Given $S^1$, the unit circle in $\Bbb R^2$, with the Lebesgue measure restricted to it denoted as $\ell(\cdot)$ (the "arc length"). Now let $R\subset S^1$ be an arbitrary measurable set with $\ell (R)=2\pi p$, then what's the minimal possible value of $p$, if extant, so that there exists a circumscribed square whose every vertex lies in $R$? (If the minimal is not attainable, then at least what's the infimum?)
For a crude estimate, clearly $p>3/4$ can guarantee the existence of such a square: let's identify $S^1$ with $J=[0,1]/\{0,1\}$ and $R$ with a subset of $J$ and $\ell(\cdot)$ with $m(\cdot)$ (the Lebesgue measure on $J$) and thus $m(R)=p$. Then the existence of the square is equivalent to that (mod $1$)$$R\cap (R+1/4)\cap (R+1/2)\cap (R+3/4)\ne \emptyset$$ , which holds if $$m(R^c\cup (R+1/4)^c\cup (R+1/2)^c\cup (R+3/4)^c)<1$$, which again holds (but not a necessary condition!) if $P(R)>3/4$.
However, as indicated in boldface above, there is no guarantee that $3/4$ is optimal.