2. In a triangle, the length of one median is half longer than the length of the side to which it is drawn. Find the angle between the other two medians.
Solution. Let $ABC$ be a triangle in which the medians $AA_1, BB_1$, and $CC_1$ intersect at point $T$, and $\overline{AA_1} = \frac{3}{2} \overline{BC}$.

From the property of medians in a triangle and the condition of the problem, we have

$$
\overline{TA_1} = \frac{1}{3} \overline{AA_1} = \frac{1}{2} \overline{BC}
$$

Thus, in triangle $CTB$, the length of the median $TA_1$ is half the length of the side to which it is drawn. Therefore, the points $B, T$, and $C$ are equally distant from point $A_1$. According to this, they lie on a circle with center at $A_1$ and radius $\frac{1}{2} \overline{BC}$. Thus $\measuredangle B T C = 90^{\circ}$, meaning the medians $BB_1$ and $CC_1$ intersect at a right angle.

![](https://cdn.mathpix.com/cropped/2024_06_05_c02eb5ea46120e551070g-1.jpg?height=445&width=368&top_left_y=1306&top_left_x=1114)