Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #19

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
absolute-valuearea-circlesarea-rectanglescaseworksymmetry-argument caseworkidentify-subproblems ↑ Prerequisites: absolute-value
📏 Long solution 💡 3 insights
Problem
The graph x² + y² = 3|x - y| + 3|x + y| encloses a region whose area equals m + nπ for integers m and n. Find m + n.

Pick an answer.

(A)
18
(B)
27
(C)
36
(D)
45
(E)
54

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Tool #7 (Subproblems) — split the plane into four wedges by the two lines y = x and y = -x. In each wedge one of |x|, |y| dominates and the absolute values open without sign-juggling. Tool #13 (Algebra) — in each wedge the equation simplifies to a circle, e.g. (x - 3)² + y² = 9. Tool #5 (Pattern) — the four wedges produce four congruent circles, related by 90° rotations around the origin. Tool #1 (Diagram) — sketching the four arcs reveals a central square with four semicircular bumps, and the area splits into a square area + four half-circle areas.

1STEP 1

Use the absolute value identity

The identity |x - y| + |x + y| = 2 max(|x|, |y|) turns the right side into 6 max(|x|, |y|), so the equation splits along y = x and y = -x.

x² + y² = 6 max(|x|, |y|)
2STEP 2

Solve the right wedge

Right wedge x ≥ |y|: here max = x, so x² + y² = 6x, i.e. (x - 3)² + y² = 9 — a circle of radius 3 centered at (3, 0).

(x - 3)² + y² = 9
3STEP 3

Trace the semicircle arc

Within x ≥ |y|, x = 3 + 3cosθ, y = 3sinθ holds only for θ ∈ [-π/2, π/2] — the right semicircle from (3, -3) to (3, 3) through (6, 0).

Right semicircle from (3, -3) to (3, 3) through (6, 0)
4STEP 4

Rotate for the other wedges

By symmetry the other three wedges give the same circle rotated 90°, 180°, 270° — four circles centered at (± 3, 0) and (0, ± 3), radius 3.

Four circles at (± 3, 0), (0, ± 3), radius 3
5STEP 5

Sketch the whole region

The arc endpoints (± 3, ± 3) are the corners of a central square of side 6, with a semicircular bump on each of its four sides.

Bounded region = (square side 6) ∪ (4 semicircles radius 3)
6STEP 6

Add the square and semicircles

Square area 6 × 6 = 36; four semicircles 4 · 12\frac{1}{2} π · 3² = 18π; total enclosed area 36 + 18π.

Area = 36 + 18π
7STEP 7

Read off m plus n

Matching m + nπ: m = 36, n = 18, so m + n = 54 — choice (E).

m + n = 36 + 18 = 54 → (E)
Answer
54
Quick sanity: the region must contain the origin (where the right side is 0 and the left is 0 — boundary). The arcs reach out to (± 6, 0) and (0, ± 6), so the enclosed region fits inside a 12 × 12 square (area 144). Our answer 36 + 18π ≈ 36 + 56.5 ≈ 92.5 is comfortably less than 144 and bigger than the inscribed square area 36 — passes the sandwich check. Also each bulge has area 9π/2 ≈ 14.1, and four of them is ≈ 56.5, matching.
💡Key takeaway

The absolute-value sum |x-y|+|x+y| is just 2 max(|x|, |y|), so the equation splits the plane into four wedges. Each wedge gives a circle of radius 3, and four of them rotate around the origin into a clover: a central square (area 36) with four semicircle bumps (area 18π). Total area 36 + 18π, so m + n = (E) 54.

  • Use the absolute value identity
  • Solve the right wedge
  • Trace the semicircle arc
  • Rotate for the other wedges
  • Sketch the whole region
  • Add the square and semicircles
  • Read off m plus n

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