3.9. In the square $A B C D$, a line is drawn through the midpoint $M$ of side $A B$, intersecting the opposite side $C D$ at point $N$. In what ratio does the line $M N$ divide the area of the square if the acute angle $A M N$ is equal to $\alpha$? Specify the possible values of $\alpha$.
3.9. Given that $ABCD$ is a square, $M \in AB, MA=MB, \angle AMN=\alpha, N \in CD$ (Fig. 3.10); we need to find $S_{AMND} : S_{BMNC}$. Let $AB=a$; then

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-504.jpg?height=373&width=431&top_left_y=1121&top_left_x=723)

Fig. 3.11 $S_{AMND}=\frac{MA+ND}{2} \cdot a=\frac{a+2ND}{4} \cdot a$, $S_{BMNC}=\frac{MB+NC}{2} \cdot a=\frac{\frac{a}{2}+a-ND}{2} \cdot a=\frac{3a-2ND}{4} \cdot a$; $\frac{S_{AMND}}{S_{BMNC}}=\frac{a+2ND}{3a-2ND}$.

Draw $NE \| AD$ and from $\triangle MNE$ find $ME=a \operatorname{ctg} \alpha$. Hence, $ND=MA-ME=\frac{a}{2}-a \operatorname{ctg} \alpha=\frac{a-2a \operatorname{ctg} \alpha}{2}$. Substituting the found value for $ND$, we get

$$
\begin{aligned}
& \frac{S_{AMND}}{S_{BMNC}}=\frac{a+a-2a \operatorname{ctg} \alpha}{3a-a+2a \operatorname{ctg} \alpha}=\frac{1-\operatorname{ctg} \alpha}{1+\operatorname{ctg} \alpha}= \\
& =\frac{\operatorname{tg} \alpha-1}{\operatorname{tg} \alpha+1}=\frac{\operatorname{tg} \alpha-\operatorname{tg} \frac{\pi}{4}}{1+\operatorname{tg} \alpha \operatorname{tg} \frac{\pi}{4}}=\operatorname{tg}\left(\alpha-\frac{\pi}{4}\right)
\end{aligned}
$$

The smallest value of the angle $\alpha$ is achieved when point $N$ coincides with vertex $D$; then $\operatorname{tg} \alpha=a: \frac{a}{2}=2$. On the other hand, $\alpha<\frac{\pi}{2}$. Thus, $\operatorname{arctg} 2 \leq \alpha<\frac{\pi}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-505.jpg?height=393&width=404&top_left_y=1013&top_left_x=58)

Fig. 3.12