Competition · AMC preparation · step 4 of 4
AMC 10 · 2019A · #19
Grade 8 arithmeticPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #15 (Reorganize): the four factors look messy, but pairing (x+1)(x+4) with (x+2)(x+3) makes both products equal to x² + 5x + (a constant) — the same hidden quantity in each. Tool #7 (Subproblems): once the products share a structure, substitute u = x² + 5x to reduce a quartic-looking expression to a quadratic. Tool #13 (Algebra): finish with a perfect square.
Pair the four factors
Pair (x+1)(x+4) = x² + 5x + 4 and (x+2)(x+3) = x² + 5x + 6 — both hide the same chunk x² + 5x.
Pair outside-with-inside — both products share x² + 5x.
6.EE.A.3Organize Information In More WaysSubstitute a single variable
Set u = x² + 5x + 5 (midpoint of +4 and +6): the product becomes (u-1)(u+1) = u² - 1.
Symmetric labels around a midpoint → classic (u-1)(u+1) = u² - 1.
Labels placed symmetrically about a midpoint turn the product into one square minus a constant.
▸ Why?
A sum times its matching difference is one square minus the other.
▸ Why?
Opening that product sends each piece against each piece, and the cross terms cancel.
Add 2019 back in
Add 2019: u² - 1 + 2019 = u² + 2018. Since u² ≥ 0, the minimum is 2018 at u = 0.
A square is ≥ 0 — so add the constant to read off the floor.
8.EE.A.2Convert To AlgebraCheck the minimum is reachable
Check u = 0 is reachable: x² + 5x + 5 = 0 has real roots x = , so 2018 is truly attained. Answer (B).
The quadratic u(x) = 0 has real roots, so the bound 2018 is achieved.
8.EE.A.2Convert To AlgebraThis AMC 10 problem only needs Grade 8 square-and-shift you already know — pair (x+1)(x+4) with (x+2)(x+3), substitute u = x² + 5x + 5 to get u² - 1 + 2019 = u² + 2018, and since u² ≥ 0 the minimum is 2018. The answer is (B).
- Pair the four factors
- Substitute a single variable
- Add 2019 back in
- Check the minimum is reachable
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