How many triangles exist whose sides are integers and the perimeter is 12?
(a) 1
(b) 3
(c) 5
(d) 7
(e) 9
For three numbers $a, b, c$ 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. Thus, $c < a + b$.

Now, adding $c$ to both sides, we have: $2c < a + b + c = 12$, that is, $2c < 12$, $\log c < 6$.

Moreover, since $3c \geq a + b + c = 12$, we have: $c \geq 4$. Therefore, $4 \leq c < 6$.

In the case of $c = 5$, we have $a + b = 7$. The possible values of $a$ and $b$ are: $a = 2$ and $b = 5$ or $a = 3$ and $b = 4$.

In the case of $c = 4$, we have $a + b = 8$, and thus we have only the solution $a = b = 4$.

Thus, we have 3 possible triangles. The correct option is (b).