18. Construct a triangle given two sides so that the medians drawn to these sides intersect at a right angle.
18. Let the lengths $a, b$ of sides $B C$ and $A C$ of a triangle be known, and we want its medians $A D$ and $B E$ to intersect at a right angle. Take a point $F$ on segment $A C$ of length $b$ such that $A F: F C=3$; then $F D \| B E$ and $\angle A D F$ should be $90^{\circ}$, i.e., point $D$ should lie on the circle with diameter $A F$ and, in addition, on the circle of radius $a / 2$ with center $C$. From this, it is clear how to construct point $D$, and then $B$. The problem has a solution if $1 / 2<b / a<2$, and in this case, the desired triangle is unique.

$\nabla$ It can be proven that in such a triangle, the square of the third side $c$ is equal to $c^{2}=\left(a^{2}+b^{2}\right) / 5$, and from this, another solution can be derived.