On the solar eclipse of May 20, 1966, the so-called greatest phase for observers in Battonya will be 0.766, and for those in Sopron, it will be 0.678. (By this, we mean that at the time of the greatest obscuration, the part of the solar disk that is not visible is the given ratio of the diameter of the solar disk that lies along the common axis of symmetry of the apparent solar and lunar disks.) Calculate the ratio of the covered area of the solar disk to its total area for these situations (considering the areas as they would appear in a suitable photograph). The angular diameters of the Sun and Moon on this day are $31^{\prime} 39^{\prime \prime}$ and $31^{\prime} 6^{\prime \prime}$, respectively.
We need to calculate the area of the common part of the circles representing the Sun and the Moon, given the two radii and the distance between their centers. Indeed, the radii can be considered proportional to the given angles of view, because 1) in the photograph, the radius of the circle is proportional to the tangent of half the angle of view, 2) for any acute angle, its radian measure lies between its sine and tangent, 3) the sine and tangent of the given angles are equal to four decimal places, so the same applies to their radian measures; furthermore, denoting the centers of the two circles by $O_{N}, O_{H}$, the radii by $R_{N}, R_{H}$, the points where the symmetry axis $O_{N} O_{H}$ intersects the common part by $S_{N}, S_{H}$, and the ratio of the maximum phase by $f$,

$$
S_{N} S_{H}=f \cdot 2 R_{N}=S_{N} O_{N}-O_{H} O_{N}+O_{H} S_{H}=R_{N}+R_{H}-O_{N} O_{H}
$$

thus the distance between the centers can indeed be calculated from the given data:

$$
O_{N} O_{H}=R_{H}-(2 f-1) R_{N}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c1ec22eadf38b56d8673g-1.jpg?height=653&width=572&top_left_y=709&top_left_x=473)

Battonya, $f=0.766$

![](https://cdn.mathpix.com/cropped/2024_05_02_c1ec22eadf38b56d8673g-1.jpg?height=735&width=569&top_left_y=695&top_left_x=1057)

Sopron, $f=0.678$

The common part is divided into two circular segments by the common chord $M_{1} M_{2}$ of the circles. The central angle corresponding to the common chord in each circle is twice the angle $\varphi_{N}$, or $\varphi_{H}$, in the triangle $O_{N} O_{H} M_{1}$. With these angles, the area of the common part, with the angles in radians, is:

$$
t=\frac{R_{N}^{2}}{2}\left(2 \varphi_{N}-\sin 2 \varphi_{N}\right)+\frac{R_{H}^{2}}{2}\left(2 \varphi_{H}-\sin 2 \varphi_{H}\right)
$$

and the angles can be calculated using the cosine rule:

$$
\begin{gathered}
\cos \varphi_{N}=\frac{R_{N}^{2}+\left(O_{N} O_{H}-R_{H}\right)\left(O_{N} O_{H}+R_{H}\right)}{2 R_{N} \cdot O_{N} O_{H}}= \\
=\frac{R_{N}-(2 f-1)\left[2 R_{H}-(2 f-1) R_{N}\right]}{2 \cdot O_{N} O_{H}}=\frac{\left[f^{2}+(1-f)^{2}\right] R_{N}-(2 f-1) R_{H}}{O_{N} O_{H}}
\end{gathered}
$$

and similarly

$$
\cos \varphi_{H}=\frac{1}{R_{H} \cdot O_{N} O_{H}}\left[R_{H}^{2}-(2 f-1) R_{H} R_{N}-2 f(1-f) R_{N}^{2}\right]
$$

The sought ratio is:

$$
k=\frac{t}{\pi R_{N}^{2}}=\frac{1}{2 \pi}\left[\left(2 \varphi_{N}-\sin 2 \varphi_{N}\right)+\left(\frac{R_{H}}{R_{N}}\right)^{2} \cdot 2 \varphi_{H}-\sin 2 \varphi_{H}\right)
$$

The angular size of the view in arcseconds for the Sun is 1899, and for the Moon 1866, which we will consider as the measures of $R_{N}$ and $R_{H}$ in a suitable unit. Then the length of $O_{N} O_{H}$ is 855.7 units in the case of Battonya, and 1190 units in the case of Sopron, and $\left(\frac{R_{H}}{R_{N}}\right)^{2}=0.9655$.[^0]

In the case of Battonya, $f=0.766, 1-f=0.234, 2 f-1=0.532$, so the angles and the value of $x-\sin x$ for the double angles are

$$
\begin{gathered}
\cos \varphi_{N}=\frac{(0.5868+0.0548) 1899-0.532 \cdot 1866}{855.7}=\frac{1219-993}{855.7} \\
\varphi_{N}=74.69^{\circ}=1.304, \quad 2 \varphi_{N}-\sin 2 \varphi_{N}=2.608-0.509=2.099
\end{gathered}
$$

similarly

$$
\varphi_{H}=79.02^{\circ}=1.379, \quad 2 \varphi_{H}-\sin 2 \varphi_{H}=2.384
$$

and thus the ratio is ²

$$
k=\frac{2.099+0.9655 \cdot 2.384}{6.283}=0.70
$$

In the case of Sopron, the intermediate results $\varphi_{N}=70.06^{\circ}$ and $\varphi_{H}=73.10^{\circ}$ yield $k=0.59$.

Péter Szeredi (Budapest, Rákóczi F. Gymnasium, 3rd grade) Sándor Kádas (Budapest, József A. Gymnasium, 3rd grade)[^1]

[^0]:    ${ }^{1}$ We wrote 4 significant digits, although the given $f$-values have only 3 significant digits, and therefore the calculated measures cannot have more than 3 correct digits; however, early rounding could lead to cumulative errors.

[^1]:    ${ }^{2}$ Rounded to 2 significant digits.