Okay, so I need to find the greatest possible perimeter of a triangle where two sides are each 8 units long, and the third side is a whole number. Hmm, let's start by recalling the triangle inequality theorem. That says that the sum of any two sides of a triangle must be greater than the third side. Right? So, if two sides are 8 each, let's call the third side 'x'. Then, according to the triangle inequality, x + 8 > 8, and 8 + 8 > x. Wait, but let me write that properly.

First, the two sides are 8 and 8, and the third side is x. So, the three conditions from the triangle inequality would be:

1. 8 + 8 > x
2. 8 + x > 8
3. 8 + x > 8

But actually, conditions 2 and 3 are the same because they both simplify to x > 0. Since the third side has to be a positive length. But since x is a whole number, that's going to be x ≥ 1. But let's focus on the first condition: 8 + 8 > x, which simplifies to 16 > x or x < 16. So, x has to be less than 16. But since x is a whole number, the maximum possible value x can take is 15. But wait, hold on. Let me check if that's correct.

Wait, the triangle inequality says that the sum of any two sides must be greater than the third. So, the two sides we have are 8 and 8, so 8 + 8 must be greater than the third side x. So, 16 > x. Therefore, x must be less than 16. Since x is a whole number, the largest possible x is 15. But we also need to make sure that the other triangle inequalities hold. For example, 8 + x > 8. That simplifies to x > 0, which is already satisfied since x is at least 1. Similarly, 8 + x > 8 is the same. So, the only restriction here is x < 16. Therefore, the maximum x can be is 15. So, the perimeter would be 8 + 8 + 15 = 31. But wait, let me confirm if a triangle with sides 8, 8, 15 is valid.

Let me check the triangle inequalities again. 8 + 8 > 15? 16 > 15, which is true. 8 + 15 > 8? 23 > 8, true. 8 + 15 > 8? Same as above. So, yes, that's a valid triangle. Therefore, the perimeter would be 31. But hold on, is there a possibility that a larger perimeter is possible? Wait, the third side can't be 16 because 8 + 8 is 16, and the third side has to be less than 16. So 15 is the maximum. Therefore, 8 + 8 + 15 = 31. That seems right. But let me think again. Maybe there's a case where the triangle can have a larger perimeter? Wait, if the third side is 15, that's the maximum allowed. So, 31 is the maximum perimeter. But just to make sure, let's think of another approach.

Alternatively, maybe using the concept of an isoceles triangle. Since two sides are equal (8 units each), the triangle is isoceles, and the third side is the base. The perimeter is 16 + x, so to maximize the perimeter, we need to maximize x. The maximum x is 15, as established before. Therefore, the answer is 31. Hmm. That seems straightforward. But maybe there's a trick here that I'm missing. Let me check another way.

Suppose we consider the triangle with sides 8, 8, and x. The perimeter is 16 + x. To maximize the perimeter, we need to maximize x. From the triangle inequality, x must be less than 16, so x can be at most 15. Therefore, the perimeter is 16 + 15 = 31. So, that's the same answer. So, seems like 31 is the correct answer. Wait, but maybe if x is 16, the triangle becomes degenerate, meaning it's a straight line, so the area is zero, but it's still technically a triangle? Wait, but by definition, a degenerate triangle is not considered a valid triangle. So, x has to be strictly less than 16. Therefore, 15 is the maximum. So, yes, 31. Therefore, the answer is 31. But let me check once more.

Suppose x is 15. Then the sides are 8, 8, 15. Let's check the triangle inequalities:

1. 8 + 8 > 15? 16 > 15, yes.

2. 8 + 15 > 8? 23 > 8, yes.

3. 8 + 15 > 8? Same as above.

All conditions are satisfied. Therefore, the triangle is valid. So, the perimeter is 8 + 8 + 15 = 31. Therefore, the greatest possible perimeter is 31 units. That seems to be the answer. But just to be thorough, let's check if there's any other possible maximum. For example, if x is 16, but as we saw, the triangle becomes degenerate. So, 15 is the maximum possible. Therefore, the answer is 31. I think that's solid.

Wait, but another thought: maybe if the triangle is not isoceles? Wait, but the problem states that two sides are each 8 units. So, the third side is different. So, regardless of the type of triangle, the two sides are 8 each, so the third side can be anything less than 16. So, 15 is the maximum. Therefore, the perimeter is 31. Yeah, that's the answer.

Alternatively, maybe the problem is a trick question where the third side is not allowed to be 15 because of some other constraint? But the problem only specifies that the third side has a whole number length. So, as long as the triangle inequality is satisfied, which we've checked, then 15 is acceptable. Therefore, 31 is the answer. So, I think that's the correct solution.
Therefore, the greatest possible perimeter of the triangle is $\boxed{31}$ units.
</think>