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

Find the area in the (x, y)-plane of the region where | |x| - 1 | + | |y| - 1 | ≤ 1.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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| ≤ 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(± x, ± y) = f(x, y) chops the problem into "area in Q1" × 4. Inside Q1 the equation simplifies to |x - 1| + |y - 1| ≤ 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.

1STEP 1

Replacing x by -x or y by -y leaves the inequality unchanged, so R is symmetric across both axes — all four quadrants match.

f(x,y) = | |x|-1 | + | |y|-1 |; f(-x,y) = f(x,-y) = f(x,y)

2STEP 2

In the first quadrant |x| = x and |y| = y, so the inequality collapses to |x - 1| + |y - 1| ≤ 1.

In Q1: |x - 1| + |y - 1| ≤ 1

3STEP 3

The boundary is a tilted square centered at (1, 1) with vertices (0, 1), (2, 1), (1, 0), (1, 2) — all inside Q1.

Vertices: (0, 1), (2, 1), (1, 0), (1, 2)

4STEP 4

Both diagonals (along y = 1 and x = 1) have length 2, so the Q1 rhombus area is $\frac{1}{2}$ · 2 · 2 = 2.

Area_Q1 =

\frac{1}{2}

· 2 · 2 = 2

5STEP 5

Reflecting Q1's square into the other three quadrants gives four non-overlapping copies, so the total area is 4 × 2 = 8.

Area(R) = 4 · Area_Q1 = 4 · 2 = 8

6STEP 6

8 matches choice (B); the others (2, 4, 12, 15) are traps from under- or over-counting the copies.

Area = 8 → (B)

Answer

Mental picture check: R is four tilted squares, one in each quadrant, centered at (± 1, ± 1), each with vertices at distance 1 from its center along the axes. Each tilted square has "side length" √(2) and area (√(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 ≤ 1 — inside. (0, 2.5) gives |0-1| + |2.5-1| = 1 + 1.5 = 2.5 not ≤ 1 — outside. Consistent with the four-diamond picture.

💡Key takeaway

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.