A hard problem from regular $n$-gon I found the following theorem in a book of mine without a proof. Could someone show me a proof of it?
Given a regular $n$-gon, with $n$ odd and vertices $v_1,\ldots,v_n$, and $C$ its circumcircle. At each $v_i$ draw a circle that is internally tangent to $C$ at $v_i$, and suppose all these tangent circles are congruent. Let $P$ be any point on the minor arc from $v_1$ to $v_n$ and let $t_i$ be the length of the tangent from $P$ to the circle tangent to $C$ at $v_i$. Then $\sum_{i=1}^n(-1)^it_i=0$.
 A: A diagram :

An attempt using complex arithmetic:
Let $R=$radius of C (circumcircle, center at the origin of the complex plane), $r=$radius of inner circles, $c_k$ their centers ($k=1\cdots n$). Further, let
$$\begin{array}{cl}
q &=&e^{i 2 \pi /n}\\
c_k &=& s \, q^k , \hskip{1cm} s = R -r\\
P &=&   R \, q^u , \hskip{1cm} u \in (0,1)\\
\end{array}
$$
For each circle $k$, let $l_k = c_k -P $ (distance from the center to the point P), let $t_k$ the  length of the tangent from P to the circle. Because the tangent is normal to the radius, we have:
$$|l_k|^2 = t_k ^2 + r^2$$
$$ t_k^2 = |s q^k - R q^u|^2 -r ^2=s^2+R^2 - s \,R (q^{k-u} + q^{u-k}) -r^2= s \, R \, \left[ 2 - (q^{k-u} + q^{u-k})\right]$$
Using $(z^y-z^{-y})^2 = z^{2y} + z^{-2y} -2$, we get
$$ t_k^2 = - s \, R \, \left[ q^{(k-u)/2} - q^{(u-k)/2}\right]^2$$
or, setting $p=q^{1/2}$ and multiplying by $q^{nk}$:
$$ (p^{n k } t_k)^2 = s \, 
R \left[ -i p^{n k } 
\left( p^{k-u} - p^{u-k} \right)\right]^2=  
s \, R \left[ p^{n k} 2 \, Im\left( p^{k-u} \right)\right]^2
$$
But $Im\left( p^{k-u} \right) \ge 0$ for $k=1\cdots n$ and $u \in (0,1)$; and $p^{nk}=(-1)^k$. Hence inside the squares on each side we have real numbers with the same sign (for each $k$), and so we can take safely the square root and sum over
$k$. On the right side we have a geometric sum:
$$\sum_{k=1}^n p^{n k} p^{k-u} = p^{-u} \sum_{k=1}^n p^{(n+1) k} = p^{-u} p^{n+1} \frac{1-p^{(n+1)n}}{1-p^{n+1}}$$
But $$p^{(n+1)n}= q^{\frac{(n+1)n}{2}}=1 $$ for $n$ odd integer, so finally
$$\sum_{k=1}^n (-1)^k t_k = 0$$
