with awang11: Placing $O$ at the origin, for any $A_i$ we must have $\vec{A_i} \cdot \vec{A_i} \in \mathbb{Z}$. Furthermore, we have $( \vec{A_i}-\vec{A_j}) \cdot (\vec{A_i}-\vec{A_j} ) \in \mathbb{Z}$ so $\vec{A_i} \cdot \vec{A_j} \in \mathbb{Q}$ for any $i,j$. Take the spiral similarity at $O$ sending $A_1$ to $(1,0)$; this scales all the dot products by $\lVert A_1 \rVert^2 \in \mathbb{Q}$ so all the dot products remain rational. Let the coordinates of $A_i$ be $(x_i, y_i)$; note $A_1 \cdot A_i =x_i \in \mathbb{Q}$, and since $\lVert A_i \rVert^2 \in \mathbb{Q}$ we must have $y_i^2 \in \mathbb{Q}$. Furthermore, for any $i,j$ we have $\vec{A_i} \cdot \vec{A_j} \in \mathbb{Q} \implies y_iy_j \in \mathbb{Q}$. Thus there exists some squarefree integer $n$ such that all the $y_i$ are rational multiples of $\sqrt{n}$. So every point $A_i$ has coordinates $(q_i,r_i\sqrt{n})$ for some $q_i,r_i \in \mathbb{Q}$, and the desired conclusion is now obvious.