Competition · AMC preparation · step 4 of 4

AMC 8 · 2009 · #20

Grade 8 geometry-2d
Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role
coordinate-geometrypythagorean-theoremsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: coordinate-geometrypythagorean-theorem
📏 Long solution 💡 4 insights 📊 Diagram
Problem
Eight dots form two parallel rows of four, with the rows 1 unit apart and adjacent dots in a row 1 unit apart. How many triangles with vertices at three of these eight dots are non-congruent (i.e., not just rotations or reflections of one already counted)?

Pick an answer.

(A)
5
(B)
6
(C)
7
(D)
8
(E)
9

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

Trying all C(8, 3) = 56 triples is wasteful. Tool #1 (Draw a Diagram) lets us drop the dots onto a coordinate grid, so each triangle's shape is just a multiset of squared side lengths from the distance formula. Tool #2 (Make a Systematic List) organizes the search by base length on the bottom row (1, 2, or 3), then sweeps the third vertex across the top row in order. Tool #3 (Eliminate Possibilities) uses the grid's two reflection symmetries — top⇔bottom and left⇔right — to skip cases that must be congruent to ones already listed, and to drop the all-same-row triples (collinear, no triangle).

1STEP 1

Set up coordinates for the dots

Drop the dots on a coordinate grid so each triangle becomes a multiset of squared side lengths — no square roots needed to compare them.

d² = (Δ x)² + (Δ y)²
2STEP 2

Rule out impossible cases

Same-row triples are collinear, so drop them; by the top⇔bottom flip we keep only triangles whose base sits on the bottom row.

cases to enumerate = {base on bottom row, apex on top row}
3STEP 3

Sort by base length

Sort by base length — 1, 2, or 3 on the bottom row. Slide the apex across the top row, and the left⇔right flip skips mirror-image repeats.

bases = {1, 2, 3}
4STEP 4

Count base 1 triangles

Base B₁B₂: apex at T₁ gives {1,1,2}, T₃ gives {1,2,5}, T₄ gives {1,5,10}; T₂ mirrors T₁ — 3 new shapes.

Base 1: {1,1,2}, {1,2,5}, {1,5,10} → 3 shapes
5STEP 5

Count base 2 triangles

Base B₁B₃: apex at T₁ gives {1,4,5} (right), T₂ gives {2,2,4} (right isosceles), T₄ gives {2,4,10}; T₃ mirrors T₁ — 3 new shapes.

Base 2: {1,4,5}, {2,2,4}, {2,4,10} → 3 shapes
6STEP 6

Count base 3 triangles

Base B₁B₄: apex at T₁ gives {1,9,10} (right), T₂ gives {2,5,9}; T₃ mirrors T₂ and T₄ mirrors T₁ — 2 new shapes.

Base 3: {1,9,10}, {2,5,9} → 2 shapes
7STEP 7

Add the three counts

Add the buckets: 3 + 3 + 2 = 8 triangles; all eight squared-side multisets differ, so no two are secretly congruent — answer (D).

total = 3 + 3 + 2 = 8 → (D)
Answer
8
The eight squared-side-length multisets are pairwise different, so the count of 8 shapes is consistent. As a second check, any triangle whose base does not lie on a row — e.g., B₁(0,0), T₂(1,1), T₃(2,1) — must also fit one of the eight buckets: this one has squared sides {2, 5, 1}, matching Triangle 2. The choices (A) 5 through (E) 9 bracket the answer tightly, and (D) 8 matches both the case sum and the AMC 8 official answer.
💡Key takeaway

This AMC 8 problem only needs the Grade 8 distance idea — squared sides (Δ x)² + (Δ y)² — plus a careful systematic list to make sure no triangle is counted twice.

  • Set up coordinates for the dots
  • Rule out impossible cases
  • Sort by base length
  • Count base 1 triangles
  • Count base 2 triangles
  • Count base 3 triangles
  • Add the three counts

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