Biot and Gay-Lussac, French natural scientists, in 1804 ascended in a balloon and reached a height of $6825 \mathrm{~m}$. Question: How far did the horizon extend on this occasion and how much $\mathrm{km}^{2}$ of land area could be seen from the balloon? (The radius of the Earth is $6377.4 \mathrm{~km}$. )
The height of the elevation is $h$, the summit point is $H$; from $H$ to the sphere, the length of any tangent is $t$, and the point of tangency is $T$. The height of the visible spherical segment is $m$, the radius of the Earth is $r$, and its center is $O$.

From the right triangle HTO:

$$
t^{2}=h(h+2 R)
$$

Furthermore, since the leg is the geometric mean between the hypotenuse and its adjacent segment,

$$
r^{2}=(r-m)(r+h)
$$

from which

$$
m=\frac{r h}{r+h}
$$

and thus the sought area:

$$
f=2 r \pi m=\frac{2 r^{2} \pi h}{r+h}
$$

Substituting the given values:

$$
t=295 \mathrm{~km}, \quad f=273188 \mathrm{~km}^{2}
$$

(Frigyes Riesz, Györ.)

The problem was also solved by: Bálint B., Bauss O., Fischer O., Freund A., Friedmann B., Geist E., Goldstein Zs., Grünhut B., Hofbauer E., Kántor N., Lichtenberg S., Schiffer H., Schneider B.