AMC 8 · 2005 · #21
Grade 7 countingProblem
How many distinct triangles can be drawn using three of the dots below as vertices?
Pick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Six dots are arranged in a $2 \times 3$ grid (two rows of three dots each). How many distinct triangles can be drawn using three of these six dots as vertices?
Givens: $6$ dots on a grid: top row at $(0,1), (1,1), (2,1)$ and bottom row at $(0,0), (1,0), (2,0)$; A triangle is formed by choosing any $3$ of the $6$ dots as vertices; Three dots that lie on the same straight line do NOT form a triangle (degenerate case); Answer choices: (A) $9$, (B) $12$, (C) $18$, (D) $20$, (E) $24$
Unknowns: The number of distinct (non-degenerate) triangles
Understand
Restated: Six dots are arranged in a $2 \times 3$ grid (two rows of three dots each). How many distinct triangles can be drawn using three of these six dots as vertices?
Givens: $6$ dots on a grid: top row at $(0,1), (1,1), (2,1)$ and bottom row at $(0,0), (1,0), (2,0)$; A triangle is formed by choosing any $3$ of the $6$ dots as vertices; Three dots that lie on the same straight line do NOT form a triangle (degenerate case); Answer choices: (A) $9$, (B) $12$, (C) $18$, (D) $20$, (E) $24$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #1 Draw a Diagram, #2 Make a Systematic List
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 $2 \times 3$ 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.
Execute — Answer: C
7.SP.C.8 Step 1 - Count all ways to choose $3$ dots from $6$.
- Use the "choose" formula: pick the $1$st, $2$nd, $3$rd in order, then divide by the $3! = 6$ orderings of the same trio.
💡 This is the universe in the complement plan: every possible triple of dots, valid triangle or not.
5.G.A.1 Step 2 - Sketch the $2 \times 3$ grid and look for any three dots that lie on a single straight line.
- With dots at $(0,0), (1,0), (2,0), (0,1), (1,1), (2,1)$, the only candidates are the horizontal rows, the vertical columns, and the diagonals.
💡 A picture of the grid makes the straight-line triples visible at a glance — the rows are the obvious culprits.
7.SP.C.8 Step 3 - List the straight lines through $3$ grid dots, in order: rows first, then columns, then diagonals.
- Each column only has $2$ dots, so no column has a $3$-dot line.
- Each diagonal also hits at most $2$ of the $6$ dots, because the grid is only $2$ rows tall.
- That leaves only the two horizontal rows.
💡 Going row $\to$ column $\to$ diagonal in order guarantees no collinear triple is missed and none is double-counted.
7.SP.C.8 Step 4 - Subtract the degenerate triples from the total.
- There are exactly $2$ collinear triples (the top row and the bottom row), and each gives a flat "triangle" that is really just a line segment.
💡 Complement counting: valid $=$ total $-$ bad. The $2$ bad triples shave the $20$ choices down to the $18$ real triangles.
7.SP.C.8 Count all ways to choose $3$ dots from $6$. Use the "choose" formula: pick the $ 5.G.A.1 Sketch the $2 \times 3$ grid and look for any three dots that lie on a single st 7.SP.C.8 List the straight lines through $3$ grid dots, in order: rows first, then column 7.SP.C.8 Subtract the degenerate triples from the total. There are exactly $2$ collinear Review
Reasonableness: 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: $\binom{3}{1} \cdot \binom{3}{2} = 3 \cdot 3 = 9$. 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.
Alternative: Tool #2 (Systematic List) directly: split by how the $3$ dots distribute across the two rows. $0$ on bottom $\to$ all $3$ on top $\to$ collinear (skip). $3$ on bottom $\to$ collinear (skip). $1$ on top, $2$ on bottom $\to 3 \cdot 3 = 9$ triangles. $2$ on top, $1$ on bottom $\to 3 \cdot 3 = 9$ triangles. Total $= 18$ — same answer (C) without needing the binomial coefficient.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting $\binom{6}{3} = 20$ unordered triples of dots and enumerating the collinear triples to subtract — the same compound-event counting that underlies Grade 7 probability.)5.G.A.1Use a pair of perpendicular number lines (axes) to define a coordinate system; understand that the first number indicates how far to travel along the horizontal axis, and the second along the vertical axis (Placing the $6$ dots at coordinates $(0,0), (1,0), (2,0), (0,1), (1,1), (2,1)$ so collinearity can be checked by inspecting rows, columns, and diagonals.)
⭐ When most picks of $3$ are valid and only a few are not, count everything and subtract the bad ones — here $\binom{6}{3} = 20$ total minus the $2$ flat rows leaves $18$ real triangles.
⭐ When most picks of $3$ are valid and only a few are not, count everything and subtract the bad ones — here $\binom{6}{3} = 20$ total minus the $2$ flat rows leaves $18$ real triangles.
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