4 in 5

Consider a general case of a quadrilateral $FGHI$ inscribed in the unit regular pentagon $ABCDE$ . Let $FG=IF = s_1$ , $FG \perp IF$ , $FG \perp GH$ , $\angle IFJ = \theta$ and $GH=s_2$ (note that $s_2>s_1$ for $\theta = 5^\circ$ as shown in the figure). Then the various angles in terms of $\theta$ are as shown. If $DF=a$ , then $EF = 1-a$ and $FJ = 1+2a \sin 18^\circ$ . By sine rule , we have:

$\frac {1-a}{\sin (54^\circ + \theta)} = \frac {s_1}{\sin 108^\circ} = \frac {1+2a\sin 18^\circ}{\sin (72^\circ -\theta)} \\ \implies s_1 = \frac {\sin 108^\circ (2\sin 18^\circ + 1)}{\sin(72^\circ - \theta) + 2\sin 18^\circ \sin(54^\circ + \theta)}$

Let $EG=b$ . By sine rule, $b = \dfrac {s_1 \sin (18^\circ - \theta)}{\sin 108^\circ}$ and

$\begin{aligned} \frac {s_2}{\sin 108^\circ} & = \frac {1-b}{\sin(36^\circ + \theta)} \\ s_2 & = \frac {\sin 108^\circ - s_1 \sin(18^\circ - \theta)}{\sin(36^\circ + \theta)} \end{aligned}$

When quadrilateral $FGHI$ is a square $s_2 = s_1 = s$ , then

$\begin{aligned} \implies s_1 & = \frac {\sin 108^\circ}{\sin(36^\circ + \theta) + \sin(18^\circ - \theta)} \\ \frac {\sin 108^\circ (2\sin 18^\circ + 1)}{\sin(72^\circ - \theta) + 2\sin 18^\circ \sin(54^\circ + \theta)} & = \frac {\sin 108^\circ}{\sin(36^\circ + \theta) + \sin(18^\circ - \theta)} \\ (2\sin 18^\circ + 1)(\sin(36^\circ + \theta) + \sin(18^\circ - \theta)) & = \sin(72^\circ - \theta) + 2\sin 18^\circ \sin(54^\circ + \theta) \end{aligned}$

Solving the equation above, we get $\theta = 0^\circ$ . And $s = \dfrac {\sin 108^\circ}{\sin 36^\circ + \sin 18^\circ} \approx 1.060497473 \implies \lfloor 10000s \rfloor = \boxed{10604}$ .

We place the regular pentagon $ABCDE$ on the plane so that $CD$ lie on the x-axis and $a$ on the positive part of y-axis. The four vertices $P$ , $Q$ , $R$ , $S$ of the square will lay on four of the pentagon’s sides. Let them be $CD$ , $AB$ , $AE$ , and $ED$ respectively. From points $Q$ , $R$ and $S$ we drop perpendicular to the axes, thus forming three congruent right triangles $\triangle PQF$ , $\triangle RQL$ and $\triangle SPG$ . Let $\angle QPF=\angle QRL=\angle PSG=\theta$ . Denote by $x$ the x-coordinate of $P$ / and by $s$ the side length of the square. Then, we have

$\begin{aligned} & P\left( x,\ 0 \right) \\ & Q\left( x+s\cdot \cos \theta ,\ s\cdot \sin \theta \right) \\ & R\left( x+s\cdot \cos \theta -s\cdot \sin \theta ,\ s\cdot \sin \theta +s\cdot \cos \theta \right) \\ & S\left( x-s\cdot \sin \theta ,\ s\cdot \cos \theta \right) \\ \end{aligned}$ The y-coordinate of vertex $A$ is the height of the pentagon, thus $A\left( 0,\ \frac{\sqrt{5+2\sqrt{5}}}{2} \right)$ .

