AMC 8 · 2009 · #20
Grade 8 geometry-2dProblem
How many non-congruent triangles have vertices at three of the eight points in the array shown below?
Pick an answer.
AMC 8 2009 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: 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)?
Givens: Two rows of $4$ equally spaced dots, the rows separated by the same spacing as the dots in a row; A triangle is formed by any $3$ chosen dots that are not all on one line; Two triangles are congruent when their three side lengths match; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Unknowns: The number of non-congruent triangles whose vertices are three of the eight dots
Understand
Restated: 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)?
Givens: Two rows of $4$ equally spaced dots, the rows separated by the same spacing as the dots in a row; A triangle is formed by any $3$ chosen dots that are not all on one line; Two triangles are congruent when their three side lengths match; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities
Trying all $\binom{8}{3} = 56$ 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$\leftrightarrow$bottom and left$\leftrightarrow$right — to skip cases that must be congruent to ones already listed, and to drop the all-same-row triples (collinear, no triangle).
Execute — Answer: D
6.G.A.3 Step 1 - Place the dots on a grid: bottom row $B_1, B_2, B_3, B_4$ at $(0,0), (1,0), (2,0), (3,0)$ and top row $T_1, T_2, T_3, T_4$ at $(0,1), (1,1), (2,1), (3,1)$.
- The squared distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2$.
- Comparing squared lengths is enough to decide congruence, so we never need to take a square root.
💡 Putting the dots on a coordinate plane turns a picture problem into a Grade 6 "polygon by coordinates" problem.
8.G.A.2 Step 2 - Eliminate impossible cases.
- Three dots from the same row are collinear and form no triangle, so cross off both all-bottom and all-top triples.
- Every other triangle has exactly two vertices on one row and one on the other.
- By the top$\leftrightarrow$bottom flip the two situations are mirror images, so we only need to enumerate triangles whose base sits on the bottom row.
💡 Cutting cases by symmetry — a reflection is a rigid motion, so reflected triangles are congruent — is a Grade 8 congruence move.
4.OA.A.3 Step 3 - Sort by base length.
- The base on the bottom row has length $1$, $2$, or $3$ (the only distinct horizontal gaps available).
- Inside each base length, use the left$\leftrightarrow$right flip to fix one base position and slide the apex $T$ across the top row.
- Two apex positions that are mirror images of each other give congruent triangles, so we only record the new shapes.
💡 Sorting the search by base length is the systematic-list move: every triangle lands in exactly one bucket and no bucket is checked twice.
8.G.B.8 Step 4 - Base $= 1$: take base $B_1 B_2$, squared length $1$.
- Slide the apex across $T_1, T_2, T_3, T_4$ and record the squared side lengths $\{|B_1 B_2|^2, |B_1 T|^2, |B_2 T|^2\}$.
- Apex at $T_1$: $\{1, 1, 2\}$ (new — Triangle 1).
- Apex at $T_2$: $\{1, 2, 1\}$ — same multiset as $T_1$, mirror image, skip.
- Apex at $T_3$: $\{1, 5, 2\}$ (new — Triangle 2).
- Apex at $T_4$: $\{1, 10, 5\}$ (new — Triangle 3).
💡 Applying the Pythagorean-distance formula to grid points is the Grade 8 "distance between two points" standard.
8.G.B.8 Step 5 - Base $= 2$: take base $B_1 B_3$, squared length $4$.
- Apex at $T_1$: $\{4, 1, 5\}$ (new — Triangle 4, a right triangle since $1 + 4 = 5$).
- Apex at $T_2$: $\{4, 2, 2\}$ (new — Triangle 5, a right isosceles triangle since $2 + 2 = 4$).
- Apex at $T_3$: $\{4, 5, 1\}$ — same as $T_1$, mirror image, skip.
- Apex at $T_4$: $\{4, 10, 2\}$ (new — Triangle 6).
💡 Same systematic sweep as base $1$; the converse of the Pythagorean theorem flags the right triangles for free.
