Let $A_1A_2A_3A_4A_5A_6A_7A_8$ be a right octagon with center $O$ and $\lambda_1$,$\lambda_2$, $\lambda_3$, $\lambda_4$ be some rational numbers for which:
$\lambda_1 \overrightarrow{OA_1}+\lambda_2 \overrightarrow{OA_2}+\lambda_3 \overrightarrow{OA_3}+\lambda_4 \overrightarrow{OA_4} =\overrightarrow{o}$.
Prove that $\lambda_1=\lambda_2=\lambda_3=\lambda_4=0$.
Pinko wrote:
Let $A_1A_2A_3A_4A_5A_6A_7A_8$ be a $ \color{red} \text{ regular }$ octagon with center $O$ and $\lambda_1$,$\lambda_2$, $\lambda_3$, $\lambda_4$ be some rational numbers for which:
$\lambda_1 \overrightarrow{OA_1}+\lambda_2 \overrightarrow{OA_2}+\lambda_3 \overrightarrow{OA_3}+\lambda_4 \overrightarrow{OA_4} =\overrightarrow{o}$.
Prove that $\lambda_1=\lambda_2=\lambda_3=\lambda_4=0$.
we have $\overrightarrow{OA_1},\overrightarrow{OA_3}$ are perpendicular then express the others in terms of them and use the fact that $a\sqrt{2}+b=0\implies a=0 $ and $b=0$ if a;b are rationals