11.7. (SFRY, 76). On a plane, a regular hexagon with side $a$ is drawn. For any value of $n \in \mathbf{N}$, greater than 1, construct a segment of length $a / n$ using only a straightedge.
11.7. Given a regular hexagon $A_{0} A_{1} B_{2} C_{2} C_{1} B_{0}$ (Fig. 48), we will construct the following points using a ruler: $A_{2}$ and $A_{8}$ at the intersection of line $A_{0} A_{1}$ with lines $C_{2} B_{2}$ and $C_{1} B_{2}$, respectively,

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-208.jpg?height=272&width=752&top_left_y=1284&top_left_x=159)

Fig. 48

$C_{8}$ at the intersection of lines $C_{1} C_{2}$ and $A_{1} B_{2}$, $B_{1}$ and $B_{3}$ at the intersection of line $B_{0} B_{2}$ with lines $A_{1} C_{1}$ and $A_{3} C_{8}$, respectively. Then we have

$$
A_{1} A_{2}=C_{1} C_{2}=a
$$

(because $A_{1} A_{8} C_{2} C_{1}$ is a parallelogram),

$$
A_{2} A_{3}=C_{1} C_{2}=a
$$

(because $\triangle A_{8} A_{3} B_{2}=\triangle C_{2} C_{1} B_{2}$),

$$
C_{2} C_{8}=A_{\mathbf{1}} A_{2}=a
$$

(because $\triangle C_{2} C_{8} B_{2}=\triangle A_{2} A_{1} B_{2}$), and finally,

$$
B_{1} B_{2}=B_{2} B_{3}=a
$$

(because $A_{1} A_{2} C_{2} C_{1}, A_{2} A_{8} C_{3} C_{2}$ are parallelograms and $B_{0} B_{2} \| A_{0} A_{1}$). Now, we will perform similar constructions for the regular hexagon $A_{1} A_{2} B_{3} C_{8} C_{2} B_{1}$, which is the result of a parallel translation of the original hexagon by the vector $\overrightarrow{A_{1} A_{2}}$, adding $1 \mathrm{k}$ to the index of each letter (some points are already constructed). We will obtain points $A_{4}, B_{4}$, $C_{4}$, then similarly points $A_{5}, B_{5}, C_{5}$, and so on. When point $B_{n}$ is constructed, we will find point $O$ at the intersection of lines $A_{0} B_{n}$ and $A_{1} B_{1}$. From the similarity of triangles $A_{0} A_{1} O$ and $A_{0} A_{n} B_{n}$, we have the segment

$$
A_{1} O=\frac{A_{n} B_{n} \cdot A_{0} A_{1}}{A_{0} A_{n}}=\frac{a \cdot a}{n a}=\frac{a}{n}
$$

which is what we needed to construct.

Remark. The following general statement holds: if a segment and a line parallel to it are given, then using only a ruler, the segment can be divided into $n$ equal parts.

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-209.jpg?height=418&width=419&top_left_y=926&top_left_x=43)

Fig. 49