△ABC has integer sides x, y, and z such that xz = 12. How many such triangles are possible?

To solve this problem, we need to consider the possible integer pairs (x, z) that multiply to 12. These pairs are (1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), and (-3, -4). However, since the lengths of the sides of a triangle cannot be negative, we discard the last three pairs.

Now, we need to consider the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

For the pair (1, 12), y must be greater than 11 and less than 13. There is only one integer, 12, that satisfies this condition.

For the pair (2, 6), y must be greater than 4 and less than 8. There are three integers, 5, 6, and 7, that satisfy this condition.

For the pair (3, 4), y must be greater than 1 and less than 7. There are five integers, 2, 3, 4, 5, and 6, that satisfy this condition.

Adding up these possibilities, there are 1 + 3 + 5 = 9 such triangles possible.