How many triangles exist whose sides are integers and the perimeter measures 12 units?
(a) 1
(b) 3
(c) 5
(d) 7
(e) 9
The correct option is (b).

For three numbers $a, b$, and $c$ to be the lengths of the sides of a triangle, each of them must be greater than the difference and less than the sum of the other two. Let $a \leq b \leq c$ be the lengths of the sides of the triangle, so that $c < a + b$. Now, adding $c$ to both sides, we have $2c < a + b + c = 12$, that is, $2c < 12$, so $c < 6$. Moreover, since $3c \geq a + b + c = 12$, we have that $c \geq 4$, so $4 \leq c < 6$.

In the case $c = 5$, we have $a + b = 7$ and the possible values of $a$ and $b$ are $a = 2$ and $b = 5$, or $a = 3$ and $b = 4$. In the case $c = 4$, we have $a + b = 8$, and therefore, the only solution is $a = b = 4$. Conclusion: we have 3 possible triangles.