During a partial solar eclipse, when the apparent diameter of the Moon and the Sun was the same, at the maximum moment, the edge of the moon disk coincided with the center of the sun disk. What was the percentage of the solar eclipse?
Solution. Let the radius of the circles be denoted by $R$, the centers by $O_{1}, O_{2}$, and the intersection points of the two circles by $M_{1}$ and $M_{2}$. The covered area consists of the sector $O_{1} M_{1} M_{2}$ and the area of the segment $M_{1} O_{1}$, doubled (the shaded regions in the diagram).

![](https://cdn.mathpix.com/cropped/2024_05_02_87ff144a1ea58ae01009g-1.jpg?height=336&width=417&top_left_y=304&top_left_x=843)

In the equilateral triangle $O_{1} M_{1} O_{2}$, $\angle M_{1} O_{1} O_{2} = 60^{\circ}$. The area of the sector $O_{1} M_{1} M_{2}$ is $\frac{1}{3}$ of the area of a circle with radius $R$, i.e., $T_{1} = \frac{R^{2} \pi}{3}$.

The combined area of the two segments is:

$$
T_{2} = 2\left(\frac{R^{2} \pi}{6} - \frac{R^{2} \sqrt{3}}{4}\right) = R^{2}\left(\frac{2 \pi - 3 \sqrt{3}}{6}\right)
$$

Thus,

$$
T_{1} + T_{2} = R^{2}\left(\frac{\pi}{3} + \frac{2 \pi - 3 \sqrt{3}}{6}\right) = R^{2}\left(\frac{4 \pi - 3 \sqrt{3}}{6}\right)
$$

The ratio of the covered area is:

$$
R^{2}\left(\frac{4 \pi - 3 \sqrt{3}}{6}\right) : R^{2} \pi = \frac{4 \pi - 3 \sqrt{3}}{6 \pi} \approx 0.391
$$

The Moon approximately covered 39% of the Sun.