AMC 10 · 2021 · #7
Grade 8 algebraPick an answer.
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.
Can it be zero?
They cannot both vanish.
Two squares can sum to 0 only if each is 0 — and the system blocking real solutions kills choice (A).
8.EE.A.2Eliminate PossibilitiesExpand and tidy
Expanding kills the cross terms.
Expanding both squares lets the cross terms ± 2xy destroy each other.
6.EE.A.3Convert To AlgebraFactor it
What remains is a product of two factors.
Spotting the common factor (x² + 1) in two groups turns a sum into a product — the key rewrite.
Spotting the common factor in two groups turns a sum into a product.
▸ Why?
A factor shared by the terms can be lifted out front, which is exactly what factoring by grouping does.
▸ Why?
Once it is a product, its size is governed by the factors, and a zero can only come from a factor.
Read the floor
Both factors are at least one.
A real number squared is never negative — the simplest possible inequality.
8.EE.A.2Convert To AlgebraCheck the floor is reached
The origin actually reaches it.
Evaluate the original expression at the conjectured optimum to confirm equality.
6.EE.A.2Guess And CheckRead the minimum
The minimum is 1.
Read off the answer choice that matches the computed minimum.
4.NF.A.2Eliminate PossibilitiesThis AMC 12 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.