1. In the plane, there is a rectangle $A B C D$, where $|A B|=a<b=|B C|$. On its side $B C$ there exists a point $K$ and on side $C D$ a point $L$ such that the given rectangle is divided into four mutually similar triangles by the segments $A K, K L$ and $L A$. Determine the value of the ratio $a: b$.
1. In the right-angled triangle $A B K$, let $\alpha=|\angle B A K|, \beta=|\angle A K B|=90^{\circ}-\alpha$ (Fig. 1). The same internal angles $90^{\circ}, \alpha, \beta$ are also present in triangles $A K L$ and $A D L$, as they are similar to triangle $A B K$ according to the problem statement.

![](https://cdn.mathpix.com/cropped/2024_04_17_01c718df6621aba2457cg-2.jpg?height=448&width=391&top_left_y=484&top_left_x=838)

Fig. 1

Let's note the (acute) angles at the common vertex $A$. Since $|\angle K A D|=90^{\circ}-\alpha=\beta$, both angles $K A L$ and $L A D$ are smaller than $\beta$, so they equal the angle $\alpha$. The right angle $B A D$ is thus divided by the rays $A K, A L$ into three equal angles of size $\alpha$, from which $\alpha=30^{\circ} (\text{and} \beta=60^{\circ})$. From the right-angled triangles $A D L$ and $A B K$, it follows that $|A K|=|A B| / \cos 30^{\circ}=2 a / \sqrt{3}$ and $|A L|=|A D| / \cos 30^{\circ}=2 b / \sqrt{3}$. Therefore, considering the condition $a1$, which contradicts the problem statement).

Award 6 points for a complete solution; 2 points for determining the angles $\alpha, \beta$, 2 points for calculating the ratio $a: b$, 2 points for the calculation or reasoning that excludes the opposite ratio (i.e., the possibility $|\angle A L K|=90^{\circ}$).