3-2. Points \(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\) divide a circle of radius 1 into six equal parts. From \(A_{1}\), a ray \(l_{1}\) is drawn in the direction of \(A_{2}\), from \(A_{2}\) - a ray \(l_{2}\) in the direction of \(A_{3}, \ldots\), from \(A_{6}\) - a ray \(l_{6}\) in the direction of \(A_{1}\). From point \(B_{1}\), taken on ray \(l_{1}\), a perpendicular is dropped onto ray \(l_{6}\), from the foot of this perpendicular a perpendicular is dropped onto \(l_{5}\), and so on. The foot of the sixth perpendicular coincides with \(B_{1}\). Find the segment \(B_{1} A_{1}\).
Solve 2. Answer: \(B_{1} A_{1}=2\).

Let \(B_{2}, B_{3}, \ldots, B_{7}\) be the feet of the perpendiculars dropped from \(l_{6}, l_{5}, \ldots, l_{1} ; x_{1}=A_{2} B_{1}, x_{2}=\) \(A_{1} B_{2}, x_{3}=A_{6} B_{3}, \ldots, x_{7}=A_{2} B_{7}\). Then \(x_{k+1}=\frac{1}{2}\left(1+x_{k}\right)\). By the condition \(x_{1}=x_{7}\). But

\[
x_{7}=\frac{1}{2}\left(1+x_{6}\right)=\frac{1}{2}\left(1+\frac{1}{2}\left(1+x_{5}\right)\right)=\frac{1}{2}\left(1+\frac{1}{2}\left(1+\frac{1}{2}\left(1+x_{5}\right)\right)\right)=\ldots
\]

therefore

\[
x_{7}=\frac{1}{2}+\frac{x_{6}}{2}=\frac{3}{4}+\frac{x_{5}}{4}=\frac{7}{8}+\frac{x_{4}}{8}=\cdots=\frac{63}{64}+\frac{x_{1}}{64}
\]

Thus, we get the equation \(\frac{63}{64}+\frac{x_{1}}{64}=x_{1}\), from which \(x_{1}=1\). It is also clear that \(B_{1} A_{1}=1+x\).