2025 AMC 10B Problem 9

The last three inequalities are symmetric in $x, y, z;$ let $p = -x + y + z,$ $q = x - y + z,$ $r = x + y - z$

Then $p, q, r \le 2$ with equal parity and $p + q + r = x + y + z \ge 2;$ count such triples

Let $p = -x + y + z,$ $q = x - y + z,$ $r = x + y - z.$ The last three inequalities say $p, q, r \le 2,$ the first says $x + y + z \ge 2,$ and $p + q + r = x + y + z.$ Since $x = \tfrac{q + r}{2}$ and so on, $p, q, r$ must all share the same parity. Now count triples with each part $\le 2,$ equal parity, and sum in $[2, 6].$ The even ones are $(2,2,2),$ the permutations of $(2,2,0),$ of $(2,0,0),$ and of $(2,2,-2),$ giving $10.$ The only odd one is $(1,1,1).$ That’s $11$ in all, and each yields a unique $(x, y, z).$ Thus, C is the correct answer.