AMC 10 · 2003 · #19

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

area-circlescircular-sectorequilateral-triangle identify-subproblemscomplementary-counting ↑ Prerequisites: area-circles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A small semicircle with diameter 1 sits on top of a large semicircle with diameter 2. The flat edge of the small semicircle is a chord of the large semicircle, and both of its endpoints lie on the large arc. The lune is the region inside the small semicircle but outside the large semicircle. Find its area.

Pick an answer.

(A)

$\frac{1}{6}\pi-\frac{\sqrt{3}}{4}$

(B)

$\frac{\sqrt{3}}{4}-\frac{1}{12}\pi$

(C)

$\frac{\sqrt{3}}{4}-\frac{1}{24}\pi$

(D)

$\frac{\sqrt{3}}{4}+\frac{1}{24}\pi$

(E)

$\frac{\sqrt{3}}{4}+\frac{1}{12}\pi$

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The crescent has curved edges on both sides, so there is no single formula for it. Tool #7 (Identify Subproblems) breaks the shape into pieces whose areas we do know: a semicircle, a circular sector, and a triangle. Tool #1 (Draw a Diagram) is what reveals the key fact — the small semicircle's flat edge is a chord of the big circle, and drawing the two big radii to its ends exposes a triangle. Tool #16 (Change Focus / Count the Complement) supplies the framing: instead of chasing the odd crescent directly, take the whole small semicircle and subtract the single sliver of it that dips inside the big circle. That sliver is a plain circular segment we can measure.

1STEP 1

Fix the two radii

Halve each diameter: the radii are R=1 and r=1/2. The small flat edge, length 1, is a chord of the large circle.

R=2/2=1, r=1/2

2STEP 2

Area of the small semicircle

A semicircle is half a circle: its area is 1/2π r², so with r=1/2 that's π/8. The lune is this minus the part inside the big semicircle.

small semicircle=1/2π r²=1/2π(1/2)²=π/8

3STEP 3

Name the bite: a circular segment

The overlap is the circular segment of the big circle above the chord, so the lune equals the small semicircle minus that segment.

lune=π/8-(segment of big circle above the chord)

4STEP 4

The chord equals the big radius, so the angle is 60 degrees

Two big radii to the chord's ends give an equilateral triangle of side 1, so the central angle is 60° and the sector is π/6.

sector=60°/360° π R²=1/6π(1)²=π/6

5STEP 5

Segment = sector minus triangle

The segment is the sector minus the triangle. The equilateral triangle of side 1 has area √3/4, so the segment is π/6-√3/4.

segment=sector-triangle=π/6-√3/4

6STEP 6

Assemble the lune

Subtract: π/8-(π/6-√3/4)=3π/24-4π/24+√3/4, so the lune is √3/4-π/24 — choice (C).

lune=π/8-π/6+√3/4=√3/4-π/24 → (C)

Answer

√(3)/4-1/24π

The lune is a thin crescent, so its area should be a bit less than the whole small semicircle and clearly positive. Numerically √3/4-π/24≈ 0.433-0.131=0.302, while the small semicircle is π/8≈ 0.393 — the lune is a little smaller, exactly as expected after cutting off a small segment. Choice (A) is negative (≈-0.09) and impossible for an area; (B) ≈ 0.171 is too small; (C) ≈ 0.302 fits.

💡Key takeaway

The crescent equals the little half-circle minus the sliver of it that overlaps the big circle; since the flat edge equals the big radius, that sliver is a 60° pie slice with its equilateral triangle removed, leaving √3/4-π/24.

Fix the two radii
Area of the small semicircle
Name the bite: a circular segment
The chord equals the big radius, so the angle is 60 degrees
Segment = sector minus triangle
Assemble the lune