A certain gas balloon with a diameter of $20 \mathrm{~m}$ is seen at an angle of $0^{\circ} 36^{\prime} 15.3^{\prime \prime}$ above us; how high did the gas balloon fly above our heads, how much illuminating gas was needed to fill the gas balloon, and what weight could have just prevented the gas balloon from rising, if the weight of $1 \mathrm{~m}^{3}$ of gas $=0.6 \mathrm{kg}$, the weight of $1 \mathrm{~m}^{3}$ of air $=1.293 \mathrm{kg}$, and if the weight of the balloon material and all other difficulties is taken as $3 \mathrm{m}^{3}$.
The distance of the center of the air balloon from the Earth is

$$
t=\frac{r}{\sin \frac{a}{2}}
$$

thus, the distance of the point of the air balloon closest to the Earth is $t_{1}=t-r$; substituting the given values, $t_{1}=1886.7 \text{ m}$. The volume of the air balloon is

$$
v=\frac{4}{3} r^{3} \pi=4188.78 \text{ m}^{3}
$$

The weight of the filled air balloon, including equipment, is:

$$
s=4188.78 \times 0.6 + 300 = 2813.27 \text{ kg}
$$

The weight of the same volume of air is $s_{1}=5416.09 \text{ kg}$, and thus the ascent can be prevented by a load equal to the weight difference $s_{1}-s$; $s_{1}-s=2602.82 \text{ kg}$.

(Schölz Károly, real gymnasium 8th grade student, Deva.)

The problem was also solved by: Bálint B., Eislitzer Gy., Fekete J., Fischer O., Friedmann B., Goldstein Zs., Grünhut B., Hofbauer E., Iványi B., Kántor N., Klein M., Kunsch M., Lichtenberg S., Riesz F., Schiffer H., Szabó I.