$18 \cdot 111$ In a trapezoid with bases 3 and 9, and legs 4 and 6, a line parallel to the bases divides the trapezoid into two trapezoids of equal perimeter. Then the ratio in which this line divides the legs is
(A) $4: 3$.
(B) $3: 2$.
(C) $4: 1$.
(D) $3: 1$.
(E) $6: 1$.
(8th American High School Mathematics Examination, 1957)
［Solution］Suppose one of the non-parallel sides of the trapezoid is divided in the ratio $m: n$, and the length of the line segment cut by the parallel line is $x$, then $m+n=6$.
By setting $3+m+x+4 \cdot \frac{m}{m+n}$

i.e. $\square$
$$
\begin{array}{l}
=9+n+x+4 \cdot \frac{n}{m+n}, \\
(m-n)+4 \cdot \frac{m-n}{m+n}=6,
\end{array}
$$

Given $m+n=6$, then
$(m-n)\left(1+\frac{4}{6}\right)=6$, so $m-n=\frac{18}{5}$.

From $m+n=6$, we can solve for $m=\frac{24}{5}, n=\frac{6}{5}$.
Thus, $m: n=4: 1$.
Therefore, the answer is $(C)$.