Pentagons Inside Pentagon

Let $AB=4a$ . Now focus on the 'star' in the middle of the big pentagon. Now, $FG=2a$ and $HG=a$ . By symmetry, $MN=KL, FK=LG$ . Consider the triangle $MHG$ , $MG=\frac{HG}{\sin 54^{\circ}}=\frac{a}{\sin 54^{\circ}}=KG$ . Then $KL = 2\times \frac{a}{\sin 54^{\circ}} -2a=MN$ .

So, $\displaystyle \left \lfloor \dfrac{1000AB}{MN} \right \rfloor = \left \lfloor \dfrac{1000\times 4a}{2a\left( \frac{1}{\sin 54^{\circ}}-1\right)} \right \rfloor =\left \lfloor 2000(\sqrt{5}+2) \right \rfloor= 8472$ .