$25 \cdot 24$ The equations of lines $l_{1}$ and $l_{2}$ are $y=m x$ and $y=n x$, respectively. Suppose the angle that $l_{1}$ makes with the horizontal (measured counterclockwise from the positive $x$-axis) is twice that of $l_{2}$, and the slope of $l_{1}$ is 4 times that of $l_{2}$. If $l_{1}$ is not horizontal, then the product $m n$ is
(A) $\frac{\sqrt{2}}{2}$.
(B) $-\frac{\sqrt{2}}{2}$.
(C) 2 .
(D) -2 .
(E) There is no unique solution according to the given conditions.
(31st American High School Mathematics Examination, 1980)
[Solution 1] As shown in the figure, $l_{1}$ and $l_{2}$ intersect the line $x=1$ at $B$ and $A$, respectively, and $C$ is the intersection of the line $x=1$ and the $x$-axis.
$\because O C=1, \quad$ the slope of $l_{2}$ is $\frac{A C}{O C}=A C$,
and the slope of $l_{1}$ is $\frac{B C}{O C}=B C$,
$\therefore \quad A C=n, \quad B C=m, \quad$ and $A B=3 n$.
Also, $\because \quad O A$ is the angle bisector, $\therefore \quad \frac{O C}{O B}=\frac{A C}{A B}$,

which means $\frac{1}{O B}=\frac{n}{3 n}$, thus $\quad O B=3$.
In the right triangle $\triangle O B C$, by the Pythagorean theorem, we have $1+(4 n)^{2}=9$, so
$$
n=\frac{\sqrt{2}}{2} \text {. }
$$

Since $m=4 n$, then $m n=4 n^{2}=2$.
Therefore, the answer is $(C)$.
[Solution 2] Let $\theta_{1}$ and $\theta_{2}$ be the inclinations of the lines $L_{1}$ and $L_{2}$, respectively, then
$$
m=\operatorname{tg} \theta_{1}, n=\operatorname{tg} \theta_{2} .
$$

Since $\theta_{1}=2 \theta_{2}$, and $m=4 n$, then
$$
4 n=m=\operatorname{tg} \theta_{1}=\operatorname{tg} 2 \theta_{2}=\frac{2 \operatorname{tg} \theta_{2}}{1-\operatorname{tg}^{2} \theta_{2}}=\frac{2 n}{1-n^{2}} \text {. }
$$

Thus,
$$
4 n\left(1-n^{2}\right)=2 n \text {. }
$$

Since $n \neq 0$, we have $2 n^{2}=1$, hence $m n=4 n^{2}=2$.
Therefore, the answer is $(C)$.