A trough, whose cross-section is an equilateral triangle, is filled with water up to the brim. We want to pour out one-fifth of the water. By how many degrees must the trough be tilted for this, so that the triangles bounding its ends move in their own vertical planes?
Solution. Since the barrel is cylindrical, the ratio of the original and the outflowing water volume is the same as the ratio of the areas of the triangles $A B C$ and $A C F$ visible in the cross-section of the barrel, where the segment $C F$ indicates the water level after the barrel is rotated by $(-\alpha)$ around $B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a8ffbca8dc6635efa8beg-1.jpg?height=391&width=1063&top_left_y=306&top_left_x=520)

The following must hold:

$$
T_{A F C}=\frac{A F \cdot m}{2}=\frac{1}{5} T_{A B C}=\frac{1}{5} \cdot \frac{a \cdot m}{2}
$$

From this, $A F=\frac{a}{5}$.

Let $T$ be the foot of the altitude from $C$ in the triangle. Then $A C T \varangle=30^{\circ}$ and $F T=\frac{a}{2}-\frac{a}{5}=\frac{3 a}{10}$. In the triangle $FTC$:

$$
\operatorname{tg}\left(30^{\circ}-\alpha\right)=\frac{\frac{3}{10} a}{m}=\frac{\frac{3}{10} a}{\frac{\sqrt{3}}{2} a}=\frac{\sqrt{3}}{5}
$$

From this, $30^{\circ}-\alpha \approx 19.1^{\circ}$, and thus $\alpha=10.9^{\circ}$. Therefore, the barrel must be tilted by $10.9^{\circ}$.

Note. Many used the false statement that the segments connecting the points dividing the opposite side into five equal parts divide the angle into five equal parts. Thus, they obtained $12^{\circ}$ as the result, which is of course incorrect. Their solution is worth 2 points.