5. We first observe that $\triangle A_{5} B_{4} A_{4} \cong$ $\triangle A_{3} B_{2} A_{2}$. Since $\angle A_{5} A_{3} A_{2}=90^{\circ}$, we have $\angle A_{2} B_{4} A_{4}=\angle A_{2} B_{4} A_{3}+$ $\angle A_{3} B_{4} A_{4}=\left(90^{\circ}-\angle B_{2} A_{2} A_{3}\right)+$ $\left(\angle B_{4} A_{5} A_{4}+\angle A_{5} A_{4} B_{4}\right)=90^{\circ}+$ $\angle B_{4} A_{5} A_{4}=120^{\circ}$. Hence $B_{4}$ belongs to the circle with center $A_{3}$ and radius $A_{3} A_{4}$, so $A_{3} A_{4}=A_{3} B_{4}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-461.jpg?height=311&width=428&top_left_y=1245&top_left_x=866)  Thus $\lambda=A_{3} B_{4} / A_{3} A_{5}=A_{3} A_{4} / A_{3} A_{5}=1 / \sqrt{3}$.
5. 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 $\lambda$.