Before a total solar eclipse - according to calculations - at 9:13, the distance between the centers of the Sun and Moon disks was $4 \frac{7}{10}$ lunar diameters. Both disks had the same apparent size and both were moving in the same apparent direction, from west to east. The Moon traveled $1 \frac{1}{16}$ lunar diameters per hour along its orbit, while the Sun moved by only $\frac{1}{12}$ of a lunar diameter in the same time.

a) At what time did the centers of the two disks coincide (total eclipse)?

b) When did the eclipse begin and

c) when did it end?
I. Solution: Our diagram shows the situation at 9:15, so $O_{1} O_{2}=4 \frac{7}{10}$ diameters.

![](https://cdn.mathpix.com/cropped/2024_05_02_afb11f43255000a942afg-1.jpg?height=158&width=722&top_left_y=276&top_left_x=688)

a) The eclipse is total when $O_{1}$ coincides with $O_{2}$. Let the number of hours that have elapsed until this happens be denoted by $x$. According to the problem,

$$
\frac{17}{16} x-\frac{1}{12} x=\frac{47}{10}
$$

from which

$$
x=\frac{47 \cdot 48}{10 \cdot 47}=4.8 \text{ hours }=4 \text{ hours } 48 \text{ minutes }
$$

b) The eclipse begins when $B_{1}$ coincides with $A_{2}$, that is,

$$
\frac{17}{16} y-\frac{1}{12} y=\frac{37}{10}
$$

from which

$$
y=\frac{37 \cdot 48}{10 \cdot 47}=3 \text{ hours } 46 \frac{34}{47} \text{ minutes }
$$

c) The eclipse ends when $A_{1}$ coincides with $B_{2}$, that is,

$$
\frac{17}{16} z-\frac{1}{12} z=\frac{57}{10}
$$

from which

$$
z=\frac{57 \cdot 48}{10 \cdot 47}=5 \text{ hours } 49 \frac{13}{37} \text{ minutes }
$$

Adding these durations to the time point of 9:13, we get that

a) the total eclipse occurs at 14:01,

b) the eclipse began at 12:59 \(\frac{34}{47}\) minutes,

c) the eclipse ended at 15:02 \(\frac{13}{47}\) minutes.

Géza Hartmann (Szombathely, Nagy Lajos g. II. o. t.)

II. Solution: Our problem can also be solved without equations, purely through arithmetic.

Since the Moon's speed relative to the Sun is

$$
\frac{17}{16}-\frac{1}{12}=\frac{47}{48} \text{ diameters/hour, }
$$

therefore
a) $O_{1}$ coincides with $O_{2}$ after \(\frac{47}{10} : \frac{47}{48} = 4\) hours \(48\) minutes,
b) $B_{1}$ coincides with $A_{2}$ after \(\frac{37}{10} : \frac{47}{48} = 3\) hours \(46 \frac{34}{47}\) minutes,
c) $A_{1}$ coincides with $B_{2}$ after \(\frac{57}{10} : \frac{47}{48} = 5\) hours \(49 \frac{13}{47}\) minutes.

Szabolcs Ujhelyi (Debrecen, Ref. g. II. o. t.)