AMC 8 · 2015 · #19

Grade 6 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Lattice or Non-Aligned Polygon Area

coordinate-geometryarea-trianglesarea-rectanglesfraction-arithmetic coordinate-geometryarea-differenceidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A triangle with vertices as $A=(1,3)$ , $B=(5,1)$ , and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle?

Pick an answer.

(A)

$frac{1}{6}$

(B)

$frac{1}{5}$

(C)

$frac{1}{4}$

(D)

$frac{1}{3}$

(E)

$frac{1}{2}$

AMC 8 2015 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A triangle has vertices $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ on a $6 \times 5$ grid. What fraction of the grid does the triangle cover?

Givens: Grid dimensions: $6$ wide by $5$ tall, so the grid has area $30$; Triangle vertices in the coordinate plane: $A=(1,3)$, $B=(5,1)$, $C=(4,4)$; Answer choices: (A) $\tfrac{1}{6}$, (B) $\tfrac{1}{5}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{1}{2}$

Unknowns: The ratio $\dfrac{\text{area of }\triangle ABC}{\text{area of the }6 \times 5\text{ grid}}$

Understand

Restated: A triangle has vertices $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ on a $6 \times 5$ grid. What fraction of the grid does the triangle cover?

Plan

Primary tool: #1 Draw a Diagram

Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems

The triangle is tilted, so its base and height are not obvious from the picture — Tool #1 (Draw a Diagram) earns its keep by letting us sketch the smallest axis-aligned rectangle that hugs the triangle. Inside that bounding rectangle, the triangle plus three right triangles in the corners fill the whole rectangle, so Tool #16 (Count the Complement) gives us a clean subtraction move: instead of finding the tilted triangle directly, find the three easy right triangles and subtract. Tool #7 (Identify Subproblems) splits the final question into two clean parts — first compute the triangle's area, then compute the requested fraction by dividing by the grid area.

Execute — Answer: A

#7 Identify Subproblems 3.MD.C.7 Step 1

Compute the area of the entire $6 \times 5$ grid.
This is the denominator of the final fraction.

$$\text{grid area} = 6 \times 5 = 30$$

💡 Treating the grid as one rectangle whose area we need is the Grade 3 area-of-a-rectangle move; setting it aside as a separate subproblem is Tool #7.

#1 Draw a Diagram 6.G.A.3 Step 2

Draw the smallest axis-aligned rectangle that encloses the triangle.
Use the extreme $x$- and $y$-coordinates of $A$, $B$, $C$: $x$ ranges from $1$ to $5$ (width $4$) and $y$ ranges from $1$ to $4$ (height $3$).

$$\text{bounding rectangle area} = (5-1) \times (4-1) = 4 \times 3 = 12$$

💡 Plotting the three points and boxing them in is a classic Grade 6 coordinate-geometry move — using coordinates to find lengths of horizontal and vertical sides.

#16 Change Focus / Count the Complement 5.G.A.2 Step 3

The bounding rectangle is filled by $\triangle ABC$ plus three right triangles in the corners.
Identify each corner triangle by pairing two vertices of $\triangle ABC$ with the rectangle corner between them.

$T_1: A(1,3), C(4,4), (1,4)$ \;\; $T_2: C(4,4), B(5,1), (5,4)$ \;\; $T_3: A(1,3), B(5,1), (1,1)$

💡 Shifting attention from the tilted triangle to the easy right triangles around it is the Tool #16 "count the complement" move; reading off the corner coordinates from the plotted points is the Grade 5 coordinate-plane skill.

#16 Change Focus / Count the Complement 6.G.A.1 Step 4

Each corner triangle has horizontal and vertical legs, so its area is $\tfrac{1}{2}\cdot\text{leg}\cdot\text{leg}$.
Compute and add.

