AMC10 2025 B

We have: \[\begin{cases} x+y+z \ge 2, \\ -x+y+z \le 2, \\ x-y+z \le 2, \\ x+y-z \le 2. \end{cases}\] Subtract the second inequality from the first: \[(x+y+z)-(-x+y+z) \ge 2-2 \quad\Rightarrow\quad 2x \ge 0 \Rightarrow x\ge 0.\]By symmetry in $x,y,z,$ we also obtain \[y\ge 0,\qquad z\ge 0.\] WLOG & by symmetry, let \[0 \le x \le y \le z.\] up to permutation. From the second inequality of we have $y+z \le 2+x.$ Using $x\le y$, we get \[y+z \le 2+y \quad\Rightarrow\quad z \le 2.\] Hence, $0 \le x \le y \le z \le 2.$ Divide into casework on $x:$ Case 1: $x=0$ Then the first two inequalities become \[y+z \ge 2,\qquad y+z \le 2 \Rightarrow y+z=2\]And because we have $0\le y\le z\le 2, (y,z)=(0,2), (1,1).$ Case 2: $x=1$ \[1+y+z \ge 2 \Rightarrow y+z\ge 1,\qquad -1+y+z \le 2 \Rightarrow y+z \le 3.\]$1\le y\le z\le 2$, so $(y,z)=(1,1)\ \text{or}\ (1,2).$ Case 3: $x=2$ \[2+y+z \ge 2 \qquad -2+y+z \le 2 \Rightarrow y+z \le 4.\]$2\le y\le z\le 2$ so the only combination $(y,z)=(2,2).$ Thus up to permutation we have $(0,0,2),\ (0,1,1),\ (1,1,1),\ (1,1,2),\ (2,2,2)$ Now count distinct ordered triples for each set of three numbers under symmetry: $3+3+3+1+1=\boxed{11}.$