A clarification on a problem on proving existence of regular polygons 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?
 A: There are two configurations of the hexagons possible: If the first is $ABCDEF$, the second may be (up to symmetry) either $BAGHIJ$ or $CGAHIJ$.
In the first case, any third circle separating $N_1=A$ and $N_2=B$ intersects the arc $AB$ in a non-vertex, hence this case is to be ruled out.
And in the second case where $N_1=A$ and $N_2=C$, say, we know for sure that $B,G$ are on $O_3$. Then the center of $O_3$ must be $A$ or $C$, and as we want $N_2$ inside the circle, it must be $C$. Then we automatically get the four intersection points $B,D, G, J$ as required.
An octagon in $O_3$ can be excluded immediuately: The lengths of edges and diagonals in $O_1$ and $O_2$ are $n$, $n\sqrt3$ and $2n$. Apart from $2n$ (i.e. a diameter), none of these lengths occurs in an octagon inscribed in $O_3$; and of course the circles cannot share a diameter (which would make them equal).
The same works for a decagon. For the dodecagon we can simple take our hexagon solution from above and subdivide it. In general, since some diagonal of the inscribed polygon must cover an angle of $60^\circ$, only $m$-gons with $6\mid m$ lead to a solution.
