12.004. In square $ABCD$, a line is drawn through the midpoint $M$ of side $AB$, intersecting the opposite side $CD$ at point $N$. In what ratio does the line $MN$ divide the area of the square if the acute angle $AMN$ is equal to $\alpha$? Specify the possible values of $\alpha$.
## Solution.

$A B C D$ is a square, $M \in A B, M A=M B, \angle A M N=\alpha, N \in C D$ (Fig. 12.7); it is required to find $S_{A M N D}: S_{B M N C}$. Let $A B=a$; then

$$
S_{A M N D}=\frac{M A+N D}{2} \cdot a=\frac{a+2 N D}{4} \cdot a, S_{B M N C}=\frac{M B+N C}{2} \cdot a=\frac{\frac{a}{2}+a-N D}{2} \cdot a=
$$

$=\frac{3 a-2 N D}{4} a ; \frac{S_{A M N D}}{S_{B M N C}}=\frac{a+2 N D}{3 a-2 N D}$. Draw $N E \| A D$ and from $\triangle M E N$ we have

$M E=a \operatorname{ctg} \alpha$. Hence $N D=M A-M E=\frac{a}{2}-a \operatorname{ctg} \alpha=\frac{a-2 a \operatorname{ctg} \alpha}{2}$. Therefore,

$$
\frac{S_{A M N D}}{S_{B M N C}}=\frac{a+a-2 a \operatorname{ctg} \alpha}{3 a-a+2 a \operatorname{ctg} \alpha}=\frac{1-\operatorname{ctg} \alpha}{1+\operatorname{ctg} \alpha}=\frac{\operatorname{tg} \alpha-1}{\operatorname{tg} \alpha+1}=\frac{\operatorname{tg} \alpha-\operatorname{tg} \frac{\pi}{4}}{1+\operatorname{tg} \alpha \operatorname{tg} \frac{\pi}{4}}=\operatorname{tg}\left(\alpha-\frac{\pi}{4}\right)
$$

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

Answer: $\operatorname{tg}\left(\alpha-\frac{\pi}{4}\right) ; \operatorname{arctg} 2<\alpha<\pi / 2$.