$T_1 = \tfrac{1}{2}(3)(1) = 1.5$ \;\; $T_2 = \tfrac{1}{2}(1)(3) = 1.5$ \;\; $T_3 = \tfrac{1}{2}(4)(2) = 4$ \;\Rightarrow\; T_1+T_2+T_3 = 7$

💡 Finding the area of a tilted triangle by decomposing the bounding rectangle into easier right triangles is the Grade 6 standard for areas of polygons in the coordinate plane.

#16 Change Focus / Count the Complement 6.G.A.1 Step 5

Subtract the three corner triangles from the bounding rectangle to get the area of $\triangle ABC$.

$$\text{Area}(\triangle ABC) = 12 - 7 = 5$$

💡 "Whole minus the complement" is the heart of Tool #16 and the coordinate-geometry area technique.

#7 Identify Subproblems 4.NF.A.1 Step 6

Form the requested fraction (triangle area over grid area) and reduce to lowest terms.

$$\dfrac{\text{Area}(\triangle ABC)}{\text{grid area}} = \dfrac{5}{30} = \dfrac{1}{6} \;\Rightarrow\; \textbf{(A)}$$

💡 Dividing both numerator and denominator by $5$ to get $\tfrac{1}{6}$ is the Grade 4 equivalent-fractions move.

[1] #7 3.MD.C.7 Compute the area of the entire $6 \times 5$ grid. This is the denominator of the

[2] #1 6.G.A.3 Draw the smallest axis-aligned rectangle that encloses the triangle. Use the ext

[3] #16 5.G.A.2 The bounding rectangle is filled by $\triangle ABC$ plus three right triangles i

[4] #16 6.G.A.1 Each corner triangle has horizontal and vertical legs, so its area is $\tfrac{1}

[5] #16 6.G.A.1 Subtract the three corner triangles from the bounding rectangle to get the area

[6] #7 4.NF.A.1 Form the requested fraction (triangle area over grid area) and reduce to lowest

Review

Reasonableness: The triangle is clearly smaller than half the grid and larger than nothing, so the answer should be a small but nonzero fraction. $\tfrac{1}{6}$ of $30$ is $5$, which matches our computed triangle area. The bounding rectangle ($12$) is itself $\tfrac{12}{30} = \tfrac{2}{5}$ of the grid, and the triangle takes up a bit less than half of that rectangle ($\tfrac{5}{12}$), so $\tfrac{2}{5} \cdot \tfrac{5}{12} = \tfrac{1}{6}$ — consistent.

Alternative: Tool #8 (Analyze the Units) suggests the Shoelace formula: $\text{Area} = \tfrac{1}{2}|x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)| = \tfrac{1}{2}|1(1-4) + 5(4-3) + 4(3-1)| = \tfrac{1}{2}|-3 + 5 + 8| = \tfrac{1}{2}(10) = 5$. Same triangle area, so the fraction is again $\tfrac{5}{30} = \tfrac{1}{6}$, confirming (A).

CCSS standards used (min grade 6)

3.MD.C.7 Relate area to multiplication and find areas of rectangles (Computing the area of the $6 \times 5$ grid as $6 \times 5 = 30$ — the denominator of the requested fraction.)
4.NF.A.1 Explain and recognize equivalent fractions (Reducing $\tfrac{5}{30}$ to $\tfrac{1}{6}$ by dividing numerator and denominator by $5$.)
5.G.A.2 Represent real-world and mathematical problems by graphing points in the first quadrant of the coordinate plane (Plotting the three vertices $A(1,3)$, $B(5,1)$, $C(4,4)$ and reading off the coordinates of the bounding rectangle's corners.)
6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find side lengths (Identifying the bounding rectangle's width $5-1=4$ and height $4-1=3$ from the extreme $x$- and $y$-coordinates.)
6.G.A.1 Find the area of polygons by composing into rectangles or decomposing into triangles (Decomposing the bounding rectangle into $\triangle ABC$ plus three right corner triangles, and computing $12 - (1.5 + 1.5 + 4) = 5$.)

⭐ This AMC 8 problem only needs Grade 6 coordinate geometry — box the tilted triangle, subtract the easy corners, then simplify the fraction!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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