Let's construct an isosceles triangle, given the base $a$, and the two medians $k$ that can be drawn from the endpoints of the base. What are the lengths of the legs of the triangle?
It is known that the centroid divides the medians in a $1: 2$ ratio. Therefore, we construct an isosceles triangle whose base is $a$ and whose legs are $\frac{2}{3} k$. The vertex $S$ opposite to $a$ in this triangle is the centroid of the desired triangle. From $S$, we drop a perpendicular to $a$, which intersects the base at $A_{1}$. On the extension of $A_{1} S$, we mark off $2 A_{1} S$, thereby obtaining the third vertex of the desired triangle; the legs of this triangle are:

$$
b=c=\sqrt{4 k^{2}-2 a^{2}}
$$

Number of solutions: 42.