8.4. From identical isosceles triangles, where the angle opposite the base is $45^{\circ}$ and the lateral side is 1, a figure was formed as shown in the diagram. Find the distance between points $A$ and $B$.
Answer: 2.

Solution: Let's denote the points $K, L, M$, as shown in the figure. We will construct an isosceles triangle $A K C$ equal to the original one. Connect vertex $C$ to other points as shown in the figure.

In the original triangles, the angle at the vertex is $45^{\circ}$. Therefore, the other two angles are each $62.5^{\circ}$. Then the angle $\angle C K L = 360^{\circ} - 62.5^{\circ} - 62.5^{\circ} - 45^{\circ} - 45^{\circ} - 62.5^{\circ} = 62.5^{\circ}$. Therefore, triangles $A K C$ and $K L C$ are equal by two sides and the angle between them. Similarly, it can be proven that triangles $L M C$ and $M B C$ are equal to the original ones. Then the angle $\angle A C B$ is $4 \times 45^{\circ} = 180^{\circ}$, which means points $A, C, B$ lie on the same line. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_c10d52ec2bf2d695bd7bg-2.jpg?height=331&width=511&top_left_y=977&top_left_x=1321)
$A B = A C + C B = 2$.

## Criteria:

![](https://cdn.mathpix.com/cropped/2024_05_06_c10d52ec2bf2d695bd7bg-2.jpg?height=377&width=552&top_left_y=1419&top_left_x=1343)

A suitable construction is made, but it is not justified that points $A, C, B$ lie on the same line (or point $C$ is the midpoint of segment $A B$, but it is not justified that the resulting triangles are equal to the original ones) - 1 point.

The solution refers to the fact that a regular octagon has a center and other obvious properties of the octagon - no more than 1 point should be deducted.