A square grid paper has a distance of $1 \mathrm{~cm}$ between its parallel lines. We draw a circle with a radius of 5 cm around the intersection of two grid lines on the paper. How many regions are formed that are bounded by arcs of the circle and segments of the grid lines (and there is no segment of a grid line inside the region)?

What is the area of each of these regions?
In the network of parallel lines, 9 lines intersect each other, and 2 lines are tangent to the circle. By traversing all these lines, we count $2 \cdot 9 \cdot 2 + 2 \cdot 2 \cdot 1 = 40$ intersection points with the circle. However, some of these points coincide, where two network lines intersect the circle at the same point. This includes the 4 points where a network line is tangent to the circle, as the radius to the point is also a network line; and the 8 points where the third line in one direction and the fourth line in the other direction from the center intersect (since $3^2 + 4^2 = 5^2$). Therefore, the network lines intersect the circle at 28 distinct points, dividing the circumference into 28 arcs. Each arc bisects an elementary square of the network, so the number of resulting regions is 56.

![](https://cdn.mathpix.com/cropped/2024_05_02_442748ea3ac4df993b76g-1.jpg?height=757&width=802&top_left_y=478&top_left_x=651)

In the area calculation, it is sufficient to consider the part of the bisected elementary squares outside the circle. Due to symmetry, we only need to determine the areas $t_{1}, t_{2}, t_{3}, t_{4}$ of the regions $A B^{\prime} B, B B^{\prime} C^{\prime} C, C C^{\prime} D^{\prime} D, D E^{\prime \prime} E$ in the diagram. We need to calculate the angles $\angle AOB, \angle AOC, \angle AOD$, and $\angle AOE$, as well as the lengths of the segments $B B^{\prime}$ and $C C^{\prime}$.

\[
\begin{aligned}
& \sin \alpha_{1} = 1 / 5 \text{, so } \alpha_{1} \approx 11.54^{\circ}, \text{ and } B B^{\prime} = 5 - \sqrt{24} \approx 0.101 \\
& \sin \alpha_{2} = 2 / 5 \text{, so } \alpha_{2} \approx 23.58^{\circ}, \text{ and } C C^{\prime} = 5 - \sqrt{21} \approx 0.417 \\
& \sin \alpha_{3} = 3 / 5 \text{, so } \alpha_{3} \approx 36.87^{\circ}, \text{ and } \alpha_{4} = 90^{\circ} - \alpha_{3} \approx 53.13^{\circ}
\end{aligned}
\]

With these, the areas of the regions $A B^{\prime} B, A C^{\prime} C, A D^{\prime} D, A E^{\prime} E$ can be found as the difference between the area of a trapezoid and a sector of a circle, and $t_{2}, t_{3}, t_{4}$ can be further calculated by subtraction:

\[
\begin{gathered}
t_{1} = A B^{\prime} B = O A B^{\prime} B - O A B = \frac{10 - \sqrt{24}}{2} - \frac{5^2 \pi \alpha_{1}}{360^{\circ}} \approx \\
\approx 5 - \sqrt{6} - 0.2182 \alpha_{1} \approx 0.034 \text{ cm}^2 ; \\
t_{1} + t_{2} = A C^{\prime} C = O A C^{\prime} C - O A C = 10 - \sqrt{21} - 0.2182 \alpha_{2} \approx \\
\approx 0.273, \text{ so } t_{2} \approx 0.239 ; \\
t_{1} + t_{2} + t_{3} = 9 - 0.2182 \alpha_{3} \approx 0.956, \quad t_{3} \approx 0.683 \\
t_{1} + t_{2} + t_{3} + t_{4} + 1 = 14 - 0.2182 \alpha_{4} \approx 2.409, \quad t_{4} \approx 0.453
\end{gathered}
\]

Thus, the areas of the parts of the bisected network squares inside the circle are approximately $0.966, 0.761, 0.317$, and $0.547 \text{ cm}^2$, rounded to the nearest thousandth.