AMC 10 · 2021 · #9

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

polynomial-factoringdifference-of-squaresperfect-squares convert-to-algebrapattern-recognition ↑ Prerequisites: polynomial-factoring

📏 Medium solution 💡 2 insights

Problem

Among all pairs of real numbers (x, y), what is the smallest possible value of the expression (xy - 1)² + (x + y)²? Both pieces are squares, so the expression is never negative; we need its exact minimum.

Pick an answer.

(A)

(B)

$~\frac{1}{4}$

(C)

$~\frac{1}{2}$

(D)

(E)

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Organize Information in More Ways

Tool #15 (Reorganize) is the heart — the original form hides the answer, but expanding and re-grouping reveals a clean product structure. Tool #13 (Algebra) executes the rewrite: expand the squares, regroup, factor as (x² + 1)(y² + 1). Tool #6 (Guess and Check) confirms the candidate (0, 0) achieves the value. Tool #3 (Eliminate) discards 0 (impossible because xy = 1 and x + y = 0 have no real solution) and the larger options.

1STEP 1

Could the minimum be 0? That needs xy = 1 and x + y = 0 at once, but then -x² = 1, impossible, so 0 is ruled out.

x + y = 0, xy = 1 → -x² = 1 (no real x)

2STEP 2

Expand both squares; the -2xy and +2xy cancel, leaving x² y² + x² + y² + 1.

(xy-1)² + (x+y)² = x² y² - 2xy + 1 + x² + 2xy + y² = x² y² + x² + y² + 1

3STEP 3

Group and factor out the common x² + 1: the sum becomes the product (x² + 1)(y² + 1).

x² y² + x² + y² + 1 = (x² + 1)(y² + 1)

4STEP 4

Since x² ≥ 0, each factor is ≥ 1, so the product satisfies (x² + 1)(y² + 1) ≥ 1.

x² + 1 ≥ 1, y² + 1 ≥ 1 → (x² + 1)(y² + 1) ≥ 1

5STEP 5

The bound is reached: at x = 0, y = 0 the original gives (0 - 1)² + 0² = 1.

(0, 0): (0 - 1)² + 0² = 1 ✓

6STEP 6

So the least value is 1 — choice (D).

min = 1 → (D)

Answer

Two independent confirmations: the algebraic factorization proves (x²+1)(y²+1) ≥ 1 for all real x, y; the point (0, 0) achieves 1, so 1 is the exact minimum. The choices smaller than 1 — {0,

\frac{1}{4}

\frac{1}{2}

} — are all ruled out by the inequality. Choice (E) 2 is too large since 1 is actually attainable.

💡Key takeaway

This AMC 10 problem only needs Grade 8 algebra you already know — expand the two squares so the ± 2xy terms cancel, factor to (x² + 1)(y² + 1), see that each factor is at least 1, and check that (0, 0) gives the minimum 1.

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