TITLE: Prove Existence of a Circle
QUESTION [30 upvotes]: 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.

REPLY [3 votes]: 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 $8$.