AMC 8 · 2005 · #21

Grade 7 counting

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

combinations-basiccomplementary-countingsystematic-enumeration complementary-countingcasework ↑ Prerequisites: combinations-basic

📏 Short solution 💡 3 insights 📊 Diagram

Problem

Six dots are arranged in a 2 × 3 grid (two rows of three dots each). How many distinct triangles can be drawn using three of these six dots as vertices?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Counting valid triangles directly would mean checking every triple for collinearity — a lot of cases. The complement is far smaller: of all triples of 3 dots, only the collinear ones are NOT triangles. So Tool #16 (Complement) fits: count every triple, then subtract the few bad ones. Tool #1 (Draw a Diagram) is the natural sidekick — sketching the 2 × 3 grid makes the only straight lines through 3 dots jump out (the two horizontal rows). Tool #2 (Systematic List) confirms there are no other 3-dot lines by walking through the diagonals.

1STEP 1

Count every way to pick 3 of the 6 dots with order ignored — the combination C(6,3) gives 20 possible trios.

C(6, 3) =

\frac{6 · 5 · 4}{3 · 2 · 1}

\frac{120}{6}

= 20

2STEP 2

Sketch the 2 × 3 grid and hunt for any 3 dots on a single straight line — the only candidates are the rows, the columns, and the diagonals.

● ● ● ; ● ● ●

3STEP 3

Each column holds only 2 dots and every diagonal hits at most 2, so the only 3-dot straight lines are the two horizontal rows.

Top row: {(0,1),(1,1),(2,1)} ; Bottom row: {(0,0),(1,0),(2,0)} ; Columns: only 2 dots each — skip ; Diagonals: hit at most 2 dots — skip

4STEP 4

Subtract the 2 collinear rows from the 20 trios, leaving 18 genuine triangles.

20 - 2 = 18 → (C)

Answer

Direct cross-check by cases. Every valid triangle must have at least one dot on the top row and at least one on the bottom row (otherwise all three are on a single row — degenerate). Case 1: 1 dot on top, 2 on bottom. Choices: C(3, 1) · C(3, 2) = 3 · 3 = 9. Case 2: 2 on top, 1 on bottom. By symmetry, also 9. Total = 9 + 9 = 18, matching answer (C). The complement method and the case method agree, which confirms the count.

💡Key takeaway

When most picks of 3 are valid and only a few are not, count everything and subtract the bad ones — here C(6, 3) = 20 total minus the 2 flat rows leaves 18 real triangles.

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