Prove Existence of a Circle There are two circles with radius $1$, $c_{A}$ and ${c}_{B}$. They intersect at two points $U$ and $V$. $A$ and $B$ are two regular $n$-gons such that $n >  3$, which are inscribed into $c_{A}$ and ${c}_{B}$ so that $U$ and $V$ are vertices of $A$ and $B$.
Then suppose a third circle, $c$, with a radius of $1$ is to be placed so that it intersects $A$ at two of its vertices $W$ and $X$ and intersects $B$ at two of its vertices $Y$ and $Z$.
Details and Assumptions:


*

*Assume that $U,V,W,X,Y,Z$ are all distinct points.

*$U$ lies outside of $c$.

*$V$ lies inside of $c$.
Given all of these details, prove that there exists a regular $2n$-gon which comprises of $W,X,Y,Z$ as its 4 vertices.


 A: Answering the easier question about why the points always lie on a regular $2n$-gon.
Leaving the more interesting question about listing the possible values of $n$ when this may happen for later (or for somebody else!).

Let us denote $\phi=2\pi/n$. We align the coordinate axes in such a way that the center of $c_A$ is at the origin $O$ and that the positive $x$-axis intersects $c_A$ at a vertex of of $A$. This implies that the points $U,V,X,W$ all have angular polar coordinates that are integer multiples of $\phi$.
Let $L_1$ (resp. $L_2$) be the line through $U$ and $V$ (resp. through $W$ and $X$). The line $L_1$ is perpendicular to the bisector of the angle between $\vec{OU}$ and $\vec{OV}$. Therefore $L_1$ points at the direction that is perpendicular to an integer multiple of $\phi/2$. The same holds for $L_2$. Thus the angle $\theta$ between $L_1$ and $L_2$ is also an integer multiple of $\phi/2$, so $\theta= k\pi/n$ for some integer $k$.
Let $s_1$ (resp. $s_2$) be the reflection w.r.t. $L_1$ (resp. $L_2$). It is well known that the composition $r=s_2\circ s_1$ is a rotation about the point of intersection $Q=L_1\cap L_2$ by the angle $2\theta=2k\pi/n=k\phi$. If this is news to you the animation below may make this clearer. There the black arrow is first reflected w.r.t. the blue line, and the red arrow is the mirror image. For its part the red arrow is then reflected w.r.t. the green line and the orange arrow is its mirror image. The animation tries to convince you that irrespective of which direction the black arrow points at, the angle between it and the red orange arrow is constant (=twice the angle between the blue and green lines). Prove this if you already haven't. It's not difficult!

It is clear that $A=s_1(B)$ and that the $s_2(A)=r(B)$ is a regular $n$-gon circumscribed by $c$. This implies that $r(c_B)=c$. Also, as the angle of rotation $2\theta$ is a multiple
of $\phi$, the sides of the regular $n$-gon $r(B)$ are parallel to those of $B$.
The figure below hopefully makes it clearer what happened. After we reflected $n$-gon $B$ (red) first w.r.t. line $L_1$ to get the $n$-gon $A$ (green) and then w.r.t. line $L_2$ the resulting $n$-gon (blue) can be gotten from $B$ also by a parallel translation. Quite irrespective of whether the circles $c_B$ and $c$ intersect on vertices of the $n$-gon $B$ or not (in the image they intentionally do not)!

