AMC 10 · 2021 · #15

Grade 8 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

systematic-enumerationcombinations-basicsymmetry-argumentfunction-evaluation caseworksymmetry-argument ↑ Prerequisites: combinations-basic

📏 Long solution 💡 3 insights

Problem

Pick FOUR distinct values from {1, 2, 3, 4, 5, 6} and label them A, B, C, D. Form two parabolas y = Ax² + B and y = Cx² + D. The two curves are an unordered pair (swapping (A, B) ⇔ (C, D) counts as the same choice). How many such labellings produce two parabolas that INTERSECT each other?

Pick an answer.

(A)

(B)

(C)

(D)

180

(E)

360

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

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): sketch two upward-opening parabolas with the same axis. The one with the larger leading coefficient is the 'narrower' one. Two upward parabolas with the same axis fail to intersect iff one is entirely above the other — which means the narrower one (larger A) also has the larger constant term (B). So they INTERSECT iff the larger leading coefficient is paired with the SMALLER constant. Tool #7 (Subproblems) splits the count into: (a) how many unordered curve-pairs total, (b) which fraction intersect. Tool #16 (Complement) lets us count non-intersecting pairs (the 'one above the other' configuration) and subtract — or just use the cleaner symmetry argument that exactly half intersect. Tool #2 (Systematic List) verifies on a small case (one specific 4-subset).

1STEP 1

Set the parabolas equal: (A - C)x² = D - B needs x² ≥ 0, so the curves intersect exactly when (A - C)(B - D) ＜ 0.

Intersect ⇔ (A - C)(B - D) ＜ 0

2STEP 2

Same-axis upward parabolas miss only when the narrower one (larger A) is also higher (larger B); they cross when the narrower starts lower.

Cross ⇔ (A ＞ C and B ＜ D) or (A ＜ C and B ＞ D)

3STEP 3

Count all unordered curve-pairs: C(6,4) = 15 subsets, each split into two ordered pairs in 12 ways, giving 15 · 12 = 180 pairs total.

Total = C(6, 4) · (C(4, 2))/2 · 2 · 2 = 15 · 12 = 180

4STEP 4

Swapping B ⇔ D sends (A - C)(B - D) to its negative, flipping the sign — every intersecting pair maps to a non-intersecting one.

B ⇔ D: (A - C)(B - D) ↦ -(A - C)(B - D)

5STEP 5

The 1-to-1 pairing splits the 180 total evenly, so 90 intersect and 90 don't.

Intersect =

\frac{180}{2}

= 90

6STEP 6

Check {1, 2, 3, 4}: list its 12 unordered curve-pairs and filter by the sign rule — 6 of the 12 intersect, matching the half-split.

{1,2,3,4}: 12 pairs, 6 intersect

7STEP 7

90 matches choice (C); 30 and 60 are too few, 180 skips the intersection filter, 360 skips the curve-swap symmetry.

90 → (C)

Answer

90 is exactly half of the total 180 unordered curve-pairs, which matches the intuition that the intersection condition is determined by the sign of (A-C)(B-D) — a symmetric coin-flip over the way we pair leading coefficients with constants. The wrong choices each correspond to a different undercounting/overcounting: 360 counts ordered curve pairs (before halving), 180 counts all pairs without the intersection filter, 60 =

\frac{180}{3}

and 30 =

\frac{90}{3}

are red herrings tied to mis-counting the within-pair assignments.

💡Key takeaway

This AMC 10 problem only needs Grade 8 understanding of parabolas plus simple counting you already know! Two upward parabolas with the same axis miss each other iff one is narrower AND higher — and intersect iff the narrower one starts LOWER. By swapping the two constants, intersecting and non-intersecting curve-pairs pair up 1-to-1, so exactly half of the 180 total pairs intersect: 90, answer (C).

More like this