Square in Square 2

Since $ABCD$ has an area of $25$ , $AB = 5$ , and since $JKLM$ has an area of $18$ , $JM = 3\sqrt{2}$ . Let $x = BE = AI$ and $y = AM = BJ$ .

Since $\triangle AMB$ is a right triangle, by Pythagorean's Theorem $y^2 + (3\sqrt{2} + y)^2 = 5^2$ which solves to $y = \frac{\sqrt{2}}{2}$ , and since $\triangle HAB \sim \triangle AMB$ by the AA similarity postulate, $\frac{x}{5} = \frac{y}{3\sqrt{2} + y}$ so that with $y = \frac{\sqrt{2}}{2}$ this solves to $x = \frac{5}{7}$ .

Therefore the ratio $\frac{AE}{BE} = \frac{5 - \frac{5}{7}}{\frac{5}{7}} = \boxed{6}$ .

We notice that the green and pink triangles are congruent. Therefore the area of one of these triangles is $A_\triangle = \dfrac {[ABCD]-[JKLM]}4 = \dfrac {25-18}4 = \dfrac 74$ . Let $AM = a$ and $BM=b$ ; then $A_\triangle = \frac 12 ab = \frac 74 \implies ab = \frac 72$ . Since the area of square $ABCD$ is $25$ , $\implies AB = 5$ . By Pythagorean theorem we have:

$\begin{aligned} a^2 + b^2 & = 25 & \small \color{#3D99F6} \text{Since }ab = \frac 72 \\ a^2 + \left(\frac 7{2a}\right)^2 & = 25 & \small \color{#3D99F6} \text{Multiply both sides by }4a^2 \\ 4a^4 + 49 & = 100a^2 & \small \color{#3D99F6} \text{Rearrange} \\ 4a^4 - 100a^2 + 49 & = 0 \\ (2a^2-1)(2a^2-49) & = 0 \end{aligned}$

Note that both $a$ and $b$ satisfy the equation. $\implies \begin{cases} a = \frac 1{\sqrt 2} \\ b = \frac 7{\sqrt 2} \end{cases}$ , since $b > a$ . Now we note that $\triangle AMB$ and $\triangle EJB$ are similar. Therefore,

$\begin{aligned} \frac {BE}{BJ} & = \frac {AB}{BM} \\ \frac {BE}{b-JM} & = \frac 5b & \small \color{#3D99F6} \text{Since }[JKLM] = 18, \implies JM = 3\sqrt 2 \\ \implies BE & = \frac {5\left(\frac 7{\sqrt 2}-3\sqrt 2\right)}{\frac 7{\sqrt 2}} = \frac 57 \end{aligned}$

Therefore, $\dfrac {AE}{BE} = \dfrac {AB-BE}{BE} = \dfrac 5{\frac 57} - 1 = \boxed 6$ .

$\text{ah}$ , $\text{df}$ and $\text{bi}$ are two letter variable names.

$\text{ah}=y=5-\frac{5 x}{\delta }$

$\text{df}=y=\frac{\delta x}{5}$

$\text{bi}=y=(5-\delta )+\frac{\delta x}{5}$

$m=\{x,y\}\text{/.}\, \text{Solve}[\text{ah}\land \text{bi}][[1]] \Rightarrow \left\{\frac{5 \delta ^2}{\delta ^2+25},\frac{5 \left(\delta ^2-5 \delta +25\right)}{\delta ^2+25}\right\}$

$l=\{x,y\}\text{/.}\, \text{Solve}[\text{ah}\land \text{df}][[1]] \Rightarrow \left\{\frac{25 \delta }{\delta ^2+25},\frac{5 \delta ^2}{\delta ^2+25}\right\}$

$\delta =(\delta \text{/.}\, \text{Solve}[\text{SquaredEuclideanDistance}[m,l]=18])[[1]] \Rightarrow \frac57$

$\frac{5-\delta }{\delta }\Rightarrow 6$

Of course, the simplier answer is $-\frac{s \left(\sqrt{2-s^2}+s\right)}{s^2-1}$ with $s$ being set to $\sqrt{\frac{18}{25}}$ giving $6$ .

With central square areas of 1, 5, 10 and 18, rational results of $\frac13$ , $1$ , $2$ and $6$ are availed, respectively.