This question is a follow-up to Maximum number of intersections between a quadrilateral and a pentagon, where it is shown that the boundaries $\partial Q,\partial P$ of a quadrilateral and a pentagon in the plane cannot intersect at more than $16$ points, since each side of $\partial Q$ meets $\partial P$ at an even number of points.
Q: Given the boundaries $\partial P_1, \partial P_2$ of two pentagons in the plane, is it possible that $$ \left|\partial P_1 \cap \partial P_2 \right| = 20?$$
Each side of $\partial P_1$ meets $\partial P_2$ at an even number of points, so equality is attained iff there is some configuration such that each side of $\partial P_1$ meets each side of $\partial P_2$ except one. $\left|\partial P_1 \cap \partial P_2\right| = 18$ is possible, as shown below,
and I believe that $\left|\partial P_1 \cap \partial P_2\right| = 20$ is impossible, but I am failing to prove it.
According to Theorem 4 from Černy et al.'s "On the number of intersections of two polygons" (2003), the number of intersections is bounded by $$5\times 5-\left\lceil \frac56 \right\rceil-5=19$$ Therefore $18$ is the maximum.
Alternatively, Theorem 5 gives the exact value $4\times 5-2=18$.
These maximal intersection numbers are hard to compute when both polygons have an odd number of vertices. Otherwise there are general formulas given by Theorem 1 and 2.
Edit: found a better result in Günther's "The maximum number of intersections of two polygons" (2012).