3.61. The tangent of the acute angle between the medians of a right triangle, drawn to its legs, is $k$. Find the angles of the triangle and the permissible values of $k$.

 untranslated text:
3.61. Тангенс острого угла между медианами прямоугольного греугольника, проведенными к его катетам, равен $k$. Найти утлы треугольника и допустимые значения $k$.
3.61. According to the condition, $\angle A B C = 90^{\circ}, A B_{1} = B_{1} C ; C A_{1} = A_{1} B$, $\operatorname{tg} \angle A O B_{1} = k$ (Fig. 3.63); it is required to find $\angle A$ and $\angle B$. Let $\angle C A A_{1} = \psi, \angle C B_{1} B = \varphi, \angle A O B_{1} = \alpha$; then $\dot{\alpha} = \varphi - \psi, \operatorname{tg} \varphi = \frac{2 a}{b}, \operatorname{tg} \psi = \frac{a}{2 b}$, where $B C = a, A C = b$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-538.jpg?height=303&width=512&top_left_y=1158&top_left_x=642)

Fig. 3.63

$$
\begin{gathered}
\operatorname{tg} \alpha = \frac{\operatorname{tg} \varphi - \operatorname{tg} \psi}{1 + \operatorname{tg} \varphi \operatorname{tg} \psi} = \frac{\frac{2 a}{b} - \frac{a}{2 b}}{1 + \frac{a^{2}}{b^{2}}} = \frac{3 a b}{2\left(b^{2} + a^{2}\right)} = \frac{3}{2\left(\frac{b}{a} + \frac{a}{b}\right)} = \\
= \frac{3}{2(\operatorname{ctg} A + \operatorname{tg} A)} = \frac{3 \operatorname{tg} A}{2\left(1 + \operatorname{tg}^{2} A\right)} = \frac{3}{4} \sin 2 A
\end{gathered}
$$

From this,

$$
\begin{aligned}
& k = \frac{3}{4} \sin 2 A ; \sin 2 A = \frac{4 k}{3} ; \angle A = \frac{1}{2} \arcsin \frac{4 k}{3} \\
& 0 < \frac{4 k}{3} \leq 1 ; 0 < k \leq \frac{3}{4} ; \angle B = \frac{\pi}{2} - \frac{1}{2} \arcsin \frac{4 k}{3}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-539.jpg?height=527&width=508&top_left_y=504&top_left_x=46)

Fig. 3.64