Triangles and Pentagon inside Pentagon

Divide the blue pentagon into 5 equal parts, as shown. Note that $72^{\circ} = \frac{2\pi}{5}$ and $54^{\circ} = \frac{3\pi}{10}$ . This means that $\frac{\tan 72^{\circ}}{\tan 54^{\circ}} = \tan \frac{2\pi}{5} \times \tan \left(\frac{\pi}{2} -\frac{3\pi}{10} \right) = \tan \frac{2\pi}{5} \tan \frac{\pi}{5} = \sqrt{5}$ .

If the blue pentagon is of length $a$ , then the area of the each of the 5 parts (of blue pentagon) is $a\times a \tan 54^{\circ}$ , while the area of each of the 5 brown triangles is $a\times a \tan 72^{\circ}$ Then $\displaystyle \left \lfloor \frac{1000N}{M}\right \rfloor = \left \lfloor \frac{1000\tan 72^{\circ} }{\tan 54^{\circ}}\right \rfloor=\left \lfloor 1000\sqrt{5}\right \rfloor = 2236$ .

See the sketches. We look at one tenth of both areas M and N. DC is their common base. So the ratio of areas will be same as the ratios of heights. For * triangle * , BC is the height. For * pentagon * it is CO.

WLOG we take BM=1, side common to two triangles, blue MBO of pentagon and gray ABM of triangles.
In triangle MBO angle BOM=36. So MO = $\dfrac{BM}{ Tan36}.$
In triangle ABM, angle MAB =18. So AM= $\dfrac{BM}{ Tan18}.$ .
So N/M=AM/MO=Tan36/Tan18 =2.236.
So the answer =2236.