Cover a rectangle

Let $ZY=1$ so the aspect ratio is $WZ=a$ . Call $AB=k$ so $AC=ak$ and call $BY=y$ so $BX=a-y$ .

By similar triangles $YC=\frac{k}{y}$ and $DZ=\frac{k}{y}(a-y)=\frac{ak}{y}-k$ . Subtracting gives $CZ=2k-\frac{ak}{y}$

Similar triangles again gives $AY=2ky-ak=k(2y-a)$ .

Since $AY+YC=AC$ we have $2ky-ak+\frac{k}{y}=ak$ or, since $k \ne 0$ simplify to $2y^{2}-2ay+1=0$

The positive solution to this quadratic is $y=\frac{a+\sqrt{a^2-2}}{2}$ . This means $AY$ can be written as $k\sqrt{a^{2}-2}$ .

Now the Pythagorean theorem on $\triangle BAY$ gives $k^{2}+[k\sqrt{a^{2}-2}]^{2}=y^{2}$ .

Solve to get $k=\frac{y}{\sqrt{a^{2}-1}}$

A final substitution of $k$ for $y$ gives a formula for $k$ in terms of the aspect ratio:

$k=\frac{a+\sqrt{a^{2}-2}}{2\sqrt{a^{2}-1}}$

I decided to let Wolfram|Alpha solve for $k=1-\frac{1}{1000000}$ and it gave me an interesting exact form which is approimately $a=\boxed{18.8296}$