When constructing a triangle from the medians of a right triangle, we find that this triangle is also a right triangle. Calculate the angles of both triangles to an accuracy of $0.1^{\circ}$.
Let the lengths of the legs and the hypotenuse of the right triangle be $a$ and $b$, and $c$, respectively. The lengths of the medians to these sides are $s_{a}, s_{b}, s_{c}$. Thus, $s_{c}=c / 2$, and $s_{a}$ is the hypotenuse of a right triangle with legs $b$ and $a / 2$, while $s_{b}$ is the hypotenuse of a right triangle with legs $a$ and $b / 2$. Therefore,

$$
s_{c}^{2}=\frac{c^{2}}{4}=\frac{a^{2}+b^{2}}{4}, \quad s_{a}^{2}=b^{2}+\frac{a^{2}}{4}=\frac{4 b^{2}+a^{2}}{4}, \quad s_{b}^{2}=a^{2}+\frac{b^{2}}{4}=\frac{4 a^{2}+b^{2}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_99491f0d558978793d1ag-1.jpg?height=582&width=770&top_left_y=457&top_left_x=667)

Accordingly, $4 s_{a}^{2}=\left(a^{2}+b^{2}\right)+3 b^{2}>4 s_{c}^{2}$, and similarly $4 s_{b}^{2}=\left(a^{2}+b^{2}\right)+3 a^{2}>4 s_{c}^{2}$. Therefore, the right triangle formed by the medians can only have one of the medians to the legs, say $s_{a}$, as its hypotenuse. (In this case, $s_{a}>s_{b}$ and $a<b$.) Thus, from (1) and the Pythagorean theorem,

$$
s_{b}^{2}+s_{c}^{2}=\frac{5 a^{2}+2 b^{2}}{4}=s_{a}^{2}=\frac{4 b^{2}+a^{2}}{4}
$$

Rearranging, we get $b^{2}=2 a^{2}$. Therefore, $c^{2}=3 a^{2}$, and

$$
s_{c}^{2}=\frac{3 a^{2}}{4}, \quad s_{a}^{2}=\frac{9 a^{2}}{4}=3 s_{c}^{2}, \quad s_{b}^{2}=\frac{6 a^{2}}{4}=2 s_{c}^{2}
$$

Thus,

$$
b=a \sqrt{2}, \quad c=a \sqrt{3}, \quad \text { and } \quad s_{b}=s_{c} \sqrt{2}, \quad s_{a}=s_{c} \sqrt{3}
$$

From this, we see that

$$
s_{c}: s_{b}: s_{a}=a: b: c
$$

The second triangle is similar to the original, so their angles are equal in pairs.

In the original triangle, $\operatorname{tg} \beta=b / a=\sqrt{2} \approx 1.414$, so with an accuracy of one degree, $\beta \approx 54.7^{\circ}, \alpha \approx 35.3^{\circ}$.

Zsuzsa Makk (Székesfehérvár, Teleki B. Secondary School II. grade)