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AMC 8 · 2019 · #21

Grade 6 geometry-2dalgebra
Archetype Coordinate Plane Figure Computation
coordinate-geometryarea-trianglessystems-of-equationslinear-equations-two-var coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-trianglessystems-of-equations
📏 Medium solution 💡 2 insights

Problem

What is the area of the triangle formed by the lines y=5y=5, y=1+xy=1+x, and y=1−xy=1-x?

(A) 4(B) 8(C) 10(D) 12(E) 16\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16

Pick an answer.

(A)
4
(B)
8
(C)
10
(D)
12
(E)
16
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Toolkit + CCSS Solution

Understand

Restated: Three lines $y = 5$, $y = 1 + x$, and $y = 1 - x$ cut out a single triangle in the coordinate plane. Find the area of that triangle.

Givens: Horizontal line $y = 5$; Line $y = 1 + x$ (slope $+1$, $y$-intercept $1$); Line $y = 1 - x$ (slope $-1$, $y$-intercept $1$); Answer choices: (A) $4$, (B) $8$, (C) $10$, (D) $12$, (E) $16$

Unknowns: The area of the triangle bounded by the three lines

Understand

Restated: Three lines $y = 5$, $y = 1 + x$, and $y = 1 - x$ cut out a single triangle in the coordinate plane. Find the area of that triangle.

Givens: Horizontal line $y = 5$; Line $y = 1 + x$ (slope $+1$, $y$-intercept $1$); Line $y = 1 - x$ (slope $-1$, $y$-intercept $1$); Answer choices: (A) $4$, (B) $8$, (C) $10$, (D) $12$, (E) $16$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The problem hands us three lines but no picture, and the question is geometric — exactly the setup that begs for Tool #1 (Draw a Diagram). A quick sketch on the coordinate plane reveals that two of the three vertices lie on the horizontal line $y = 5$, which makes that segment a perfect horizontal base. Tool #7 (Identify Subproblems) then splits the work into three clean pieces: (a) find the three vertices, (b) read off the base length and the height from the picture, (c) apply the triangle area formula. Breaking the problem this way avoids any need for a distance formula or coordinate-geometry algebra beyond solving simple one-step equations.

Execute — Answer: E

#1 Draw a Diagram 5.G.A.1 Step 1
  • Sketch the three lines.
  • $y = 5$ is a horizontal line.
  • $y = 1 + x$ passes through $(0, 1)$ going up-right at $45^\circ$.
  • $y = 1 - x$ passes through $(0, 1)$ going up-left at $45^\circ$.
  • The picture shows a triangle with a flat top sitting on the line $y = 5$ and a sharp bottom vertex near $(0, 1)$.
Lines: $y = 5,\; y = 1 + x,\; y = 1 - x$

💡 Plotting points and lines on a coordinate grid is the Grade 5 'use a pair of perpendicular number lines' skill.

#7 Identify Subproblems 6.EE.B.7 Step 2
  • Find the top-right vertex where $y = 5$ meets $y = 1 + x$.
  • Substitute $y = 5$ into the slanted line and solve for $x$.
$$5 = 1 + x \;\Rightarrow\; x = 4 \;\Rightarrow\; (4,\, 5)$$

💡 Solving a one-step equation of the form $px = q$ (here $x = 4$) is exactly the Grade 6 equation-solving standard.

#7 Identify Subproblems 6.NS.C.6 Step 3
  • Find the top-left vertex where $y = 5$ meets $y = 1 - x$.
  • Substitute $y = 5$ and solve.
$$5 = 1 - x \;\Rightarrow\; x = -4 \;\Rightarrow\; (-4,\, 5)$$

💡 Locating $(-4, 5)$ on the grid uses the Grade 6 understanding that negative coordinates name points to the left of the $y$-axis.

