AMC 10 · 2024 · #19

Grade 8 arithmetic

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Lattice Constraint Analysis

slope-interceptcoordinate-geometryratio-proportiongcdlogical-deduction caseworkidentify-subproblemspattern-recognition ↑ Prerequisites: slope-interceptcoordinate-geometryratio-proportion

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Build a 3 × 4 table. The rows are slope types: zero slope, nonzero rational slope, irrational slope. The columns are lattice-point counts on a nonvertical line: exactly 0, exactly 1, exactly 2, or two-or-more. Mark each of the 12 cells "Possible" if some line of that slope type contains exactly that many lattice points, otherwise "Not Possible". How many cells are "Possible"?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Use Matrix Logic

The problem literally hands you a 3 × 4 grid to fill — Tool #4 (Use Matrix Logic) is the structural fit: build the table row by row, derive each cell from a rule, count the ticks. Tool #7 (Identify Subproblems) splits the table into three slope-type rows; each row gets one shared key fact ("1 lattice point → infinitely many" for rational slopes; " ≤ 1 lattice point" for irrational slopes), and the four columns are filled from that fact. Tool #1 (Draw a Diagram) gives the proof that rational-slope lines with one lattice point have infinitely many: from (x₀, y₀) take the integer step (q, p) — picture a tilted staircase whose every step is a lattice point.

1STEP 1

Two lattice points on a nonvertical line give a rational slope — so an irrational slope allows at most one lattice point.

m =

\frac{y₂ - y₁}{x₂ - x₁}

∈ Q → m irrational implies ≤ 1 lattice point

2STEP 2

A rational slope $\frac{p}{q}$ through one lattice point hits another at every integer step (q, p) — so one lattice point forces infinitely many.

(x₀ + kq, y₀ + kp) ∈ line for every k ∈ Z

3STEP 3

Zero slope y = b: non-integer b → 0 points, integer b → infinitely many, and 1 or 2 impossible; Row 1 Possible: 2.

y = b: (0, No, No, ≥ 2) = (✓, ×, ×, ✓)

4STEP 4

Nonzero rational slope: y = $\frac{1}{2}$ x + $\frac{1}{3}$ gives 0, y = $\frac{2}{3}$ x gives infinitely many, middle columns blocked; Row 2 Possible: 2.

y =

\frac{p}{q}

x + b: (0, No, No, ≥ 2) = (✓, ×, ×, ✓)

5STEP 5

Irrational slope: y = √2 x + 0.5 gives 0, y = √2 x gives exactly 1, but 2 or more is impossible; Row 3 Possible: 2.

y = (irrat.)x + b: (0, 1, No, No) = (✓, ✓, ×, ×)

6STEP 6

Add the row tallies: 2 + 2 + 2 = 6 Possible cells → (C).

Total Possible = 2 + 2 + 2 = 6 → (C)

Answer

The filled grid: [ & 0 & 1 & 2 & ≥ 2 ; slope 0 & ✓ & × & × & ✓ ; nonzero rat. & ✓ & × & × & ✓ ; irrational & ✓ & ✓ & × & × ] Six ticks, six "Possible". The two structural rules — "rational slope: 1 → ∞" and "irrational slope: ≤ 1 ever" — close out the × cells crisply, and each ✓ cell has a named witness line. Answer (C) = 6 matches.

💡Key takeaway

This AMC 10 problem only needs the Grade 8 rule "slope between two lattice points is rational" — that single fact closes nine of the twelve cells, leaving exactly six "Possible"!

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