Ratio of Areas!

A square with sides parallel to the $x-$ and $y-$ axes can be broken up into $8$ congruent triangles so that $4$ of those triangles make up a square with slanted sides.

Therefore, the ratio of the area of one square in $A_1$ to the area of its connected square in $A_2$ is $\frac{8}{4} = 2$ .

Since each square in $A_1$ has one connected square in $A_2$ , this ratio is maintained throughout the whole diagram, so $\frac{A_1}{A_2}$ is also $\boxed{2}$ .

Let the area of the largest square with sides parallel to the $x$ - and $y$ -axes be $a_1$ , the second largest be $a_2$ , the next be $a_3$ and so on. Similarly, $b_1$ , $b_2$ , $b_3$ , $\cdots$ for the squares with slanted sides. Due to symmetry, we have $\dfrac {a_{n+1}}{a_n} = \dfrac {b_{n+1}}{b_n} = k$ , where $k$ is a constant less than 1. Then

$\begin{cases} A_1 = a_1 + a_2 + a_3 + a_4 + \cdots & = a_1 \left(1 + k + k^2 + k^3 + \cdots \right) \\ A_2 = b_1 + b_2 + b_3 + b_4 + \cdots & = b_1 \left(1 + k + k^2 + k^3 + \cdots \right) \end{cases}$

Therefore, $\dfrac {A_1}{A_2} = \dfrac {a_1}{b_2} = \dfrac {a^2}{\left(\frac a{\sqrt 2} \right)^2} = \boxed 2$ .