\section*{Problem 5 - V1104}

In Moscow, the tallest television tower in the world is being built, which will have a height of approximately \(500 \mathrm{~m}\) upon completion.

How far can a point on the Earth's surface be from the top of the tower at most, so that it is still visible from there (without considering the refraction of light)?

Radius of the Earth \(R=6370 \mathrm{~km}\).
}

A right-angled triangle is formed, where the hypotenuse is exactly the sum of the Earth's radius and the height of the TV tower. We set \(c=6370.5 \mathrm{~km}\). The known cathetus is the Earth's radius, and we write it as \(a=6370 \mathrm{~km}\). It is a right-angled triangle because the sought-after cathetus \(b\) lies tangentially to the Earth's great circle.

According to the Pythagorean theorem,

\[
b^{2}=c^{2}-a^{2}=(6370.5 \mathrm{~km})^{2}-(6370 \mathrm{~km})^{2}=6370.25 \mathrm{~km}^{2}
\]

and thus

\[
b \approx 79.8 \mathrm{~km}
\]

This means that from the TV tower, one can see approximately \(79.8 \mathrm{~km}\) far.