The equations of line $AB$ is

$y=-\left( \tan 36{}^\circ \right)x+\frac{\sqrt{5+2\sqrt{5}}}{2}\Leftrightarrow y=-\sqrt{5-2\sqrt{5}}x+\frac{\sqrt{5+2\sqrt{5}}}{2}$

The equations of line $AE$ is

$y=\left( \tan 36{}^\circ \right)x+\frac{\sqrt{5+2\sqrt{5}}}{2}\Leftrightarrow y=\sqrt{5-2\sqrt{5}}x+\frac{\sqrt{5+2\sqrt{5}}}{2}$

The equations of line $DE$ is

$y=\left( \tan 108{}^\circ \right)\left( x+\frac{1}{2} \right)x\Leftrightarrow y=-\sqrt{5+2\sqrt{5}}\left( x+\frac{1}{2} \right)$

Since $Q\in AB$ , $R\in AE$ and $S\in DE$ , we get the system $\left\{ \begin{aligned} s\cdot \sin \theta & =-\sqrt{5-2\sqrt{5}}\left( x+s\cdot \cos \theta \right)+\frac{\sqrt{5+2\sqrt{5}}}{2} \\ s\cdot \sin \theta +s\cdot \cos \theta & =\sqrt{5-2\sqrt{5}}\left( x-s\cdot \sin \theta +s\cdot \cos \theta \right)+\frac{\sqrt{5+2\sqrt{5}}}{2} \\ s\cdot \cos \theta & =-\sqrt{5+2\sqrt{5}}\left( x-s\cdot \sin \theta +\frac{1}{2} \right) \\ \end{aligned} \right.$ Settting $y=s\cdot \sin \theta$ and $z=s\cdot \cos \theta$ , we come to a linear 3x3 system:

$\left\{ \begin{aligned} y & =-\sqrt{5-2\sqrt{5}}\left( x+z \right)+\frac{\sqrt{5+2\sqrt{5}}}{2} \\ y+z & =\sqrt{5-2\sqrt{5}}\left( x-y+z \right)+\frac{\sqrt{5+2\sqrt{5}}}{2} \\ z & =-\sqrt{5+2\sqrt{5}}\left( x-y+\frac{1}{2} \right) \\ \end{aligned} \right.$ which solves to

$x\approx 0.402113032590307,\text{ }y\approx 1.00859303213020,\text{ }z\approx 0.327711741622372$ Now we have,
$\begin{aligned} & {{s}^{2}} ={{s}^{2}}\cdot \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right) \\ & ={{\left( s\cdot \sin \theta \right)}^{2}}+{{\left( s\cdot \cos \theta \right)}^{2}} \\ & ={{y}^{2}}+{{z}^{2}} \\ & \approx {{\left( 1.00859303213020 \right)}^{2}}+{{\left( 0.327711741622372 \right)}^{2}} \\ & =1.12465489005875 \\ \end{aligned}$ Hence, $s=\sqrt{1.12465489005875}\approx 1.06049747291484$ .

For the answer, $\left\lfloor 10000s \right\rfloor =\boxed{10604}$ .

While I can't prove it, I'm pretty certain the square must be oriented symmetrically to the pentagon, i.e. a side of the square must be parallel to a side of the pentagon, as in the figure. In fact, I suspect this is the only way to inscribe a square. If someone knows, I'd be grateful to learn more.

Symmetry implies $\triangle JBH$ is isosceles, and thus $JB = BH.$ Let $LB$ bisect $\angle ABC.$

Since $\angle B = 108^\circ,\enspace\angle SHB = 36^\circ ,\enspace\angle FHC = (180^\circ - 90^\circ -36^\circ) = 54^\circ.$

Similarly $\angle HFC = (180^\circ - 54^\circ - 108^\circ) = 18^\circ.$

Let $s = JH$ and $x = JB.$

Using the law of sines on $\triangle HFC,\hspace{1cm} \dfrac{s}{\sin 108^\circ} = \dfrac{1 - x}{\sin 18^\circ}.$

From the right $\triangle JLB,\enspace s = 2x\cdot\cos 36^\circ$

Substituting for $s,\hspace{1cm}\dfrac{2x\cos 36^\circ}{\sin 108^\circ} = \dfrac{1-x}{\sin 18^\circ}$

Let $k = \dfrac{\cos 36^\circ\sin 18^\circ}{\sin 108^\circ} \implies x = \dfrac{1}{2k+1}$

Evaluating , $\enspace x = \frac{1}{2}(5 - \sqrt{5} - 2\sqrt{5 - 2\sqrt{5}}) \approx 0.655423$

Evaluating , $\enspace s = \sqrt{5} - \sqrt{\frac{1}{2}(5 - \sqrt{5})} \approx 1.060497$

So the requested answer is $\boxed {10604}$