AMC 10 · 2021 · #9

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition

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

📏 Medium solution 💡 2 insights

Problem

What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$ ?

Pick an answer.

(A)

(B)

$~\frac{1}{4}$

(C)

$~\frac{1}{2}$

(D)

(E)

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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

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

#3 Eliminate Possibilities 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.

$x + y = 0,\;\;xy = 1\;\Rightarrow\;-x^2 = 1$ (no real $x$)

💡 Two squares can sum to $0$ only if each is $0$ — and the system blocking real solutions kills choice (A).

#13 Convert to Algebra 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.

$$(xy-1)^2 + (x+y)^2 = x^2 y^2 - 2xy + 1 + x^2 + 2xy + y^2 = x^2 y^2 + x^2 + y^2 + 1$$

💡 Expanding both squares lets the cross terms $\pm 2xy$ destroy each other.

#15 Organize Information in More Ways 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.

$$x^2 y^2 + x^2 + y^2 + 1 = (x^2 + 1)(y^2 + 1)$$

💡 Spotting the common factor $(x^2 + 1)$ in two groups turns a sum into a product — the key rewrite.

#13 Convert to Algebra 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 $\ge 1$.

$$x^2 + 1 \ge 1,\;\; y^2 + 1 \ge 1\;\Rightarrow\;(x^2 + 1)(y^2 + 1) \ge 1$$

💡 A real number squared is never negative — the simplest possible inequality.

#6 Guess and Check 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: $(0 \cdot 0 - 1)^2 + (0 + 0)^2 = 1 + 0 = 1$ ✓.

$$(0, 0):\;\;(0 - 1)^2 + 0^2 = 1\;\checkmark$$

💡 Evaluate the original expression at the conjectured optimum to confirm equality.

#3 Eliminate Possibilities 4.NF.A.2 Step 6

The minimum is $1$, matching choice (D).

$$\min = 1\;\Rightarrow\;\textbf{(D)}$$

💡 Read off the answer choice that matches the computed minimum.

[1] #3 8.EE.A.2 First check whether the minimum could be $0$. If $(xy - 1)^2 + (x + y)^2 = 0$, b

[2] #13 6.EE.A.3 Expand the two squares: $(xy - 1)^2 = x^2 y^2 - 2xy + 1$ and $(x + y)^2 = x^2 +

[3] #15 6.EE.A.4 Reorganize: group $x^2 y^2 + y^2$ and $x^2 + 1$. Factor $y^2$ out of the first p

[4] #13 8.EE.A.2 Every real square satisfies $x^2 \ge 0$, so $x^2 + 1 \ge 1$ and $y^2 + 1 \ge 1$.

[5] #6 6.EE.A.2 Check that the lower bound $1$ is actually hit. Take $x = 0, y = 0$: $x^2 + 1 =

[6] #3 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.2 Use 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.3 Apply 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.4 Identify 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.2 Write, 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.2 Compare 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 $1$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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