18.9 A segment of length 1 is divided into four segments. Then the four segments can form a quadrilateral if and only if each segment satisfies which of the following conditions?
(A) Equals $\frac{1}{4}$.
(B) Equals or is greater than $\frac{1}{8}$ and less than $\frac{1}{2}$.
(C) Is greater than $\frac{1}{8}$ or less than $\frac{1}{2}$.
(D) Is greater than $\frac{1}{8}$ or less than $\frac{1}{4}$.
(E) Is less than $\frac{1}{2}$.
(19th American High School Mathematics Examination, 1968)
[Solution] First, prove the following fact:
Four line segments can form a quadrilateral $\Leftrightarrow$ each of the four segments is less than the sum of the other three.
$\Rightarrow$ This is obvious.
$\Leftarrow$ If the lengths of the four segments are denoted as $S_{1}, S_{2}, S_{3}, S_{4}$, and if any one segment is less than the sum of the other three, then there must be two segments whose sum is not greater than the sum of the other two. Without loss of generality, assume
$$
S_{1}+S_{2} \leqslant S_{3}+S_{4} .
$$

This means there exists a triangle with side lengths $S_{1}+S_{2}, S_{3}, S_{4}$, which can be considered a special case of a quadrilateral.
From the above conclusion, if $S_{1}, S_{2}, S_{3}, S_{4}$ can form a quadrilateral, then
$$
S_{1}1-\frac{1}{2}=\frac{1}{2} \text {, }
$$

Thus,
$$
S_{1}<\frac{1}{2}<S_{2}+S_{3}+S_{4} \text {. }
$$

Similarly, we can derive the corresponding inequalities for the other segments. This means that segments of lengths $S_{1}, S_{2}, S_{3}, S_{4}$ can form a quadrilateral.
Therefore, the answer is $(E)$.