Let's construct a triangle, given the sides $a$ and $b$, and knowing that the medians corresponding to these sides are perpendicular to each other.
I. Solution: Imagine the problem is solved. The labeling is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_8d2389ceccdf995439ecg-1.jpg?height=339&width=654&top_left_y=210&top_left_x=730)

Denote the endpoints of the medians, i.e., the midpoints of sides $a$ and $b$, by $A_{1}$ and $B_{1}$, respectively. The triangle $B_{1} A_{1} C_{\triangle}$ is a 1:2 reduction of triangle $A B C_{\triangle}$ from the similarity center $C$. The median $B_{1} A^{\prime}$ of the smaller triangle is parallel to the median $A A_{1}$, hence $B A^{\prime} \perp A B_{1}$. Therefore, the point $B_{1}$ lies on the Thales circle drawn over $B A^{\prime} = \frac{3}{4} a$ and also on the circle of radius $\frac{b}{2}$ drawn around $C$. The condition for solvability is that these two circles intersect at two different points, i.e., $\frac{a}{4} < \frac{b}{2} < a$, or $\frac{a}{2} < b < 2a$.

Zsombok Zoltán (Budapest, IV. Konyves Kálmán g. I. o.t.)

II. Solution: Due to the perpendicularity of the medians, the centroid $S$ lies on the Thales circle drawn over $B A_{1}$. Since $A_{1} A = 3 \cdot A_{1} S$, the point $A$ is one of the geometric places of this circle magnified by a ratio of $3:1$ from point $A_{1}$, and the other geometric place is the circle of radius $b$ drawn around $C$.

Katona Péter (Budapest, IX., Apáczai Csere g. II.o.t.)