AMC 8 · 2009 · #20

Grade 8 geometry-2d

Archetype Casework by Position / Vertex Type

coordinate-geometrypythagorean-theoremsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: coordinate-geometrypythagorean-theorem

📏 Long solution 💡 4 insights 📊 Diagram

Problem

How many non-congruent triangles have vertices at three of the eight points in the array shown below?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#1 Draw a Diagram 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.

$$d^2 = (\Delta x)^2 + (\Delta y)^2$$

💡 Putting the dots on a coordinate plane turns a picture problem into a Grade 6 "polygon by coordinates" problem.

#3 Eliminate Possibilities 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.

$$\text{cases to enumerate} = \{\text{base on bottom row, apex on top row}\}$$

💡 Cutting cases by symmetry — a reflection is a rigid motion, so reflected triangles are congruent — is a Grade 8 congruence move.

#2 Make a Systematic List 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.

$$\text{bases} = \{1, 2, 3\}$$

💡 Sorting the search by base length is the systematic-list move: every triangle lands in exactly one bucket and no bucket is checked twice.

#2 Make a Systematic List 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).

$$\text{Base } 1: \;\{1,1,2\},\ \{1,2,5\},\ \{1,5,10\} \;\Rightarrow\; 3 \text{ shapes}$$

💡 Applying the Pythagorean-distance formula to grid points is the Grade 8 "distance between two points" standard.

#2 Make a Systematic List 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).

$$\text{Base } 2: \;\{1,4,5\},\ \{2,2,4\},\ \{2,4,10\} \;\Rightarrow\; 3 \text{ shapes}$$

💡 Same systematic sweep as base $1$; the converse of the Pythagorean theorem flags the right triangles for free.

#2 Make a Systematic List 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.

$$\text{Base } 3: \;\{1,9,10\},\ \{2,5,9\} \;\Rightarrow\; 2 \text{ shapes}$$

💡 The full $B_1 B_4$ base has more left-right symmetry, so two of the four apex positions are mirror duplicates.

#2 Make a Systematic List 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.

$$\text{total} = 3 + 3 + 2 = 8 \;\Rightarrow\; \textbf{(D)}$$

💡 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.

[1] #1 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

[2] #3 8.G.A.2 Eliminate impossible cases. Three dots from the same row are collinear and form

[3] #2 4.OA.A.3 Sort by base length. The base on the bottom row has length $1$, $2$, or $3$ (the

[4] #2 8.G.B.8 Base $= 1$: take base $B_1 B_2$, squared length $1$. Slide the apex across $T_1,

[5] #2 8.G.B.8 Base $= 2$: take base $B_1 B_3$, squared length $4$. Apex at $T_1$: ${4, 1, 5}

[6] #2 8.G.B.8 Base $= 3$: take base $B_1 B_4$, squared length $9$. Apex at $T_1$: ${9, 1, 10\

[7] #2 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.3 Solve 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.3 Draw 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.2 Understand 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.8 Apply 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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