The Middle Red Square

Here is a solution, not by me:

Let $AB=1$ , $\angle D_1BB_1=\alpha$ , $BB_1=x$ ,

then $EF=BB_1 sin \alpha = x sin \alpha$

In $\triangle B_1CD$ , $sin \alpha = \frac{CD}{B_1D}$ = $\frac{1}{\sqrt{1^{2}+(1-x)^{2}}}$

$\therefore$ Area of square $EFGH = EF^{2} = x^{2} sin^{2} \alpha = \frac{x^{2}}{1^{2}+(1-x)^{2}} = \frac{1}{41}$

$20x^{2}+x-1=0$

$(5x-1)(4x+1)=0$

$\therefore x=\frac{1}{5}, \boxed{k=5}$

.

Let the side of $ABCD$ be $l$ , then $DD_1=\dfrac{l}{k}$ .

It follows that $AD_1=l \left(1-\dfrac{1}{k}\right)$ and, by the pythagorean theorem ,

$BD_1=l \sqrt{2+\dfrac{1}{k^2}-\dfrac{2}{k}}$

Since $\triangle ABD_1$ and $\triangle DD_1K$ are similar

$\dfrac{AB}{BD_1} = \dfrac{KD_1}{DD_1}$

$\dfrac{l}{\sqrt{2+\dfrac{1}{k^2}-\dfrac{2}{k}}} = \dfrac{KD_1}{\dfrac{l}{k}} \iff KD_1 = \dfrac{l}{k\sqrt{2+\dfrac{1}{k^2}-\dfrac{2}{k}}}$

Since $GH=KD_1$ , the ratio of the area of the red square to the area of $ABCD$ is

$r=\dfrac{(KD_1)^2}{l^2} = \dfrac{1}{2k^2-2k+1}$

In this case the ratio is $\dfrac{1}{41}$

$\dfrac{1}{2k^2-2k+1}=\dfrac{1}{41}$

$\begin{aligned} k^2-k-20=0 & \iff (k-5)(k+4)=0\\ & \iff k=5,k=-4 \end{aligned}$

Obviously we discard the negative solution, so the answer is $k=\boxed{5}$