AMC 10 · 2021 · #19

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

absolute-valuearea-circlesarea-rectanglescaseworksymmetry-argument caseworkidentify-subproblems ↑ Prerequisites: absolute-value

📏 Long solution 💡 3 insights

Problem

The area of the region bounded by the graph of $x^2+y^2 = 3|x-y| + 3|x+y|$ is $m+n\pi$ , where $m$ and $n$ are integers. What is $m + n$ ?

Pick an answer.

(A)

~18

(B)

~27

(C)

~36

(D)

~45

(E)

~54

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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

#7 Identify Subproblems 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 $y = -x$.

$$x^2 + y^2 = 6 \max(|x|, |y|)$$

💡 $|x - y| + |x + y|$ measures twice the bigger of $|x|$ or $|y|$.

#13 Convert to Algebra 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 $(3, 0)$.

$$(x - 3)^2 + y^2 = 9$$

💡 Completing the square moves the equation from algebraic form to "circle" form.

#1 Draw a Diagram 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 $(6, 0)$.

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

💡 Half of the small circle lives inside the wedge; the other half is in the central square.

#5 Look for a Pattern 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 $(0, 6), (-6, 0), (0, -6)$.

Four circles at $(\pm 3, 0), (0, \pm 3)$, radius $3$

💡 Rotational symmetry $90°$ — same algebra in every wedge.

#1 Draw a Diagram 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.

Bounded region = (square side $6$) $\cup$ (4 semicircles radius $3$)

💡 Square in the middle, four half-pancakes glued to the four sides.

#7 Identify Subproblems 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: $36 + 18\pi$.

Area $= 36 + 18\pi$

💡 Square area $36$ plus four half-circles of radius $3$ gives $18\pi$.

#13 Convert to Algebra 4.NBT.B.4 Step 7

Matching $m + n\pi$ form: $m = 36$, $n = 18$, so $m + n = 36 + 18 = 54$ — choice (E).

$$m + n = 36 + 18 = 54 \Rightarrow \textbf{(E)}$$

💡 Read off $m$ and $n$ from the area and add.

[1] #7 6.NS.C.7 Identity check: $|x - y| + |x + y| = 2 \max(|x|, |y|)$. (Quick proof: if $|x| \g

[2] #13 8.G.B.7 Wedge 1 — right wedge $x \ge |y|$. Here $\max(|x|, |y|) = x$, so the equation be

[3] #1 5.G.A.2 Inside the wedge $x \ge |y|$, parametrize the circle as $x = 3 + 3\cos\theta$, $

[4] #5 8.G.A.3 By symmetry (Tool #5), the other three wedges give identical circles rotated by

[5] #1 7.G.B.6 Sketch the four semicircles. Their endpoints are $(\pm 3, \pm 3)$, the four corn

[6] #7 7.G.B.4 Area of the central square: $6 \times 6 = 36$. Area of the four semicircles: $4

[7] #13 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.7 Understand 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.7 Apply 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.2 Represent 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.3 Describe 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.6 Solve 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.4 Know 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.4 Fluently 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 $m + n = \textbf{(E) }54$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 8)

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