China Mathematical Competition

For by eliminating $y$ and simplifying it, we get $$(3+4k^2)x^2+8kmx+4m^2-48=0.$$ Define $A(x_1,y_1)$, $B(x_2,y_2)$. Then $x_1+x_2=-\frac{8km}{3+4k^2}$. $${\Delta }_1=(8km{)}^2-4(3+4k^2)(4m^2-48)>0.\stackrel{◯}{1}$$ For by eliminating $y$ and simplifying it, we get $$(3-k^2)x^2-2kmx-m^2-12=0.$$ Define $C(x_3,y_3)$, $D(x_4,y_4)$. Then $x_3+x_4=\frac{2km}{3-k^2}$. $${\Delta }_2=(-2km{)}^2+4(3-k^2)(m^2+12)>0.\stackrel{◯}{2}$$ From $\overrightarrow{AC}+\overrightarrow{BD}=0$ we get $(x_4-x_2)+(x_3-x_1)=0$, which implies that $x_1+x_2=x_3+x_4$. Then $$-\frac{8km}{3+4k^2}=\frac{2km}{3-k^2}.$$ Therefore, $km=0$ or $-\frac{4}{3+4k^2}=\frac{1}{3-k^2}$ (discarded). Then a possible solution is either $k=0$ or $m=0$. When $k=0$, from ① and ② we have $-2\sqrt{3}<m<2\sqrt{3}$. As $m$ is an integer, it can be $-3,-2,-1,0,1,2,3$. When $m=0$, from ① and ② we have $-\sqrt{3}<k<\sqrt{3}$. As $k$ is an integer, it can be $-1,0,1$. Combining the results above, we conclude that there are nine lines in total satisfying the given conditions.