AMC 10 · 2021 · #9
Grade 8 arithmeticProblem
What is the least possible value of (xy−1)2+(x+y)2 for real numbers x and y?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The expression is $E(x, y) = (xy - 1)^2 + (x + y)^2$; $x$ and $y$ range over all real numbers; Both summands are squares of real numbers, hence $\ge 0$; Choices: (A) $0$, (B) $\frac{1}{4}$, (C) $\frac{1}{2}$, (D) $1$, (E) $2$
Unknowns: The minimum value of $E(x, y)$ and the $(x, y)$ that achieves it
Understand
Restated: 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.
Givens: The expression is $E(x, y) = (xy - 1)^2 + (x + y)^2$; $x$ and $y$ range over all real numbers; Both summands are squares of real numbers, hence $\ge 0$; Choices: (A) $0$, (B) $\frac{1}{4}$, (C) $\frac{1}{2}$, (D) $1$, (E) $2$
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #13 Convert to Algebra, #6 Guess and Check, #3 Eliminate Possibilities
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 $(x^2 + 1)(y^2 + 1)$. 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.
Execute — Answer: D
8.EE.A.2 Step 1 - First check whether the minimum could be $0$.
- If $(xy - 1)^2 + (x + y)^2 = 0$, both squares must be $0$, forcing $xy = 1$ and $x + y = 0$.
- From $x + y = 0$, $y = -x$, so $xy = -x^2$.
- Then $-x^2 = 1$ has no real solution.
- So (A) $0$ is impossible.
💡 Two squares can sum to $0$ only if each is $0$ — and the system blocking real solutions kills choice (A).
6.EE.A.3 Step 2 - Expand the two squares: $(xy - 1)^2 = x^2 y^2 - 2xy + 1$ and $(x + y)^2 = x^2 + 2xy + y^2$.
- Add them — the $-2xy$ and $+2xy$ cancel.
💡 Expanding both squares lets the cross terms $\pm 2xy$ destroy each other.
6.EE.A.4 Step 3 - Reorganize: group $x^2 y^2 + y^2$ and $x^2 + 1$.
- Factor $y^2$ out of the first pair: $y^2 (x^2 + 1) + (x^2 + 1) = (x^2 + 1)(y^2 + 1)$.
- The expression is the product of two simple non-negative pieces.
💡 Spotting the common factor $(x^2 + 1)$ in two groups turns a sum into a product — the key rewrite.
8.EE.A.2 Step 4 - Every real square satisfies $x^2 \ge 0$, so $x^2 + 1 \ge 1$ and $y^2 + 1 \ge 1$.
- Multiplying two real numbers each $\ge 1$ gives a product $\ge 1$.
💡 A real number squared is never negative — the simplest possible inequality.
6.EE.A.2 Step 5 - Check that the lower bound $1$ is actually hit.
- Take $x = 0, y = 0$: $x^2 + 1 = 1$ and $y^2 + 1 = 1$, product $= 1$.
- Plug back into the original: $(0 \cdot 0 - 1)^2 + (0 + 0)^2 = 1 + 0 = 1$ ✓.
💡 Evaluate the original expression at the conjectured optimum to confirm equality.
4.NF.A.2 Step 6 The minimum is $1$, matching choice (D).
💡 Read off the answer choice that matches the computed minimum.
8.EE.A.2 First check whether the minimum could be $0$. If $(xy - 1)^2 + (x + y)^2 = 0$, b 6.EE.A.3 Expand the two squares: $(xy - 1)^2 = x^2 y^2 - 2xy + 1$ and $(x + y)^2 = x^2 + 6.EE.A.4 Reorganize: group $x^2 y^2 + y^2$ and $x^2 + 1$. Factor $y^2$ out of the first p 8.EE.A.2 Every real square satisfies $x^2 \ge 0$, so $x^2 + 1 \ge 1$ and $y^2 + 1 \ge 1$. 6.EE.A.2 Check that the lower bound $1$ is actually hit. Take $x = 0, y = 0$: $x^2 + 1 = 4.NF.A.2 The minimum is $1$, matching choice (D). Review
Reasonableness: Two independent confirmations: the algebraic factorization proves $(x^2+1)(y^2+1) \ge 1$ for all real $x, y$; the point $(0, 0)$ achieves $1$, so $1$ is the exact minimum. The choices smaller than $1$ — $\{0, \tfrac{1}{4}, \tfrac{1}{2}\}$ — are all ruled out by the inequality. Choice (E) $2$ is too large since $1$ is actually attainable.
Alternative: Tool #6 (pure Guess and Check) by trying simple symmetric pairs. $(0,0)\!\to\!1$; $(1,-1)\!\to\!(\!-1\!-\!1)^2 + 0^2 = 4$; $(1,1)\!\to\!0^2 + 2^2 = 4$. The smallest so far is $1$. This is suggestive but not a proof — the factorization path is the rigorous one.
CCSS standards used (min grade 8)
8.EE.A.2Use square root and cube root symbols to represent solutions (Recognizing $x^2 \ge 0$ for any real $x$ — the foundation of $x^2 + 1 \ge 1$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(xy-1)^2 + (x+y)^2$ and cancelling the $\pm 2xy$ cross terms.)6.EE.A.4Identify when two expressions are equivalent (Recognizing $x^2 y^2 + x^2 + y^2 + 1 = (x^2 + 1)(y^2 + 1)$ via the common factor $(x^2 + 1)$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating the original expression at $(x, y) = (0, 0)$ to confirm equality.)4.NF.A.2Compare two fractions with different numerators and different denominators (Selecting the matching numerical choice $1$ from the five options.)
⭐ This AMC 10 problem only needs Grade 8 algebra you already know — expand the two squares so the $\pm 2xy$ terms cancel, factor to $(x^2 + 1)(y^2 + 1)$, see that each factor is at least $1$, and check that $(0, 0)$ gives the minimum $1$.
⭐ This AMC 10 problem only needs Grade 8 algebra you already know — expand the two squares so the $\pm 2xy$ terms cancel, factor to $(x^2 + 1)(y^2 + 1)$, see that each factor is at least $1$, and check that $(0, 0)$ gives the minimum $1$.
More like this
Same archetype — closest grade level first.