12.030. An equilateral triangle is intersected by a line passing through the midpoint of one of its sides and forming an acute angle $\alpha$ with this side. In what ratio does this line divide the area of the triangle
Solution.

Let $A B=B C=A C=a, \angle M E C=\alpha, A E=E C$ (Fig. 12.33). From $\triangle M K E$, $\angle M K E=90^{\circ}, E K=M K \cdot \operatorname{ctg} \alpha$. In $\triangle M K C, \angle M K C=90^{\circ}, \quad K C=$ $=M K \cdot \operatorname{ctg} 60^{\circ}=\frac{M K}{\sqrt{3}}, E C=E K+K C=M K\left(\operatorname{ctg} \alpha+\frac{1}{\sqrt{3}}\right)$,

$M K\left(\operatorname{ctg} \alpha+\frac{1}{\sqrt{3}}\right)=\frac{a}{2} ; M K=\frac{a}{2\left(\operatorname{ctg} \alpha+\frac{1}{\sqrt{3}}\right)}, S_{\triangle E C M}=\frac{1}{2} \cdot M K \cdot E C=$ $=\frac{a^{2}}{8\left(\operatorname{ctg} \alpha+\frac{1}{\sqrt{3}}\right)} \cdot S_{A B M E}=S_{\triangle A B C}-S_{\triangle E C M}, S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}$, then $\frac{S_{A B M E}}{S_{E C M}}=$ $=\frac{\frac{2 a^{2} \sqrt{3}}{2 \cdot 4}-\frac{a^{2}}{8\left(\operatorname{ctg} \alpha+\frac{1}{\sqrt{3}}\right)}}{\frac{a^{2}}{8} \cdot\left(\operatorname{ctg} \alpha+\frac{1}{\sqrt{3}}\right)}=\frac{2 \sqrt{3}-\frac{1}{\operatorname{ctg} \alpha+\frac{1}{\sqrt{3}}}}{\operatorname{ctg} \alpha+\frac{1}{\sqrt{3}}}=2 \sqrt{3} \operatorname{ctg} \alpha+1=\frac{2 \sqrt{3} \cos \alpha+\sin \alpha}{\sin \alpha}$. Answer: $\frac{2 \sqrt{3} \cos \alpha+\sin \alpha}{\sin \alpha}$.