AMC 10 · 2023 · #13

Grade 8 geometry-2d

absolute-valuecoordinate-geometrysymmetry-argumentarea-rectangles symmetry-argumentcaseworkidentify-subproblems ↑ Prerequisites: absolute-valuecoordinate-geometry

📏 Long solution 💡 3 insights

Problem

What is the area of the region in the coordinate plane defined by

$| | x | - 1 | + | | y | - 1 | \le 1$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 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.

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Toolkit + CCSS Solution

Understand

Restated: Find the area in the $(x, y)$-plane of the region where $\bigl|\,|x| - 1\,\bigr| + \bigl|\,|y| - 1\,\bigr| \le 1$.

Givens: Region $R = \{(x, y) : \bigl|\,|x|-1\,\bigr| + \bigl|\,|y|-1\,\bigr| \le 1\}$; Inner absolute values around $|x|$ and $|y|$ (nested twice); Answer choices: (A) $2$, (B) $8$, (C) $4$, (D) $15$, (E) $12$

Unknowns: Area of $R$

Understand

Restated: Find the area in the $(x, y)$-plane of the region where $\bigl|\,|x| - 1\,\bigr| + \bigl|\,|y| - 1\,\bigr| \le 1$.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems, #3 Eliminate Possibilities

Tool #1 (Draw a Diagram): the region is in the coordinate plane, so sketching beats algebra. Tool #9 (Easier Problem): the bare inequality $|x| + |y| \le 1$ is the familiar tilted unit square (a diamond) of area $2$; building $R$ from one of these in each quadrant is much easier than expanding the nested absolute values blindly. Tool #7 (Identify Subproblems): the symmetry $f(\pm x, \pm y) = f(x, y)$ chops the problem into "area in Q1" $\times 4$. Inside Q1 the equation simplifies to $|x - 1| + |y - 1| \le 1$, a single tilted unit square centered at $(1, 1)$ — compute that, multiply by $4$. Tool #3 (Eliminate): the choices ($2, 4, 8, 12, 15$) all differ by huge multiples; once you know the Q1 area is $2$, the only viable answer is $8$.

Execute — Answer: B

#7 Identify Subproblems 8.G.A.1 Step 1

Spot the symmetry.
Replacing $x$ by $-x$ leaves $|x|$ unchanged, hence leaves the whole inequality unchanged.
Same with $y \mapsto -y$.
So $R$ is symmetric across both axes — copies of the Q1 piece tile all four quadrants identically.

$$f(x,y) = \bigl|\,|x|-1\,\bigr| + \bigl|\,|y|-1\,\bigr|; \; f(-x,y) = f(x,-y) = f(x,y)$$

💡 Reflections across the axes are rigid motions that preserve $f$, so the region's pieces in each quadrant are congruent — Grade 8 reflection facts.

#9 Solve an Easier Related Problem 7.NS.A.1 Step 2

Restrict to the first quadrant ($x, y \ge 0$).
Then $|x| = x$ and $|y| = y$, so the inequality collapses to $|x - 1| + |y - 1| \le 1$.

$$\text{In Q1: } |x - 1| + |y - 1| \le 1$$

💡 Killing the outer absolute values in Q1 reduces a nested mess to one familiar tilted-square inequality — Tool #9's "strip away complexity" move.

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

Sketch the Q1 region.
The boundary $|x - 1| + |y - 1| = 1$ is a tilted square (taxicab unit circle) centered at $(1, 1)$.
The four vertices come from making one of $|x-1|, |y-1|$ equal $1$ and the other $0$: $(1 \pm 1, 1)$ and $(1, 1 \pm 1)$, i.e.
$(0, 1), (2, 1), (1, 0), (1, 2)$.
All four lie in the closed first quadrant ($x, y \ge 0$), so the tilted square sits entirely inside Q1.

$$\text{Vertices: } (0, 1), (2, 1), (1, 0), (1, 2)$$

💡 Plotting the four vertices makes the diamond visible — Grade 6 polygons on the coordinate plane.

