Slovenian Mathematical Olympiad, 2008

Since $D$ is the midpoint of $AB$ and $|AD| = 3$, we have $|AB| = 6$. Let $E$ be the midpoint of $BC$ and let $F$ be the midpoint of $AC$. The sides of the triangle $ADT$ satisfy Pythagoras's theorem, so $ADT$ is a right triangle. The ratio in which the centre of gravity divides the median is $2:1$ and since $|AT| = 4$ we have $|AE| = 6$. Using Pythagoras's theorem for the triangle $ABE$ we can find $|BE| = \sqrt{|AB|^2 + |AE|^2} = \sqrt{36+36} = 6\sqrt{2}$. This implies $|BC| = 2|BE| = 12\sqrt{2}$. Since $E$ and $F$ are the midpoints of $BC$ and $AC$, the segment $EF$ is parallel to $AB$.

Since $EA$ is perpendicular to $AB$, $EA$ is also perpendicular to $EF$. Thus, $AEF$ is a right triangle. We have already shown that $|AE| = 6$, and we have $|EF| = \frac{|AB|}{2} = 3$. We use Pythagoras's theorem once more to find
$$|FA| = \sqrt{|EF|^2 + |AE|^2} = \sqrt{9+36} = 3\sqrt{5}.$$
The length of the side $AC$ is $|AC| = 2|AF| = 6\sqrt{5}$.