Blue/Red

Label the diagram as follows, and let the side of the blue square be $b$ , the side of the red square be $1$ , and $AG$ be $p$ :

Since $\triangle AJG \sim \triangle AEI$ , $\cfrac{AG}{JG} = \cfrac{AI}{EI}$ , or $\cfrac{p}{1} = \cfrac{p + 1 + b}{b}$ , which rearranges to $p = \cfrac{b + 1}{b - 1}$ .

Since $\triangle AFH \sim \triangle FLK$ , $\cfrac{AH}{FH} = \cfrac{FK}{LK}$ , or $\cfrac{p + 1}{b} = \cfrac{\frac{1}{2}(b - 1)}{1}$ , which rearranges to $p = \frac{1}{2}b(b - 1) - 1$ .

That means $p = \cfrac{b + 1}{b - 1} = \frac{1}{2}b(b - 1) - 1$ , which rearranges to $b(b - 3)(b + 1) = 0$ , and solves to $b = 3$ for $b > 0$ .

Therefore, the ratio of the area of the blue square to the area of one of the red squares is $\cfrac{b^2}{1^2} = \boxed{9}$ .

Let the side length of the red square and blue square be $1$ and $x$ respectively, and $AH=a$ . Since $\triangle AGH$ and $\triangle AEK$ are similar, we have:

$\frac {AH}{GH} = \frac {AK}{EK} \implies a = \frac {a+1+x}x \implies a = \frac {x+1}{x-1}$

Similarly, $\triangle AFL$ and $\triangle ADJ$ are similar,

$\begin{aligned} \frac {AL}{FL} & = \frac {AJ}{DJ} \\ \frac {a+1}x & = \frac {a+1+\overline{LJ}}{1+x} = \frac {a+1+\frac {x-1}2}{1+x} \\ \frac {a+1}x & = \frac {2a+x+1}{2(x+1)} \\ 2(a+1)(x+1) & = 2ax + x^2 + x \\ 2ax + 2a + 2x + 2 & = 2ax + x^2 + x \\ 2\blue a + 2x+2 & = x^2 + x & \small \blue{\text{Note that }a = \frac {x+1}{x-1}} \\ \frac {2\blue{(x+1)}}\blue{x-1} + 2(x+1) & = x(x+1) & \small \blue{\text{Since }x+1 > 0} \\ \frac 2{x-1} & = x-2 \\ 2 & = x^2 - 3x + 2 \\ x(x-3) & = 0 & \small \blue{\text{Since }x > 0} \\ x & = \boxed 3 \end{aligned}$