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

Compute the Q1 area.
The tilted square has diagonals along the lines $y = 1$ (from $(0,1)$ to $(2,1)$) and $x = 1$ (from $(1,0)$ to $(1,2)$), each of length $2$.
A square (or rhombus) has area $\tfrac{1}{2} d_1 d_2$.

$$\text{Area}_{Q1} = \tfrac{1}{2} \cdot 2 \cdot 2 = 2$$

💡 Diagonal-product formula for a rhombus — Grade 6 polygon area.

#7 Identify Subproblems 8.G.A.1 Step 5

Multiply by $4$ for the four quadrants.
The Q2, Q3, Q4 copies are obtained by reflecting Q1's tilted square across the axes, giving four tilted squares centered at $(\pm 1, \pm 1)$ — none of them overlap because each sits inside its own quadrant.
Total area is $4 \times 2 = 8$.

$$\text{Area}(R) = 4 \cdot \text{Area}_{Q1} = 4 \cdot 2 = 8$$

💡 Four congruent non-overlapping copies — add their areas — Grade 8 use of rigid motions to combine.

#3 Eliminate Possibilities 6.G.A.1 Step 6

Match to the choices: $8$ is (B).
The other choices ($2, 4, 12, 15$) are common traps — $2$ forgets to multiply by $4$, $4$ doubles only twice, $12$ and $15$ overshoot.

$$\text{Area} = 8 \;\Rightarrow\; \textbf{(B)}$$

💡 Picking the matching choice is the multiple-choice closeout.

[1] #7 8.G.A.1 Spot the symmetry. Replacing $x$ by $-x$ leaves $|x|$ unchanged, hence leaves th

[2] #9 7.NS.A.1 Restrict to the first quadrant ($x, y \ge 0$). Then $|x| = x$ and $|y| = y$, so

[3] #1 6.G.A.3 Sketch the Q1 region. The boundary $|x - 1| + |y - 1| = 1$ is a tilted square (t

[4] #7 6.G.A.1 Compute the Q1 area. The tilted square has diagonals along the lines $y = 1$ (fr

[5] #7 8.G.A.1 Multiply by $4$ for the four quadrants. The Q2, Q3, Q4 copies are obtained by re

[6] #3 6.G.A.1 Match to the choices: $8$ is (B). The other choices ($2, 4, 12, 15$) are common

Review

Reasonableness: Mental picture check: $R$ is four tilted squares, one in each quadrant, centered at $(\pm 1, \pm 1)$, each with vertices at distance $1$ from its center along the axes. Each tilted square has "side length" $\sqrt{2}$ and area $(\sqrt{2})^2 = 2$. Four of them give $8$. Coordinate sanity-check at a boundary point: $(0, 1)$ gives $|0-1| + |1-1| = 1$ — on the boundary. $(0.5, 1)$ gives $|0.5-1| + 0 = 0.5 \le 1$ — inside. $(0, 2.5)$ gives $|0-1| + |2.5-1| = 1 + 1.5 = 2.5 \not\le 1$ — outside. Consistent with the four-diamond picture.

Alternative: Tool #10 (Physical / build) by case-split: in Q1 set $u = x - 1, v = y - 1$ (translate). The region becomes $|u| + |v| \le 1$ — the unit taxicab ball, area $2$. By the $\pm x, \pm y$ symmetry, total area $= 4 \times 2 = 8$. Same answer with explicit substitution rather than vertex calculation.

CCSS standards used (min grade 8)

8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations (Recognizing that reflections across the $x$- and $y$-axes leave the region invariant, so all four quadrant pieces are congruent.)
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers (Removing the outer absolute values in Q1 by using $|x| = x$ and $|y| = y$ when $x, y \ge 0$.)
6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Plotting the four vertices $(0, 1), (2, 1), (1, 0), (1, 2)$ of the tilted square in Q1.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Applying the rhombus area formula $\tfrac{1}{2} d_1 d_2$ to the tilted unit square in Q1.)

⭐ This AMC 10 problem only needs Grade 8 "reflections preserve the shape" you already know — the four quadrants each hold one identical tilted square of area $2$, and four of them give a total area of $8$.