AMC 10 · 2021 · #19
Grade 8 geometry-2dProblem
The area of the region bounded by the graph ofx2+y2=3∣x−y∣+3∣x+y∣is m+nπ, where m and n are integers. What is m+n?
Pick an answer.
AMC 10 2021 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.
Toolkit + CCSS Solution
Understand
Restated: The graph $x^2 + y^2 = 3|x - y| + 3|x + y|$ encloses a region whose area equals $m + n\pi$ for integers $m$ and $n$. Find $m + n$.
Givens: Equation $x^2 + y^2 = 3|x - y| + 3|x + y|$; Enclosed area is $m + n\pi$ with $m, n$ integers; Choices: (A) $18$, (B) $27$, (C) $36$, (D) $45$, (E) $54$
Unknowns: $m + n$
Understand
Restated: The graph $x^2 + y^2 = 3|x - y| + 3|x + y|$ encloses a region whose area equals $m + n\pi$ for integers $m$ and $n$. Find $m + n$.
Givens: Equation $x^2 + y^2 = 3|x - y| + 3|x + y|$; Enclosed area is $m + n\pi$ with $m, n$ integers; Choices: (A) $18$, (B) $27$, (C) $36$, (D) $45$, (E) $54$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #5 Look for a Pattern, #13 Convert to Algebra
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. $(x - 3)^2 + y^2 = 9$. 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.
Execute — Answer: E
6.NS.C.7 Step 1 - Identity check: $|x - y| + |x + y| = 2 \max(|x|, |y|)$.
- (Quick proof: if $|x| \ge |y|$, both $x + y$ and $x - y$ have $|x|$-sign, so the sum is $|2x| = 2|x|$; symmetric if $|y| \ge |x|$.) So the right-hand side becomes $6 \max(|x|, |y|)$, and the equation splits along the lines $y = x$ and $y = -x$.
💡 $|x - y| + |x + y|$ measures twice the bigger of $|x|$ or $|y|$.
8.G.B.7 Step 2 - Wedge 1 — right wedge $x \ge |y|$.
- Here $\max(|x|, |y|) = x$, so the equation becomes $x^2 + y^2 = 6x$, i.e.
- $(x - 3)^2 + y^2 = 9$.
- This is a circle of radius $3$ centered at $(3, 0)$.
💡 Completing the square moves the equation from algebraic form to "circle" form.
5.G.A.2 Step 3 - Inside the wedge $x \ge |y|$, parametrize the circle as $x = 3 + 3\cos\theta$, $y = 3\sin\theta$.
- The wedge condition $x \ge |y|$ is satisfied exactly for $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, which traces the right semicircle from $(3, -3)$ to $(3, 3)$ bulging to $(6, 0)$.
💡 Half of the small circle lives inside the wedge; the other half is in the central square.
8.G.A.3 Step 4 - By symmetry (Tool #5), the other three wedges give identical circles rotated by $90°, 180°, 270°$ around the origin.
- Their centers are $(0, 3), (-3, 0), (0, -3)$ and each contributes its own semicircle of radius $3$ bulging outward to $(0, 6), (-6, 0), (0, -6)$.
💡 Rotational symmetry $90°$ — same algebra in every wedge.
7.G.B.6 Step 5 - Sketch the four semicircles.
- Their endpoints are $(\pm 3, \pm 3)$, the four corners of a central square with vertices $(3, 3), (-3, 3), (-3, -3), (3, -3)$ and side length $6$.
- The bounded region is this central square plus four semicircular bumps sticking out, one on each side.
💡 Square in the middle, four half-pancakes glued to the four sides.
7.G.B.4 Step 6 - Area of the central square: $6 \times 6 = 36$.
- Area of the four semicircles: $4 \cdot \tfrac{1}{2} \pi \cdot 3^2 = 18\pi$.
- Total enclosed area: $36 + 18\pi$.
💡 Square area $36$ plus four half-circles of radius $3$ gives $18\pi$.
4.NBT.B.4 Step 7 Matching $m + n\pi$ form: $m = 36$, $n = 18$, so $m + n = 36 + 18 = 54$ — choice (E).
💡 Read off $m$ and $n$ from the area and add.
6.NS.C.7 Identity check: $|x - y| + |x + y| = 2 \max(|x|, |y|)$. (Quick proof: if $|x| \g 8.G.B.7 Wedge 1 — right wedge $x \ge |y|$. Here $\max(|x|, |y|) = x$, so the equation be 5.G.A.2 Inside the wedge $x \ge |y|$, parametrize the circle as $x = 3 + 3\cos\theta$, $ 8.G.A.3 By symmetry (Tool #5), the other three wedges give identical circles rotated by 7.G.B.6 Sketch the four semicircles. Their endpoints are $(\pm 3, \pm 3)$, the four corn 7.G.B.4 Area of the central square: $6 \times 6 = 36$. Area of the four semicircles: $4 4.NBT.B.4 Matching $m + n\pi$ form: $m = 36$, $n = 18$, so $m + n = 36 + 18 = 54$ — choice Review
Reasonableness: 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 $(\pm 6, 0)$ and $(0, \pm 6)$, so the enclosed region fits inside a $12 \times 12$ square (area $144$). Our answer $36 + 18\pi \approx 36 + 56.5 \approx 92.5$ is comfortably less than $144$ and bigger than the inscribed square area $36$ — passes the sandwich check. Also each bulge has area $\tfrac{9\pi}{2} \approx 14.1$, and four of them is $\approx 56.5$, matching.
Alternative: Tool #10 (Physical Representation) — sketch graph paper, scale $1$ unit = $1$ cm, and draw the four circles by compass at radius $3$ centered at $(\pm 3, 0), (0, \pm 3)$. Cut along the outer envelope. Lay the cut-out flat: you can see the central square + four semicircle caps directly. Count squares for the square part (36 unit squares) and use $\pi r^2 / 2 = 4.5\pi$ per semicircle for the bumps, total $36 + 18\pi$.
CCSS standards used (min grade 8)
6.NS.C.7Understand ordering and absolute value of rational numbers (Simplifying $|x - y| + |x + y| = 2 \max(|x|, |y|)$ by reasoning about the signs of $x - y$ and $x + y$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Recognizing $(x - 3)^2 + y^2 = 9$ as a circle of radius $3$ via the distance-squared formula.)5.G.A.2Represent real-world and mathematical problems by graphing points (Locating the semicircle endpoints $(\pm 3, \pm 3)$ and the outer arc points $(\pm 6, 0), (0, \pm 6)$ on the coordinate plane.)8.G.A.3Describe the effect of dilations, translations, rotations, and reflections on coordinates (Using $90°$ rotational symmetry to copy one wedge's analysis to the other three wedges.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Decomposing the enclosed region into a square plus four semicircles before computing area.)7.G.B.4Know the formulas for area and circumference of a circle (Computing the area of each semicircle as $\tfrac{1}{2}\pi r^2 = \tfrac{9\pi}{2}$ and summing the four.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding $m + n = 36 + 18 = 54$ to get the final answer.)
⭐ 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 $18\pi$). Total area $36 + 18\pi$, so $m + n = \textbf{(E) }54$.
⭐ 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 $18\pi$). Total area $36 + 18\pi$, so $m + n = \textbf{(E) }54$.
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