#7 Identify Subproblems 6.EE.B.7 Step 4
  • Find the bottom vertex where the two slanted lines meet.
  • Set $1 + x = 1 - x$, giving $2x = 0$, so $x = 0$ and $y = 1$.
$$1 + x = 1 - x \;\Rightarrow\; x = 0,\; y = 1 \;\Rightarrow\; (0,\, 1)$$

💡 Setting two expressions equal and solving for the variable is the Grade 6 'write and solve equations' move.

#1 Draw a Diagram 6.G.A.3 Step 5
  • Read the base and height straight from the diagram.
  • The two top vertices $(-4, 5)$ and $(4, 5)$ share the height $y = 5$, so the segment between them is a horizontal base.
  • The bottom vertex $(0, 1)$ sits directly below the midpoint, so the height is the vertical drop from $y = 5$ to $y = 1$.
$$\text{base} = 4 - (-4) = 8,\quad \text{height} = 5 - 1 = 4$$

💡 Reading side lengths off a polygon drawn from its vertex coordinates is the Grade 6 'draw polygons in the coordinate plane' standard.

#7 Identify Subproblems 6.G.A.1 Step 6

Apply the triangle area formula with the base and height just found.

$$\text{Area} = \tfrac{1}{2} \times 8 \times 4 = 16 \;\Rightarrow\; \textbf{(E)}$$

💡 Finding the area of a triangle from its base and height is the Grade 6 triangle-area standard.

[1] #1 5.G.A.1 Sketch the three lines. $y = 5$ is a horizontal line. $y = 1 + x$ passes through
[2] #7 6.EE.B.7 Find the top-right vertex where $y = 5$ meets $y = 1 + x$. Substitute $y = 5$ in
[3] #7 6.NS.C.6 Find the top-left vertex where $y = 5$ meets $y = 1 - x$. Substitute $y = 5$ and
[4] #7 6.EE.B.7 Find the bottom vertex where the two slanted lines meet. Set $1 + x = 1 - x$, gi
[5] #1 6.G.A.3 Read the base and height straight from the diagram. The two top vertices $(-4, 5
[6] #7 6.G.A.1 Apply the triangle area formula with the base and height just found.

Review

Reasonableness: The triangle is isosceles with a horizontal base of length $8$ and a height of $4$, so its area is half of an $8 \times 4 = 32$ rectangle, which gives $16$. That matches choice (E). The answer is also plausible from the picture: the triangle fits snugly inside the rectangle with corners $(\pm 4, 1)$ and $(\pm 4, 5)$ — that rectangle has area $32$, and the triangle is clearly half of it.

Alternative: Tool #13 (Convert to Algebra) — use the Shoelace formula on the three vertices $(4, 5), (-4, 5), (0, 1)$: $\text{Area} = \tfrac{1}{2} \,|\, 4(5 - 1) + (-4)(1 - 5) + 0(5 - 5) \,| = \tfrac{1}{2}\,|16 + 16 + 0| = 16$. Same answer, but much heavier machinery than the simple base-times-height picture.

CCSS standards used (min grade 6)

  • 5.G.A.1 Use a pair of perpendicular number lines forming a coordinate system (Sketching the three lines on the coordinate plane so the triangle and its base on $y = 5$ become visible.)
  • 6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving the one-step equations $5 = 1 + x$ and $1 + x = 1 - x$ to find the $x$-coordinates of two of the triangle's vertices.)
  • 6.NS.C.6 Understand a rational number as a point on the number line (Locating the vertex $(-4, 5)$, which requires knowing that $-4$ is a point to the left of the $y$-axis on the coordinate plane.)
  • 6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Reading the base length $|4 - (-4)| = 8$ and the height $|5 - 1| = 4$ directly off the triangle drawn from its three vertex coordinates.)
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Applying $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$ to the triangle with base $8$ and height $4$.)

⭐ This AMC 8 problem only needs Grade 6 coordinate-plane and triangle-area skills you already know!

⭐ This AMC 8 problem only needs Grade 6 coordinate-plane and triangle-area skills you already know!

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