2003 AMC 10A Problem 7

Let the sides be integers $a \le b \le c$ with $a + b + c = 7$. The triangle inequality $a + b > c$ is equivalent to $c < \tfrac{7}{2}$, so $c \le 3$. Since three sides of at most $3$ must total $7$, and $2 + 2 + 2 = 6$ is too small, the longest side is exactly $c = 3$.

Then $a + b = 4$ with $a \le b \le 3$: either $(a, b) = (1, 3)$ or $(2, 2)$.

Both triples $(1, 3, 3)$ and $(2, 2, 3)$ satisfy $a + b > c$, so there are exactly $2$ non-congruent triangles.

The answer is $\boxed{\textbf{(B)}\ 2}$.

For integer-sided triangles with a given perimeter, the only real constraint is the triangle inequality, and it is cleanest to apply to the longest side: it must be less than half the perimeter. Bounding the largest side first shrinks the search to a couple of cases.

• Listing every way to write 7 as three positive integers and forgetting the triangle inequality, which admits (1,1,5) and (1,2,4).
• Counting orderings such as $(1,3,3)$ and $(3,1,3)$ separately; congruent triangles count once.