2025 AMC 10B Problem 9

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 $\boxed{\textbf{(C)}\ 11}$.

Systems of linear inequalities in several integer variables are counted by first making the region finite, and the fastest way to do that is to combine the inequalities rather than analyse them one at a time. Adding two of the last three, for instance $(x-y+z) + (x+y-z) \le 4$, gives $x \le 2$ directly, and pairing the first with each of the others gives the lower bounds; rewriting everything through $S$ packages both steps at once. Once the region is confined to a $3 \times 3 \times 3$ box, enumeration is safe, and exploiting the full symmetry in $x$, $y$, $z$ cuts the $27$ checks down to three short cases.

Derive the bounds by direct pairing instead: adding the third and fourth inequalities gives $2x \le 4$, so $x \le 2$, and similarly $y, z \le 2$; and from the second, $y + z \le 2 + x$, so $2 \le x + y + z \le 2x + 2$ and $x \ge 0$, similarly for $y$ and $z$. Then check all $27$ triples in $\{0,1,2\}^3$ by hand; the $16$ failures are the four with $S \le 1$ and the twelve that break one of the last three inequalities.

• Reading -x-y-z <= -2 as x+y+z <= 2, or dropping it once the bounds are found; the other three conditions alone leave 15 triples, choice (D).
• Deriving only the upper bounds $x, y, z \le 2$ and then allowing negative coordinates, which makes the region look infinite.
• Counting the five shapes $\{0,0,2\}$, $\{0,1,1\}$, $\{1,1,1\}$, $\{1,1,2\}$, $\{2,2,2\}$ instead of the ordered triples they represent.