AMC 8 · 2024 · #18

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

area-circlesratio-proportionfraction-arithmetic area-differenceratio-proportion ↑ Prerequisites: area-circlesfraction-arithmetic

📏 Long solution 💡 3 insights 📊 Diagram

Problem

Three circles share the same center O, with radii 1, 2, and 3. The whole inner ring (between the radius-1 and radius-2 circles) is shaded. Inside the outer ring (between the radius-2 and radius-3 circles), only a pie-slice with central angle ∠ BOC = x degrees is shaded. We are told the shaded area equals the unshaded area, and we must find x.

Pick an answer.

(A)

108

(B)

120

(C)

135

(D)

144

(E)

150

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shaded region is a compound shape (the inner ring plus a sector of the outer ring), so the right move is Tool #7 — split it into two pieces we already know how to compute, then add. We use Tool #1 to keep the picture straight (the inner ring is fully shaded, the outer ring only partially). After computing x, Tool #3 confirms it equals one of (A)-(E). We deliberately avoid Tool #13 (Algebra) — once we see that the outer ring needs to supply only 1.5π out of its 5π, a simple fraction-of-a-circle argument ( $\frac{1.5}{5}$ = $\frac{3}{10}$ of 360^°) gives the angle without ever writing an equation in x.

1STEP 1

By A = π r², the three circles have areas π, 4π, and 9π, so the whole figure has total area 9π.

A₁ = π, A₂ = 4π, A₃ = 9π → Total = 9π

2STEP 2

Split into two rings: the inner ring is 4π - π = 3π (all shaded), the outer ring is 9π - 4π = 5π (only partly shaded).

A_inner ring = 4π - π = 3π, A_outer ring = 9π - 4π = 5π

3STEP 3

Shaded = unshaded means each is half of 9π, so the shaded area must be 4.5π.

Shaded =

\frac{1}{2}

· 9π = 4.5π

4STEP 4

The inner ring gives 3π free, so the sector needs 4.5π - 3π = 1.5π, which is $\frac{1.5\pi}{5\pi}$ = $\frac{3}{10}$ of the outer ring.

Sector area needed = 4.5π - 3π = 1.5π,

\frac{1.5\pi}{5\pi}

\frac{3}{10}

5STEP 5

A sector's central angle is the same fraction of 360^° as of the area, so $\frac{3}{10}$ of 360^° gives ∠ BOC = 108^°.

∠ BOC =

\frac{3}{10}

· 360^° = 108^°

6STEP 6

Among 108, 120, 135, 144, 150, our 108^° is exactly choice (A); bigger angles would shade more than 4.5π.

108^° → (A)

Answer

108

Plug x = 108^° back into the shaded-area expression: 3π +

\frac{108}{360}

· 5π = 3π +

\frac{3}{10}

· 5π = 3π + 1.5π = 4.5π, which is exactly half of 9π. So shaded equals unshaded, as required. The angle is also reasonable by feel: the outer ring (5π) is bigger than the inner ring (3π), so we only need a modest slice of it — about

\frac{1}{3}

of the full circle, and indeed 108^° is slightly less than 120^° =

\frac{1}{3}

· 360^°.

💡Key takeaway

This AMC 8 problem only needs the Grade 7 circle-area formula A = π r² you already know!

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