3-5. On the extensions of the sides \(A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}\) of a regular \(n\)-gon ( \(n \geqslant 5\) - note by the editor) \(A_{1} A_{2} \ldots A_{n}\), 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 \(A_{n} A_{1}\).
Solution 5. Let \(x_{k}=A_{k+1} B_{k}, \alpha\) be the exterior angle of a regular \(n\)-gon, and \(a\) 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 \(x_{1}=a_{1}+b_{1} x_{2}=a_{2}+b_{2} x_{3}=\) \(\cdots=a_{n}+b_{n} x_{1}\), where \(b_{n}=(\cos \alpha)^{n} \neq 1\). Therefore, \(x_{1}\) (and also \(x_{k}\) for any \(k\)) is uniquely determined. It is also clear that we will obtain a solution if we set \(x_{1}=x_{2}=\cdots=x_{n}=x\), where \(x=(a+x) \cos \alpha\), i.e., \(x: a=\cos \alpha:(1-\cos \alpha)\)

\section*{XX Olympiad (1957)}

\section*{First Round}

\section*{7th Grade}