A strange side length

Let the angle between 2 radii joining the centre of the circle to 2 adjacent vertices of the polygon be x.

Applying the cosine rule we get: $(side length)^{2} = 1^{2} + 1^{2} - 2 (1) (1) cos(x)$

$(\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}})^{2}$ = 2 - 2 cos(x)

$(2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}})$ = 2 - 2 cos (x)

2 cos (x) = $(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}})$

Squaring both sides we get:

4 $cos^{2} (x)$ = $(2+\sqrt{2+\sqrt{2+\sqrt{3}}})$

4 $cos^{2} (x)$ - 2 = $(\sqrt{2+\sqrt{2+\sqrt{3}}})$

2 cos (2x) = $(\sqrt{2+\sqrt{2+\sqrt{3}}})$

Squaring both sides again we get:

4 $cos^{2} (2x)$ = $2+\sqrt{2+\sqrt{3}}$

4 $cos^{2} (2x)$ - 2 = $\sqrt{2+\sqrt{3}}$

2 cos (4x) = $\sqrt{2+\sqrt{3}}$

Squaring both sides again we get:

4 $cos^{2} (4x)$ = $2+\sqrt{3}$

4 $cos^{2} (4x)$ - 2 = $\sqrt{3}$

2 cos (8x) = $\sqrt{3}$

cos (8x) = $\frac{1}{2} \sqrt{3}$

8x = 30 degrees

x = 30/8 degrees

Number of sides = $\frac{360}{30/8}$ = $\frac{(360)(8)}{30}$ = 96 sides

$Answer = \boxed{96}$