Two circular disks, each with a radius of 2 units, lie on top of each other in such a way that together they cover an area of $6 \pi$ square units. Let's confine the distance between the centers of the two disks between two fractions, where the denominators are 100, the numerators are integers, and the difference between the numerators is 1.
Let's denote the centers of the circles by $A, B$, their intersection points by $C, D$, and the angle $ACB$ by $x$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e45e86805be695adb467g-1.jpg?height=655&width=865&top_left_y=220&top_left_x=619)

The sought distance

$$
AB = 2 AC \sin \frac{x}{2} = 4 \sin \frac{x}{2}
$$

The area of the common part of the two circles is

$$
T = AC^2 (\pi - x) - 2 T_{ABC} = 4 (\pi - x) - 4 \sin x
$$

This is how much smaller the area covered together is compared to twice the area of one circle:

$$
\begin{gathered}
4 (\pi - x) - 4 \sin x = 8 \pi - 6 \pi \\
x = \frac{\pi}{2} - \sin x
\end{gathered}
$$

It is easy to see that $ABx_8 \quad \text{and} \quad 1.5708 - 0.7446 = 0.8262 < x_9
$$

Thus, the root lies between these two values, somewhere where $\sin x_{10} = 0.74$. Here (based on the tables of cyclometric functions) $x_{10} = 0.8331$, which is smaller at $\left(\pi / 2 - \sin x_{10}\right)$ than $x_{9}$ at $\left(\pi / 2 - \sin x_{9}\right)$. Therefore,

$$
x_8 = 0.83 < x < x_{10} = 0.8331
$$

It is not advisable to directly read the appropriate bounds for $\sin x / 2$ from the table; a more accurate estimate is provided by the

$$
\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}
$$

relationship. From this, we get that

$$
\begin{gathered}
2.600 = 8(1 - 0.675) < 16 \sin^2 \frac{x}{2} < 8 \left(1 - \sqrt{1 - 0.74^2}\right) < 8(1 - 0.672) = 2.624 \\
1.612 < AB < 1.620
\end{gathered}
$$

Thus, the two sought fractions are 161/100 and $162 / 100$.