cubics, so we can find a second cubic $E$ passing through the $x_i$ and not proportional to $C$. Let $x_9$ be the ninth intersection of $C$ and $E$.
I claim that $x_9$ is also on $C_1$. Proof: Consider $C_1$ as a loop in $\mathbb{RP}^2$, and let $\tilde{C}_1$ be the preimage in the universal cover $S^2$. The loop $C_1$ is not contractible so, if we travel all the way around $C_1$, we will go from $(x,y,z)$ to $(-x,-y,-z)$ in $S^2$. Since $E$ is an odd degree polynomial, $E(x,y,z) = - E(-x,-y,-z)$. So $E$ must change sign an odd number of times along $C_1$. We already know that it changes sign at $x_1$, ..., $x_8$; the only place to get an additional sign change is at $x_9$. (Note that this argument shows that $C_1 \cap D$ is always odd, counting with multiplicity. I think that a careful count will show that $C_1 \cap D_1$ is odd, and the other three possibilities are always even.)
Now, $E$ may not have the right topology. But, for $\epsilon$ sufficiently small, the curve $D= C+\epsilon E$ also passes through $x_1$, ..., $x_9$, and it does have the right topology.
