A place with a geographic latitude of $46^{\circ} 22^{\prime}$, at what time does the sun rise and set on May 19th, when the sun's declination is: $\delta=19^{\circ} 52^{\prime} ; \mathrm{s}$ how fast does that place move due to the Earth's rotation?
Let the height of the sun above the horizon be denoted by $m$, its declination by $\delta$, and its altitude as perceived from the observer's location, which is equal to the geographic latitude, by $\varphi$, and its hour angle by $\omega$. Then $R-m$, $R-\varphi$, and $R-\delta$ form a spherical triangle, in which $\omega$ is the angle opposite to $R-m$. According to the second fundamental formula of spherical trigonometry, we then have: $\sin m=\sin \varphi \sin \delta+\cos \varphi \cos \delta \cos \omega$, from which

$$
\cos \omega=\frac{\sin m-\sin \varphi \sin \delta}{\cos \varphi \cos \delta}
$$

But at sunrise and sunset, the sun's height above the horizon is $m=0$, and thus from (1) we get:

$$
\cos \omega=-\operatorname{tg} \varphi \operatorname{tg} \delta
$$

or

$$
\begin{gathered}
\cos (2 R-\omega)=\operatorname{tg} \varphi \operatorname{tg} \delta \\
\log \cos (2 R-\omega)=\log \operatorname{tg} \varphi+\log \operatorname{tg} \delta \\
=0 \cdot 02073+9 \cdot 55791-10 \\
=9 \cdot 57864-10 \\
2 R-\omega=67^{\circ} 43^{\prime} 40^{\prime \prime} \\
\omega=112^{\circ} 16^{\prime} 20^{\prime \prime}
\end{gathered}
$$

Since every $15^{\circ}$ of rotation corresponds to 1 hour, we have

$$
\omega=7^{\circ} 29^{\prime} 5^{\prime \prime}
$$

which value also indicates the time of sunset; the time of sunrise is therefore

$$
-\omega=4^{\circ} 30^{\prime} 55^{\prime \prime}
$$

The speed of the place in question due to the Earth's axial rotation is

$$
\begin{gathered}
c=\frac{2 r \pi \cos \varphi}{86400} \mathrm{~ms}^{-1} \\
=\frac{40 \cdot 10^{6} \cos \varphi}{86400} \mathrm{~ms}^{-1} \\
=\frac{10^{5} \cos \varphi}{216} \mathrm{~ms}^{-1} \\
\log c=5+\log \cos \varphi-\log 216 \\
=5+9 \cdot 83887-2 \cdot 33447-10 \\
=2 \cdot 50440 \\
c=319 \cdot 45 \mathrm{~ms}^{-1}
\end{gathered}
$$

(Sándor Kugel, 8th grade student, Losoncz)

The problem was also solved by: Gergely Jorga, 8th grade student, Arad.