Solution:
The area of sector $O P_{2} P_{4}$ is one sixth the area of the circle because its angle is $60^{\circ}$. The desired area is just that of the sector subtracted by the area of equilateral triangle $O X Y$.

Note that the altitude of this triangle is the distance from $O$ to $P_{1} P_{5}$, which is $\frac{1}{2}$. Thus, the side length of the triangle is $\frac{\sqrt{3}}{3}$, implying that the area is $\frac{\sqrt{3}}{12}$. Thus, we find that $A=\frac{\pi}{6}-\frac{\sqrt{3}}{12}$. Thus, $120 A=20 \pi-10 \sqrt{3} \approx 62.8-17.3$, which has floor 45 .