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Problem 9 Easy

How many ordered triples of integers (x, y,z) satisfy the following system of inequalities?

- x - y - z \leqslant - 2

- x + y + z \leqslant 2

x - y + z \leqslant 2

x + y - z \leqslant 2

  • A.

    4

  • B.

    8

  • C.

    11

  • D.

    15

  • E.

    17

Answer:C
Solution

System of inequalities:

x + y + z \geq 2

-x + y + z \leq 2

x - y + z \leq 2

x + y - z \leq 2

Combining the first and second inequalities, we get x \geq 0. By symmetry, x, y, z \geq 0.

Suppose 0 \leq x \leq y \leq z (WLOG).

From the second inequality, y + z \leq 2 + x \leq 2 + y, which gives z \leq 2.

Thus, 0 \leq x \leq y \leq z \leq 2.

Case 1: x = 0, then from the first and second inequalities, y + z = 2. Solutions: (y, z) = (0, 2) or (1, 1).

Case 2: x = 1, then from the second inequality, y + z \leq 3. Since 1 \leq y \leq z \leq 2, solutions: (y, z) = (1, 1) or (1, 2).

Case 3: x = 2, then (x, y, z) = (2, 2, 2).

In total:

(1, 1, 1) and (2, 2, 2) each give 1 solution

(0, 0, 2), (0, 1, 1), (1, 1, 2) each give 3 solutions

Total number = 1 \times 2 + 3 \times 3 = 11

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