Slovenija 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 $|AC|=2|AF|=6\sqrt{5}$.