8.G.B.8 Step 6 - Base $= 3$: take base $B_1 B_4$, squared length $9$.
- Apex at $T_1$: $\{9, 1, 10\}$ (new — Triangle 7, a right triangle since $1 + 9 = 10$).
- Apex at $T_2$: $\{9, 2, 5\}$ (new — Triangle 8).
- Apex at $T_3$: $\{9, 5, 2\}$ — same as $T_2$, mirror image, skip.
- Apex at $T_4$: $\{9, 10, 1\}$ — same as $T_1$, mirror image, skip.
💡 The full $B_1 B_4$ base has more left-right symmetry, so two of the four apex positions are mirror duplicates.
4.OA.A.3 Step 7 - Total the buckets.
- The systematic list found $3 + 3 + 2 = 8$ non-congruent triangles, and the eight side-length multisets $\{1,1,2\}, \{1,2,5\}, \{1,5,10\}, \{1,4,5\}, \{2,2,4\}, \{2,4,10\}, \{1,9,10\}, \{2,5,9\}$ are all different, so no two of the eight are secretly congruent.
💡 Adding the case counts is the Grade 4 multi-step-word-problem finish: each case is a clean subtotal, and the answer is just their sum.
6.G.A.3 Place the dots on a grid: bottom row $B_1, B_2, B_3, B_4$ at $(0,0), (1,0), (2,0 8.G.A.2 Eliminate impossible cases. Three dots from the same row are collinear and form 4.OA.A.3 Sort by base length. The base on the bottom row has length $1$, $2$, or $3$ (the 8.G.B.8 Base $= 1$: take base $B_1 B_2$, squared length $1$. Slide the apex across $T_1, 8.G.B.8 Base $= 2$: take base $B_1 B_3$, squared length $4$. Apex at $T_1$: ${4, 1, 5} 8.G.B.8 Base $= 3$: take base $B_1 B_4$, squared length $9$. Apex at $T_1$: ${9, 1, 10\ 4.OA.A.3 Total the buckets. The systematic list found $3 + 3 + 2 = 8$ non-congruent trian Review
Reasonableness: 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., $B_1(0,0), T_2(1,1), T_3(2,1)$ — 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.
Alternative: Tool #11 (Eliminate Possibilities) on the choices. The base-$1$ case alone produces three clearly different shapes ($\{1,1,2\}, \{1,2,5\}, \{1,5,10\}$), and the base-$2$ case adds an isosceles right triangle $\{2,2,4\}$ — that is already $4$ distinct shapes, so $(A) 5$ is barely tight. One more shape from base $2$ ($\{1,4,5\}$) and another ($\{2,4,10\}$) raise the count to $6$, eliminating $(A)$ and $(B)$. Two more shapes from base $3$ ($\{1,9,10\}$ and $\{2,5,9\}$) finish at $8$, ruling out $(C)$ and (since no further base length exists) $(E)$ as well.
CCSS standards used (min grade 8)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Splitting the count into three base-length buckets and adding the subtotals $3 + 3 + 2 = 8$.)6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Assigning coordinates $(0,0)$ through $(3,1)$ to the eight dots so that each triangle is a polygon in the coordinate plane.)8.G.A.2Understand that a two-dimensional figure is congruent to another using transformations (Using the top$\leftrightarrow$bottom and left$\leftrightarrow$right reflections (rigid motions) to declare mirror-image triangles congruent and prune the search.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Computing each squared side length $(\Delta x)^2 + (\Delta y)^2$ and comparing the multisets to decide non-congruence.)
⭐ This AMC 8 problem only needs the Grade 8 distance idea — squared sides $(\Delta x)^2 + (\Delta y)^2$ — plus a careful systematic list to make sure no triangle is counted twice.
⭐ This AMC 8 problem only needs the Grade 8 distance idea — squared sides $(\Delta x)^2 + (\Delta y)^2$ — plus a careful systematic list to make sure no triangle is counted twice.
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