AMC 10 · 2004 · #13

Grade 8 countinggeometry-2d
Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position
combinations-basiccoordinate-geometrydouble-counting systematic-enumerationcasework ↑ Prerequisites: combinations-basic
📏 Long solution 💡 2 insights
Problem
S is the set of nine points whose two coordinates each come from {-1,0,1} — a 3 × 3 square array of lattice points. Count the distinct lines of the plane that contain at least two of these nine points.

Pick an answer.

(A)
8
(B)
20
(C)
24
(D)
27
(E)
36
How to solve
Strategy Make a Systematic List

The danger in this problem is double counting: the diagonal b=a carries three points and would be found three separate times if you listed pairs. The fix is to list the objects you are actually counting — lines — and to sort them by a feature that no two copies of the same line can disagree about. Direction is that feature. Because both coordinate gaps are at most 2, only a handful of directions are possible at all, so Tool #2 (Make a Systematic List) becomes a finite, checkable sweep: for each direction, list the lines with that direction. Tool #1 (Draw a Diagram) makes the grid concrete while sweeping. Tool #16 (Change Focus) supplies an independent cross-check afterwards — count pairs instead of lines, then repair the overcount — so two different routes must agree before the count is trusted.

1STEP 1

Plot the nine points

Plot the nine points: four corners, four edge midpoints and the centre.

S={(-1,1),(0,1),(1,1),(-1,0),(0,0),(1,0),(-1,-1),(0,-1),(1,-1)}
2STEP 2

Only eight directions are possible

The coordinate differences are tiny, so only eight directions are possible.

Δ a,Δ b∈{-2,-1,0,1,2} → slope∈{0, ±1/2, ± 1, ± 2} or vertical
3STEP 3

Sweep the vertical, horizontal, and slope ± 1 buckets

Rows, columns and the two diagonal directions give 12 lines.

3 (vertical)+3 (horizontal)+3 (slope 1)+3 (slope -1)=12
4STEP 4

Sweep the slope ± 2 and ±1/2 buckets

The four steeper and shallower directions add 8 more, corner to far midpoint.

b=2a± 1, b=-2a± 1, b=1/2a±1/2, b=-1/2a±1/2 → 4 × 2=8
5STEP 5

Total, then cross-check by counting pairs

Counting pairs and correcting for the three-point lines confirms 20, choice (B).

12+8=20; C(9, 2)=36, 36-8·(3-1)=36-16=20 → (B)
Answer
20
The count must sit strictly between 8 and 36. It is at most 36 because there are only 36 pairs of points, and strictly less because some pairs share a line — so (E) 36 is impossible. It is more than 8 because the 8 three-point lines are only the crowded ones, and pairs like (-1,-1) with (0,1) lie on none of them, so (A) 8 is impossible. The overcount repair fixes the value exactly: each of the 8 three-point lines is counted 3 times among the 36 pairs, costing 2 extra apiece, and 36-16=20. Choice (C) 24 is what you get by forgetting the two diagonals in the repair (36-2 · 6=24), and (D) 27 matches no consistent count at all. Two independent routes both land on 20.
💡Key takeaway

Sort the lines by direction instead of listing pairs of points: the grid allows only eight directions, and inside each one there are just a few places a line can sit.

  • Plot the nine points
  • Only eight directions are possible
  • Sweep the vertical, horizontal, and slope ± 1 buckets
  • Sweep the slope ± 2 and ±1/2 buckets
  • Total, then cross-check by counting pairs

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