TITLE: Is the circle in the square best at avoiding random lines?
QUESTION [20 upvotes]: This question is inspired by a recent one (and takes a great deal from the answers there). Given a convex subset $\Delta$ of the unit square, let $p(\Delta)$ be the probability that a random line does not intersect $\Delta.$ The exact model for the selection of the lines is that a random point is chosen uniformly on the boundary and then a second is chosen randomly from the union of the other three sides. Consider the circle $\Delta_O$ of radius $\frac1{\sqrt{2\pi}}$ centered at $(\frac12,\frac12),$ I can show that $p(\Delta_0)=0.34470989\dots.$
Let $S$ be the family of convex subsets of the square with area $\frac12.$

What are the best upper and lower bounds you can give on $\sup_{S}p(\Delta)?$
In particular, is there a specific $\Delta \in S$ with $p(\Delta) \gt p(\Delta_O)?$

Here is a certain octagon $\Delta_8$ along with the circle $\Delta_0$ described above.  
I find that the octagon  is the best of its kind but it is not quite as good as the circle. I get that  $p(\Delta_8)=0.31984\dots.$ The lower left corner is at about $(0.32606,0.11670).$
Will Sawin in a comment to the question mentioned above observes that an optimal region will have no straight sides and also no corners. Any set $t\Delta_O+(1-t)\Delta_8$ with $0 \lt t \lt 1$ will be in $S.$ I don't know if these are worth investigating very deeply since some other octagon might be better for the purpose. Such a convex combination would still have $8$ corners but the segments would be curved. To deal with the corners on this, or some other $\Delta$ with corners, we could, for a very small positive $\epsilon,$  take all points within $\epsilon$ of a point of $\Delta.$ This would enlarge the area a bit so we would then shrink the smoothed figure to restore the correct area. This smoothing will fix corners but not straight sides (of length exceeding $\epsilon.$)
It seems clear that an optimal region has the same $8$ lines of reflective symmetry as the square. I'm not claiming to have a proof of that, nor am I asking for one.
In an answer to the linked question Will Sawin also makes an impressive start on a proof on finding the optimal set using the Calculus of Variations. I recommend reading it. The question here is more humble, just beat the circle. Perhaps there is a $16$-gon that succeeds or gets close. If so, it could perhaps be massaged to do even better.
An upper bound  is $p(\Delta) \lt \frac23.$ It is given by Christian Remling. He calls it crude, and I do think it is pretty far from optimal, but it is also the record that I know of so far and the proof sketched is very nice.

REPLY [2 votes]: Indeed it seems that the circle is not optimal although the improvements I've found only raise the probability slightly and are visually hard to distinguish from a circle. Here is the circle in red and a region $\Delta_h$ bounded by $4$ hyperbolas. Recall that the circle has $p(\Delta_0)=0.34470989\dots.$ It turns out that  $p(\Delta_h)=0.34494\dots.$ 
It was suggested that the optimum might be a shape bounded by $4$ hyperbolas near the corners and $4$ parabolas near the middles of the sides. The best I could do with that is $\Delta_{hp}$ shown here with $p(\Delta_{hp})=.34544686\dots.$

The parabola at the bottom is roughly $y=1.221614(x-\frac12)^2+.1045232.$ The transition from the parabola to the hyperbola to its left occurs between $(0.26957, 0.16938)$ and  $(0.267287, .170719)$ (the respective endpoints of the polygonal approximations) with the middle of the hyperbola at $(0.213614,0.213614).$
Some disclaimers: I am not certain this is correct. I expected that the contact point of a tangent line rolling around would transition from hyperbola to parabola exactly when its intersection of the line with the square passes a corner and begins to intersect opposite sides. I believe this would maintain a constant area between the tangent and the square.However I don't see that happening here. Also,
 The later digits may not be exact, I tried to truncate before a small calculated digit. These are actually polygons obtained by sampling (at equally radially spaced points). The area may come out to be less than $\frac12$ in which case the figure is dilated appropriately. This will map parabolas to parabolas but deform hyperbolas. However replacing the deformed sections by a best hyperbola approximation changes the area only slightly. Iterating a few times stabilizes. I did not do this adjusting to  $\Delta_h$ so it may be cheated out of a bit.
I don't know if this is really the optimal shape. I could investigate small perturbations of the polygon but have not. I am pretty sure that something similar to a quarter circle in a corner OR a parabola along with most of the top OR  the top, with the upper portions of the two sides connected by the arc of a conic could not do as well. I'm not saying it would be hard to rule those out. I just haven't rigorously done so.