2025 AMC 10B Problem 9

There could be infinitely many triples, so first find a small box that all the answers must lie in. Adding two inequalities together often makes variables cancel, so try that.

Step 1: upper bounds. Add the third and fourth inequalities:
$$(x - y + z) + (x + y - z) \le 2 + 2, \quad\text{so}\quad 2x \le 4, \quad x \le 2.$$
The same trick with other pairs gives $y \le 2$ and $z \le 2$.

Step 2: lower bounds. Add the first and second:
$$(-x - y - z) + (-x + y + z) \le -2 + 2, \quad\text{so}\quad -2x \le 0, \quad x \ge 0.$$
Likewise $y \ge 0$ and $z \ge 0$.

So each of $x, y, z$ is $0$, $1$ or $2$. That is only $27$ triples to think about.

Step 3: rewrite the rules using the sum. Let $S = x + y + z$.
- The first inequality says $S \ge 2$.
- The second says $-x + y + z \le 2$. Since $-x + y + z = S - 2x$, this means $S - 2x \le 2$, that is, $x \ge \frac{S - 2}{2}$.
- The third and fourth say the same for $y$ and $z$.

So the rules are: the sum is at least $2$, and every number is at least $\frac{S-2}{2}$.

Step 4: go through the possible sums.

The total is $6 + 1 + 3 + 0 + 1 = 11$.

The answer is $\boxed{\textbf{(C)}\ 11}$.

With several inequalities, adding two of them can cancel variables and give a simple bound on one letter. Once every letter is stuck in a small range, a careful list is safe. Organizing the list by the sum $x + y + z$ works well here because every rule can be written in terms of that sum.

Write $S = x + y + z$. Multiplying the first inequality by $-1$ turns it into
$$S \ge 2.$$
The other three can be rewritten using $S$ as well: $-x+y+z = S - 2x$, $x-y+z = S - 2y$, and $x+y-z = S - 2z$. So they say
$$S - 2x \le 2, \qquad S - 2y \le 2, \qquad S - 2z \le 2,$$
and all three hold precisely when the largest of the left sides does, that is when
$$S \le 2 + 2m, \qquad m = \min(x,y,z).$$

The whole system is therefore
$$2 \le S \le 2 + 2m.$$

This already bounds everything. From $2 \le 2 + 2m$ we get $m \ge 0$, so $x, y, z \ge 0$. And if $M = \max(x,y,z)$, the other two coordinates are each at least $m$, so $S \ge M + 2m$, giving $M \le S - 2m \le 2$. Hence
$$0 \le x, y, z \le 2.$$

Because the system is symmetric in $x$, $y$, $z$, count by the value of $m$.

$m = 0$. The condition becomes $2 \le S \le 2$, so $S = 2$ with at least one coordinate $0$. The triples with entries in $\{0,1,2\}$ summing to $2$ are the arrangements of $(0,0,2)$ and of $(0,1,1)$: $3 + 3 = 6$ triples, and all of them do contain a $0$.

$m = 1$. All entries lie in $\{1,2\}$ with at least one $1$, and $S \le 4$. Writing $k$ for the number of $2$s, $S = 3 + k \le 4$ forces $k \le 1$. That gives $(1,1,1)$ and the three arrangements of $(1,1,2)$: $4$ triples.

$m = 2$. Then $x = y = z = 2$, $S = 6 \le 2 + 4 = 6$: $1$ triple.

Altogether $6 + 4 + 1 = 11$ triples, namely
$$(0,0,2),\ (0,2,0),\ (2,0,0),\ (0,1,1),\ (1,0,1),\ (1,1,0),\ (1,1,1),\ (1,1,2),\ (1,2,1),\ (2,1,1),\ (2,2,2).$$

The answer is $\boxed{\textbf{(C)}\ 11}$.