AMC 10 · 2021 · #7

Grade 9 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

symmetric-polynomialscompleting-the-squareperfect-squaressystematic-enumeration symmetry-argumentcasework ↑ Prerequisites: symmetric-polynomials

📏 Long solution 💡 3 insights

Problem

Three integers are fed into a cyclic expression. Five conditions are on offer. Find the one that by itself forces the expression to equal one — not for a lucky choice of numbers, but for every integer triple obeying it.

Pick an answer.

(A)

$x>y \text{ and } y=z$

(B)

$x=y-1 \text{ and } y=z-1$

(C)

$x=z+1 \text{ and } y=x+1$

(D)

$x=z \text{ and } y-1=x$

(E)

x+y+z=1

How to solve

Strategy Organize Information in More Ways

Substituting five conditions into x(x-y)+y(y-z)+z(z-x) one at a time is five separate messes, and for a condition that survives, one substitution proves nothing — a sufficient condition has to hold for infinitely many triples. So the first move is Tool #15 (Organize Information in More Ways): expand the expression, double it, and watch it collapse into (x-y)²+(y-z)²+(z-x)². That single rewrite converts the whole problem into a statement about the gaps between the three numbers. Tool #4 (Introduce a Variable) then gives those gaps names, which exposes a hidden constraint the three of them always satisfy. Tool #2 (Make a Systematic List) finishes the theory: over the integers there is only one way for three squares to add to 2, so the set of winning triples is completely described. With that description in hand, Tool #3 (Eliminate Possibilities) turns each answer choice into a one-line gap check instead of an algebra grind.

1STEP 1

Decide what actually has to be proved

Decide what actually has to be proved.

condition ⟹ x(x-y)+y(y-z)+z(z-x)=1 for every integer triple obeying it

2STEP 2

Expand the cyclic expression

Expand the cyclic expression.

x(x-y)+y(y-z)+z(z-x) = x² - xy + y² - yz + z² - zx = x²+y²+z²-xy-yz-zx

3STEP 3

Double it and squares appear

Doubling reveals a sum of squares.

2(x²+y²+z²-xy-yz-zx) = (x-y)²+(y-z)²+(z-x)², so the equation becomes (x-y)²+(y-z)²+(z-x)² = 2

4STEP 4

Name the three gaps

Name the three gaps.

a=x-y, b=y-z, c=z-x ⟹ a+b+c=0 always, and the equation asks for a²+b²+c²=2

5STEP 5

List the only integer gap pattern

Only one gap pattern is possible.

a²+b²+c²=2 ⟹ {a²,b²,c²}={0,1,1} ⟹ {a,b,c}={0,1,-1}

6STEP 6

Knock out the two loose conditions

Counterexamples kill the loose conditions.

(x,y,z)=(5,0,0): 5 · 5 + 0 · 0 + 0 ·(-5) = 25 ≠ 1; (x,y,z)=(3,-1,-1): 3 · 4 + (-1) · 0 + (-1)(-4) = 16 ≠ 1

7STEP 7

Measure the gaps in the last three

What survives is two variables equal and one exactly larger by one.

(B) (-1,-1,2)→ 6→ 3; (C) (-1,2,-1)→ 6→ 3; (D) (-1,1,0)→ 2→ 1 t(t-(t+1))+(t+1)((t+1)-t)+t(t-t) = -t + (t+1) + 0 = 1 → (D)

Answer

x=z and y-1=x

Choice (D) survives concrete tests on both signs. With x=z=4 and y=5: 4(4-5)+5(5-4)+4(4-4) = -4+5+0 = 1. With x=z=-2 and y=-1: -2(-2+1)+(-1)(-1+2)+(-2)(-2+2) = 2-1+0 = 1. The sum-of-squares rewrite also checks out numerically on that first triple: the gaps are -1, 1, 0, their squares add to 2, and half of 2 is 1. The eliminated choices fail in two distinct ways, which is a good sign the analysis is real rather than lucky. (A) and (E) fail because they leave a gap unbounded — each is true for infinitely many triples, only some of which work, and (A) is the sharper trap since it does succeed whenever x=y+1. (B) and (C) fail differently: they are rigid, they force three consecutive integers, and they make the expression the constant 3 every single time. A condition that always produces 3 is not a near miss but a clean elimination.

💡Key takeaway

When an expression depends only on the differences between the variables, rewrite it as a sum of squares of those differences — a small total then leaves only a handful of possible gaps, and each answer choice becomes something you read rather than compute.

Decide what actually has to be proved
Expand the cyclic expression
Double it and squares appear
Name the three gaps
List the only integer gap pattern
Knock out the two loose conditions
Measure the gaps in the last three