Let $s_3$ is the reflection w.r.t. the line $L_3$ passing through $Y$ and $Z$. If $O_B$ is the center of the circle $c_B$, then the angle $\beta=\angle YO_BZ$ is an integer multiple of $\phi$, say $\beta=\ell\phi$. The angle between $YZ$ and $O_BY$ is thus 
$$\gamma=\frac\pi2-\frac\beta2=\frac\pi2-\frac{\ell\pi}n=\frac\pi{2n}(n-2\ell).$$
This means that the angle between $L_3$ and the extension of any edge of $B$ is an integer multiple of $\pi/2n=\phi/4$. Thus, under the reflection $s_3$ the directions of
those edges change by an integer multiple of $\phi/2$. Therefore the regular $n$-gon
$s_3(B)$ is either parallel to $r(B)$ itself or parallel to a version rotated by $\phi/2$ (depending on the parity of $n-2\ell\equiv n\pmod2$). Because $s_3(B)=s_3(s_1(A))$ the $n$-gons $s_3(B)$ and $A$ are parallel. With a little imagination you see in the above figure that the green 11-gon is "half a tick" off synch from the blue and red 11-gons that are parallel to each other.
As both regular $n$-gons, $s_3(B)$ and $r(B)$ are circumscribed by $c$, $r(B)$
contains $W$ and $X$ as vertices, and $s_3(B)$ contains $Y$ and $Z$ as its vertices, the claim follows from this.

Extras that may or may not help in simplifying the above argument or finding the solutions:
Because we get $c$ by rotating $c_B$ about $Q$, the point $Q$ must be equidistant from the centers of $c_B$ and $c$. In other words, $Q$ is on the line $YZ$ (so the lines $L_1$, $L_2$ and $L_3$ intersect at the same point $Q$). 
Recalling that $s_3$ is the reflection w.r.t. $YZ$, then $s_3\circ r$ is yet another reflection (as an orientation reversing rigid motion of the plane), call it $s_4$. Clearly $s_4(Q)=Q$ and $s_4(c_B)=c_B$, so $s_4$ must be the reflection w.r.t. line joining $Q$ and the center of $c_B$.
A: I tried to solve the problem in a more general setting. However, despite my (hard!) effort this is NOT an answer. The reason I am posting it here is because it might shed some further light on this problem and on its general version, which I find very entertaining and interesting. So, if my reasoning could help someone find a nice solution, here it is.
Let $A_m$ and $A_n$ be two regular polygons of $m$ and $n$ sides respectively, each inscribed in a circle of radius $1$. Let's say that they "fit together to order $d$" (invented notation) if $d$ divides both $m$ and $n$.
This corresponds visually to putting the polygons one over the other in such a way that the centers of the corrisponding circumscribed circles coincide and letting one of the vertices of one polygon coincide with a vertex of the other one. Then $d$ is the number of vertices in the picture which belong to both polygons.
The possible "fitting orders" for $A_m$ and $A_n$ are then $1=d_1,d_2,d_3,\dots,d_r=\gcd(m,n)$ where the $d_i$'s are all common divisors of $m$ and $n$. Let's denote this set as $D_{m,n}$.
Now, for the geometrical setting, let's take a circle $C$ with radius $1$ and center $O$ and two chords $\overline{A_1B}_1$ and $\overline{A_2B}_2$. Let $M_1$ and $M_2$ be the midpoint of $\overline{A_1B}_1$, resp. $\overline{A_2B}_2$, and let $\theta$ be the angle $M_1\widehat{O}M_2$, as in the picture below:

Reflecting $C$ about $\overline{A_1B}_1$ gives another circle $C_1$; doing the same with $\overline{A_2B}_2$ gives a circle $C_2$. These circles might or might not intersect; let's suppose they do. Then they will meet at two points $P$ and $Q$.

Reflecting $P$ about the two chords, we get two points $P_1$ and $P_2$ which lay on the original circle $C$. Doing the same with $Q$, we get two other points $Q_1$ and $Q_2$. By symmetry, one can see that $\overline{PQ}=\overline{P_1Q}_1=\overline{P_2Q}_2$ and also $P\widehat{O}_1Q=P\widehat{O}_2Q=P_1\widehat{O}Q_1=P_2\widehat{O}Q_2$, where $O_1$ and $O_2$ are the centers of $C_1$ and $C_2$.
Here's the picture for $P_1$ and $Q_1$:

So, we started with two chords on a circle and ended up with two other chords of equal length in some random position on the circle. Getting back to polygons, given three polygons $A_m,A_n$ and $A_p$ we can choose two chords of $A_m$ (i.e. segments whose endpoints are vertices of the polygon) by selecting $d_1\in D_{m,n}$ and $d_2\in D_{m,p}$ and by deciding what is their position on the polygon. This choice will determine two other chords which, in general will be chords of the circumscribed circle of $A_m$ but not of $A_m$ itself;
the challenge is determining what values of $n$ and $p$ lead to chords which are also chords of $A_m$. A necessary condition for this is that the angle $P\widehat{O}_1Q$ must be of the form $2\pi/d^*$ where $d^*$ is a common divisor to $m,n$ and $p$. This angle $\alpha^*$ is uniquely determined by the distance $\overline{O_1O_2}=L$, because $L/2=\cos(\alpha^*/2)$. On the other side, we have (by the law of cosines)
\begin{equation}
L^2=\overline{OO}_1^2+\overline{OO}_2^2-2\overline{OO}_1\ \overline{OO}_2\cos\theta
\end{equation}
But we have also
\begin{equation}
\overline{OO}_1=2\cos\left(\frac{\pi}{d_1}\right)\qquad
\overline{OO}_2=2\cos\left(\frac{\pi}{d_2}\right)
\end{equation}
so
\begin{equation}
(L/2)^2=\cos^2(\pi/d_1)+\cos^2(\pi/d_2)-2\cos(\pi/d_1)\cos(\pi/d_2)\cos\theta=
\cos^2(\pi/d^*)
\end{equation}
We have that $\theta$ as well has to be a multiple of $2\pi/m$, so we can set $\theta = 2\pi/d$ with $d$ divisor of $m$. 
In the end the problem is finding $d_1,d_2,d^*$ and $d$ such that
\begin{equation}
\cos^2(\pi/d_1)+\cos^2(\pi/d_2)-2\cos(\pi/d_1)\cos(\pi/d_2)\cos(\pi/d)=
\cos^2(\pi/d^*)
\end{equation}
This will yeld all the possible combinations of three polygons which will "fit together" in some way. I don't know how to solve this, but I thought to share it with you in case someone could see the solution.
A: Here's a straightforward proof of the existence of the $2n$-gon.
Ignoring polygons for the time being, take points $W$ and $X$ on $\bigcirc A$ and points $Y$ and $Z$ on $\bigcirc B$, such that $W$, $X$, $Y$, $Z$ all lie on $\bigcirc C$, where all three circles are congruent.

