What are the conditions for a triangle constructed from the medians of a right-angled triangle, as sides, to also be a right-angled triangle?
I. Solution: Let the sides of the right-angled triangle be $as_{b}>s_{c}$.

If the triangle formed by the medians is also a right-angled triangle, then the Pythagorean theorem applies to it as well:

$$
s_{a}^{2}=s_{b}^{2}+s_{c}^{2}
$$

which means

$$
b^{2}+\frac{a^{2}}{4}=a^{2}+\frac{b^{2}}{4}+\frac{a^{2}+b^{2}}{4}
$$

from which we get

$$
b^{2}=2 a^{2}, \quad \text { i.e., } \quad b=a \sqrt{2}
$$

Since our reasoning can also be carried out in reverse order, this necessary condition is also sufficient.

Therefore, a triangle constructed from the medians of a right-angled triangle, as its sides, is a right-angled triangle if and only if the ratio of the legs of the original right-angled triangle is: $1: \sqrt{2}$.

Gavajda Pál (Bp. I. Petőfi g. I. o. t.)

II. Solution: Extend the right-angled triangle $A C B$ to form a rectangle $A C B D$, and let the midpoints of segments $A C, C B, B D$, and $B A$ be denoted by $E, F, G, H$ respectively (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_a9ecd16b87357d394246g-1.jpg?height=488&width=617&top_left_y=1149&top_left_x=754)

Since it is clear that $F G=C H=s_{c}, G A=B E=s_{b}$, and $A F=s_{a}$, the sides of $\triangle A F G$ are the medians of $\triangle A B C$. The angle at vertex $G$ opposite to $s_{a}$ in this triangle is a right angle if and only if

$$
\triangle B G F \sim \triangle D A G
$$

indeed, in this case, the angles marked with one arc complement the angles marked with two arcs to form a right angle.

Thus, in this case, the ratio of the legs is

$$
\frac{a}{2}: \frac{b}{2}=\frac{b}{2}: a
$$

from which we get

$$
b=a \sqrt{2}
$$

Tusnády Gábor (Sátoraljaújhely, Kossuth g. II. o. t.)