, 2005

Solution:

The answer is (B). Indeed, from the triangle inequality we have
$$a < b + c \Longleftrightarrow a + a < a + b + c = 1 \Longleftrightarrow a < \frac{1}{2}$$
and the other two symmetric relations obtained by swapping $a, b$ and $c$ in the inequalities above. Therefore, the following three statements are equivalent:
$$\begin{gathered} a, b, c \text{ are the sides of a triangle } \\ a < b + c, \quad b < a + c, \quad c < a + b \\ a < \frac{1}{2}, \quad b < \frac{1}{2}, \quad c < \frac{1}{2} \end{gathered}$$
On the other hand, it is easy to construct counterexamples showing how the other three conditions are not equivalent to the one required:
- If $a = b = c = \frac{1}{3}$ (equilateral triangle), (A) does not hold but it is possible to construct a triangle;
- If $a = b = \frac{1}{2}$ and $c = 0$, (C) holds but it is not possible to construct a triangle;
- If $a = b = \frac{3}{8}$ and $c = \frac{2}{8}$, (D) does not hold but it is possible to construct a triangle.