Define $w$, $x$, $y$, $z$ to be the measures of angles made, respectively, by radial segments $\overline{AW}$, $\overline{AX}$, $\overline{BY}$, $\overline{BZ}$ with $\overleftrightarrow{AB}$. (These four measures, as described, are ambiguous; that's okay.) Because $\square AWCX$ and $\square BYCZ$ are rhombi —and therefore also parallelograms— radial segments $\overline{CX}$, $\overline{CW}$, $\overline{CZ}$, $\overline{CY}$ make comparable angles with the line through $C$ parallel to $\overleftrightarrow{AB}$.
Clearly, then, the measure of any $\angle PCQ$, with $P$ and $Q$ in $\{W, X, Y, Z\}$, is some combination of $\pm w$, $\pm x$, $\pm y$, $\pm z$, and $\pm \pi$. (Allowing $\pm \pi$ eliminates any problems with our ambiguously-defined $w$, $x$, $y$, $z$.) Most importantly: If $w$, $x$, $y$, $z$, $\pi$ are multiples of a common value, then those $\angle PCQ$s will be, as well.
Recalling the original context of the problem, we have that $\bigcirc A$ and $\bigcirc B$ meet at vertices of inscribed $n$-gons. Necessarily, $\overleftrightarrow{AB}$ is a line of symmetry for the compound polygonal figure, either passing through the polygons' vertices, or perpendicularly-bisecting the polygons' edges, or both. In any case, radial segments to the vertices of each polygon make angles with $\overleftrightarrow{AB}$ that are multiples of $\pi/n$.
Therefore, if $W$, $X$, $Y$, $Z$ are themselves vertices of these $n$-gons, then $w$, $x$, $y$, $z$ (and $\pi$!) are multiples of $\pi/n$, as are the $\angle PCQ$s. As $\pi/n = 2\pi/(2n)$ is the central angle between neighboring vertices of a $2n$-gon, we have our result.
Note: If $n$ is even, and if the polygons overlap in such a way that they have vertices on the line of symmetry $\overleftrightarrow{AB}$, then $w$, $x$, $y$, $z$ (and $\pi$!) are multiples of $2\pi/n$, so that $W$, $X$, $Y$, $Z$ are vertices of an $n$-gon about $C$.
A: This geometric problem can be expressed in purely algebraic terms, to be precise
it all amounts to a certain $\mathbb Q$- linear relation between roots of unity,
and this theme is a classic that has already been studied by Mann, Schoenberg,
Conway and others. It follows from what we show below that $n \leq 15$.
Denote by $\Omega_A,\Omega_B,\Omega_C$ the centers of $c_A,c_B,c$ respectively,
and $z_A,z_B,z_C$ the corresponding complex numbers (or "affixes" as they are called in French). Similarly, denote by $z_U,z_V,z_W,z_X,z_Y,z_Z$ the affixes of $U,V,W,X,Y,Z$. Let
$\zeta$ be a primitive $n$-th root of unity.
We may assume without loss of generality that $z_A=0$. Then there are four integers
$u,v,w,x\in[0,n-1]$ such that $z_U=\zeta^{u},z_V=\zeta^{v},z_W=\zeta^{w},z_X=\zeta^{x}$. For the same reason,
there are two  integers
$y,z\in[0,n-1]$ such that $z_Y=z_B-\zeta^{y},z_Z=z_B-\zeta^{w}$.
Since $\Omega_AU\Omega_BV$, $\Omega_AX\Omega_CW$ and $\Omega_BY\Omega_CZ$ are parallelograms, we deduce $z_B=\zeta^{u}+\zeta^{v},z_C=z_W+z_X-z_A=z_Y+z_Z-z_B$.
Since $X,U,V,W$ are four distinct points on $c$, we see that $x,u,v,w$ are pairwise distinct. Similarly, $y,u,v,z$ are pairwise distinct, and $x,y,z,w$ are pairwise distinct.
In the end, $x,y,z,u,v,w$ are all distinct. 
Combining all those equalities, we see that
$$
\begin{array}{lcl}
z_B &=& \zeta^u +\zeta^v \\
z_Y &=& \zeta^u +\zeta^v-\zeta^y \\
z_Z &=& \zeta^u +\zeta^v-\zeta^z \\
z_C &=& \zeta^x+\zeta^w = \zeta^u +\zeta^v-\zeta^y-\zeta^z
\end{array}\tag{1}
$$
So everything reduces to the relation
$$
\zeta^x +\zeta^y+\zeta^z+\zeta^w-(\zeta^u+\zeta^v)=0 \tag{2}
$$
The above is an example of what I call an $U$-relation. It is defined
as a double uple $((a_1,a_2,\ldots,a_r),(\xi_1,\xi_2,\ldots,\xi_r))$ 
