11 On the extensions of the sides $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}$ of a regular $n$-gon $(n \geq 5) A_{1} A_{2} \ldots A_{n}$, construct points $B_{1}, B_{2}$, $\ldots, B_{n}$ such that $B_{1} B_{2}$ is perpendicular to $A_{1} A_{2}, B_{2} B_{3}$ is perpendicular to $A_{2} A_{3}, \ldots, B_{n} B_{1}$ is perpendicular to $A_{n} A_{1}$.
Let $x_{k}=A_{k+1} B_{k}, \alpha$ be the exterior angle of a regular $n$-gon, $a$ be the length of its side. Then $x_{1}=\left(a+x_{2}\right) \cos$ $\alpha, x_{2}=\left(a+x_{3}\right) \cos \alpha, \ldots$,

$x_{n}=\left(a+x_{1}\right) \cos \alpha$. Then $x_{1}=a_{1}+b_{1} x_{2}=a_{2}+b_{2} x_{3}=\ldots=a_{n}+b_{n} x_{1}$, where $b_{n}=(\cos \alpha)^{n} \neq 1$. Therefore, $x_{1}$ (and also

$x_{k}$ for any $k$) is uniquely defined. It is also clear that we will obtain a solution if we set $x_{1}=x_{2}=\ldots=x_{n}=$ $x$, where $x=(a+x) \cos \alpha$, that is, $x=\frac{a \cos \alpha}{1-\cos \alpha}$.