A pentagon and an inscribed square

For simplicity, let $a=1$ . Then, the length of all diagonals of the regular pentagon is the golden ratio $\varphi =\frac{1+\sqrt{5}}{2}$ . If we extend the sides $AB$ and $DC$ of the pentagon, a golden triangle is formed, $\triangle NBC$ .
The length of the base of this triangle is $BC=1$ , hence its other two sides have length $NB=NC=\varphi$ .

$EN$ is an axis of symmetry of the compound shape, thus, $M$ is the midpoint of $BC$ and $\angle M=90{}^\circ$ .
By Pythagorean theorem, on $\triangle EMB$ and $\triangle NMB$ , we get $EM=NM=\sqrt{{{\varphi }^{2}}-\frac{1}{4}}$ $\Rightarrow EN=2\sqrt{{{\varphi }^{2}}-\frac{1}{4}}=\sqrt{4{{\varphi }^{2}}-1}$ Finally, triangles $\triangle ELK$ and $\triangle EAD$ are similar, as well as $\triangle ALI$ and $\triangle AEN$ , hence,

$\begin{aligned} & \frac{b}{\varphi }=\frac{LK}{AD}=\frac{EL}{EA}=\frac{EA-AL}{EA}=1-\frac{AL}{EA}=1-\frac{b}{EN}=1-\frac{b}{\sqrt{4{{\varphi }^{2}}-1}} \\ & \Rightarrow \frac{b}{\varphi }+\frac{b}{\sqrt{4{{\varphi }^{2}}-1}}=1 \\ & \Rightarrow b=\frac{1}{\frac{1}{\varphi }+\frac{1}{\sqrt{4{{\varphi }^{2}}-1}}} \\ & \Rightarrow b\approx 1.0604974 \\ \end{aligned}$ For the answer,
$\left\lfloor {{10}^{5}}r \right\rfloor =\left\lfloor {{10}^{5}}\frac{b}{1} \right\rfloor =\boxed{106049}.$