satsifying the identity $\sum_{k=1}^{r}a_k\xi_k=0$ where the $a_k$ are nonzero
integers and the $\xi_k$ are distinct roots of unity. I call the integer $r$ 
the length of the $U$-relation. There are three natural
operations on $U$-relations : permutation, rotation and multiplication
by a constant. I mean by that that
$((a_{\sigma(1)},a_{\sigma(2)},\ldots,a_{\sigma(r)}),\
(\xi_{\sigma(1)},\xi_{\sigma(2)},\ldots,\xi_{\sigma(r)}))$
is still an $U$-relation when $\sigma$ is a permutation of the integers between
$1$ and $r$, $((a_1,a_2,\ldots,a_r),(\alpha\xi_1,\alpha\xi_2,\ldots,\alpha\xi_r))$
is still an $U$-relation when $\alpha$ is a root of unity, and
$((ka_1,ka_2,\ldots,ka_r),(\xi_1,\xi_2,\ldots,\xi_r))$
is still an $U$-relation when $k\in{\mathbb Z},k\neq 0$. The regular $U$-relation
in length $r$ is the $U$-relation $((1,1,\ldots,1),(1,\eta,\eta^2,\ldots,\eta^{r-1}))$
where $\eta$ is a primitive $r$-th root of unity. An $U$-relation
$((a_1,a_2,\ldots,a_r),(\xi_1,\xi_2,\ldots,\xi_r))$  is said to be reducible
if there is a partition $I\cup J$ of $[1,r]$ into two non-empty parts such that
$\sum_{k\in I}a_k\xi_k=\sum_{k\in J}a_k\xi_k=0$. It is easy to see that a regular
$U$-relation is irreducible iff its length is prime.
Theorem For any $k$, up to permutation, rotation and multiplication by
a constant there are only finitely many irreducible $U$-relations of length $k$. In length $\leq 7$, the only non-regular irreducible $U$-relations are 
$$
((\epsilon_1,\epsilon_2,\epsilon_3,\epsilon_4,\epsilon_5,\epsilon_6),
(-\epsilon_1\alpha,-\epsilon_2\alpha^2,\epsilon_3\beta,\epsilon_4\beta^2,\epsilon_5\beta^3,\epsilon_6\beta^4)), \tag{3}
$$
where all the $\epsilon_k(1\leq k\leq 6)$ are $\pm 1$, $\alpha$ is a primitive third root of unity and $\beta$ is a primitive fifth root
of unity.
Proof of theorem : see Henry B. Mann, "On linear relations between roots of 
unity", Mathematika 12(1965), pp.107-117.
Corollary. If we denote by $(t_1,t_2,t_3,t_4,t_5,t_6)$, the second part of (3), 
up to permutation and rotation there are
exactly $\binom{6}{2} \times \frac{4!}{8}=15\times6=90$ solutions to (2) (with $x,y,z,u,v,w$ pairwise distinct), all of which come from (3) : they can be described by
$$
\lbrace \zeta^u,\zeta^v \rbrace=
 \lbrace -t_i,-t_j \rbrace,
 \lbrace \zeta^x,\zeta^y,\zeta^z,\zeta^w \rbrace=
 \bigg\lbrace t_k \ \bigg| \ k\neq i, k\neq j \bigg\rbrace, \ \ 1\leq i \lt j \leq 6 \tag{4}
$$
Proof of corollary By theorem, all irreducible $U$-relations of length $<6$
are regular, and so have all their coefficients of the same sign. So if (2) were not
an irreducible relation, it would necessarily follows that $\zeta^u+\zeta^v=0$, so 
$U$ and $V$ are diametrically opposed, hence $c_B=c$ which is impossible. So (2) must
be irreducible. Since it has coefficients of differents signs, it does not come
from the regular $U$-relation in length $6$. So it must come from (3), which yields
(4).
To count the solutions, note that there are $\binom{6}{2}$ possible
values for $\lbrace i,j\rbrace$ in (4), and also that in counting the
possibilities for $(\zeta^x,\zeta^y,\zeta^z,\zeta^w)$, we obtain
repetitions by interchanging $x$ and $w$, by interchanging $y$ and $z$,
or by interchanging $\lbrace x,w\rbrace$ and $\lbrace y,z\rbrace$. The subgroup
of ${\mathfrak S}(x,y,z,w)$ fixing a given configuration has cardinality $8$.
A: Looks like false to me...
For $n=4$, which seems to satisfy $n>3$ requirement, the $A$ and $B$ squares can not intersect at opposite vertices (because they would be the same, $A=B$), so they'll have to share one side. Then there's no $X$ or $Y$ vertex between $U$ and $V$ where $C$ octagon might cross $A$ and $B$...
