So I was looking at this thread: How can I prove the existence of an octagon/decagon/dodecagon?, and while this question intrigues me I also have no way to solve it, and it was unanswered in the original thread due to lack of clarity of the question. So I will attempt to clarify the problem.
Let there be two intersecting circles, say, call them $O_1$ and $O_2$, both with radius length $n$. Let the intersection points be $N_1$ and $N_2$.
Then inscribe a regular hexagon in both $O_1$ and $O_2$ such that $N_1$ and $N_2$ are vertices of both hexagons.
The first part of the question is this: is there a circle, let's call it $O_3$, with radius $n$ that passes through $P_1, P_2, P_3,$ and $P_4$ such that $P_1$ and $P_2$ are vertices of the hexagon in $O_1$, and $P_3$ and $P_4$ are vertices of the hexagon in $O_2$, and that $N_1$ is outside the circle, and $N_2$ is inside the circle.
The second part of the question is this: can a regular octagon/decagon/dodecagon be inscribed in $O_3$ such that $P_1, P_2, P_3,$ and $P_4$ are among its vertices?
The question writer also claims possibility with pentagons, where you prove/disprove existence of hexagon/octagon/decagon, and heptagons, where you prove/disprove existence of decagon/dodecagon/14-gon.
My thoughts on this problem was to use coordinates and attempt to bash out coordinates of $P_1, P_2, P_3,$ and $P_4$, but I quickly saw that I was going nowhere fast.
Hopefully I clarified the problem enough.
I'm especially curious on how the pentagon and heptagon would work now; the hexagon to me seems like a "special case" where it is simpler than the pentagon/heptagon to prove/disprove existence. The hexagon can have $O_3$'s center at $N_2$, which greatly simplifies the problem. How would it work for a pentagon/heptagon?