Hexagon inside pentagon

Starting with a unit pentagon centered at the origin in standard orientation, its vertices $\{P_i \}$ can be computed, as well as the normals to its sides $\{n_i \}$ . In particular , if the radius of the pentagon is $R$ then,

$R = \dfrac{1}{2 \sin \frac{\pi}{5} }$

Focusing on the right half of the pentagon, we take

$P_1= R ( \sin \frac{\pi}{5}, - \cos \frac{\pi}{5} )$

$P_2 = R ( \sin \frac{3 \pi}{5} , - \cos \frac{3 \pi}{5} )$

the normals to the right sides of the pentagon are

$n_1 = ( \sin \frac{2 \pi}{5} , - \cos \frac{ 2 \pi}{5} )$

$n_2 = ( \sin \frac{4 \pi}{5} , - \cos \frac{ 4 \pi}{5} )$

The hexagon is positioned as shown in the diagram above. Its center is at $(0, Y)$ , and its side length is $s$ .

The right point on the hexagon has the following coordinates

$Q_1 = ( s, Y )$

and the upper right point has coordinates

$Q_2 = ( \frac{1}{2} s , Y + \dfrac{\sqrt{3}}{2} s )$

Imposing the condition that these two points lie on the sides of the pentagon, yields

$n_1 . (Q_1 - P_1) = 0$ , and

$n_2 . (Q_2 - P_2) = 0$

these equations are linear equations in $s$ and $Y$ , and can be solved readily to yield,

$s \approx 0.71458037072$ and $Y \approx -0.02778046$

Hence, the answer is $\lfloor 10^4(0.71458037072) \rfloor = \boxed{7145}$