Problem 1129

11.6. Without loss of generality, we can assume that the lengths of the sides of the given octagon $A_{1} A_{2} \ldots A_{8}$ are rational numbers (otherwise, we will prove the required statement for a similar octagon $A_{1}^{\prime} A_{2}^{\prime} \ldots A_{\mathbf{8}}^{\prime}$, where $A_{1}^{\prime} A_{2}^{\prime}=1$, and thus the other sides are rational; as a result, the statement will be proven for the original octagon as well). Consider the vectors

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the sum of which is 0. Since all angles of the octagon are equal and their sum is $6 \cdot 180^{\circ}$, each angle is $(3/4) \cdot 180^{\circ}$, and the angles between vectors $a_{i}$ and $a_{i+1}\left(a_{9}=a_{1}\right)$ are $(1 / 4) \cdot 180^{\circ}=45^{\circ}$ (Fig. 47). Project all vectors onto an axis parallel to, for example, vector $a_{1}$, and let $x$ be the length of the projection of the sum $a_{i}+a_{5}$, and $y$ be the length of the projection of the sum $a_{2}+a_{4}+a_{6}+a_{8}$. Since the projection of the sum $a_{8}+a_{7}$ is 0 (because $a_{3} \perp a_{1}, a_{7} \perp a_{1}$), we have $x-y=0$. On the other hand, the length of the projection of each of the vectors $\boldsymbol{a}_{2}, \boldsymbol{a}_{4}, \boldsymbol{a}_{6}, \boldsymbol{a}_{8}$ is a rational number multiplied by $\cos 45^{\circ}=\sqrt{2} / 2$.

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Fig. 47

Therefore, we have $x=y=z \sqrt{2}$, where $x, z \in \mathbf{Q}$, from which it follows that $x=0$ and $a_{5}=-a_{1}$. Similarly, it can be shown that

$$a_{6}=-a_{2}, a_{7}=-a_{3}, a_{8}=-a_{4}$$

Thus, we obtain

$$A_{1} A_{2}=A_{5} A_{6}, A_{2} A_{8}=A_{6} A_{7}, A_{8} A_{4}=A_{7} A_{8}, A_{4} A_{5}=A_{8} A_{1}$$

which is what we needed to prove.