China Mathematical Competition

For $\begin{cases} y = kx + m, \\ \frac{x^2}{16} + \frac{y^2}{12} = 1 \end{cases}$ 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 $x_1 + x_2 = -\frac{8km}{3+4k^2}$.
$$\Delta_1 = (8km)^2 - 4(3 + 4k^2)(4m^2 - 48) > 0. \quad \textcircled{1}$$
For $\begin{cases} y = kx + m, \\ \frac{x^2}{4} - \frac{y^2}{12} = 1 \end{cases}$ 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 $x_3 + x_4 = \frac{2km}{3 - k^2}$.
$$\Delta_2 = (-2km)^2 + 4(3-k^2)(m^2+12) > 0. \quad \textcircled{2}$$
From $\vec{AC} + \vec{BD} = 0$ 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.