### Instruction: Triangle $ABC$ has three different integer side lengths. Side $AC$ is the longest side and side $AB$ is the shortest side. If the perimeter of $ABC$ is 384 units, what is the greatest possible difference $AC - AB$?### Response: To find the greatest possible difference $AC - AB$, I want to make $AC$ as long as possible and $AB$ as short as possible, subject to the constraints of the problem.
I know that the sum of any two sides of a triangle must be greater than the third side, so I can use that to bound $AC$ and $AB$.
Let $x = AC$, $y = BC$, and $z = AB$.
Then I have $x + y + z = 384$ and $x > y > z$.
I also have $x < y + z$, $y < x + z$, and $z < x + y$ by the triangle inequality.
I want to maximize $x - z$, so I will try to make $x$ close to $y + z$ and $z$ close to $x - y$.
This means I should look for values of $y$ that are close to half of 384, or 192.
The closest integer to 192 that is not equal to 192 is 191, so I will try $y = 191$.
Then I have $x + z = 384 - 191 = 193$.
Since $x > y > z$, I have $x > 191$ and $z < 191$.
To make $x$ as large as possible, I will try $x = 192$.
Then I have $z = 193 - 192 = 1$.
This satisfies $x > y > z$ and $x < y + z$, but I need to check the other inequalities.
I have $y < x + z$, which is true since $191 < 192 + 1$.
I also have $z < x + y$, which is true since $1 < 192 + 191$.
So this is a valid triangle with the maximum difference $x - z = 192 - 1 = 191$.