1982 IMO 第 5 题

A5 (NET 2) ${ }^{\mathrm{IMO}}$ Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be a regular hexagon. Each of its diagonals $A_{i-1} A_{i+1}$ is divided into the same ratio $\frac{\lambda}{1-\lambda}$, where $0<\lambda<1$, by a point $B_{i}$ in such a way that $A_{i}, B_{i}$, and $B_{i+2}$ are collinear ( $i \equiv$ $1, \ldots, 6(\bmod 6))$. Compute $\lambda$.

A5 (NET 2) ${ }^{\mathrm{IMO}}$ 设 $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ 为正六边形。它的每条对角线 $A_{i-1} A_{i+1}$ 被点 $B_{i}$ 分成相同的比率 $\frac{\lambda}{1-\lambda}$，其中 $0<\lambda<1$，使得 $A_{i}、B_{i}$ 和 $B_{i+2}$ 共线( $i \equiv$ $1, \ldots, 6(\bmod 6))$。计算 $\lambda$。