2003 AMC 10A Problem 7

Three lengths make a triangle only if the two shorter ones together are longer than the longest one. The longest side is the one that can cause trouble, so pin it down first.

1. The longest side has to be less than the other two together. All three add to $7$, so the longest side is less than half of $7$, which is $3.5$. As a whole number, it is at most $3$.
2. It cannot be smaller than $3$: three sides of at most $2$ add to at most $2 + 2 + 2 = 6$, which is not $7$. So the longest side is exactly $3$.
3. The other two sides add to $7 - 3 = 4$, and neither is more than $3$. The options are $1 + 3$ and $2 + 2$.
4. Check both triangles:
- Sides $1, 3, 3$: $1 + 3 = 4 > 3$. Works.
- Sides $2, 2, 3$: $2 + 2 = 4 > 3$. Works.
5. For comparison, splits like $1, 1, 5$ or $1, 2, 4$ fail, because $1 + 1 < 5$ and $1 + 2 < 4$: the short sides cannot reach.

There are exactly $2$ triangles.

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

For whole-number triangles with a fixed perimeter, the only rule to check is the triangle inequality, and it is easiest to check on the longest side: that side must be less than half the perimeter. Fixing the longest side first shrinks the list to a couple of cases you can check by hand.

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 $\boxed{\textbf{(B)}\ 2}$.