2025 AMC 10B Problem 9

P1-10~5 minPrint

Basic Counting Systems of Equations Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

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

$\begin{align*} -x-y-z&\le -2\\ -x+y+z&\le 2\\ x-y+z&\le 2\\ x+y-z&\le 2 \end{align*}$

Show solution without answering

Key insight

Add the inequalities in pairs to trap x, y and z between 0 and 2. Then list the few triples, sorted by their sum.

Solution

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.

Sum $S$	Each number must be at least	Triples
$0$ or $1$	(not allowed, $S \ge 2$ )	none
$2$	$0$	$(2,0,0)$ , $(1,1,0)$ in any order: $3 + 3 = 6$
$3$	$\frac12$ , so at least $1$	$(1,1,1)$ : $1$
$4$	$1$	$(2,1,1)$ in any order: $3$
$5$	$\frac32$ , so at least $2$	three $2$ s add to $6$ , not $5$ : none
$6$	$2$	$(2,2,2)$ : $1$

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

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques