AMC 10 · 2003 · #15

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)

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

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

Halving the diameters gives radii 1 and 1/2, and the chord has length 1.

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

2STEP 2

Area of the small semicircle

The small half-disk has area π/8, the starting amount.

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

3STEP 3

Name the bite: a circular segment

The part inside the large semicircle is a circular segment above the chord, so the lune is π/8 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

The chord equals the radius, so the triangle is equilateral and the sector is π/6.

sector=60°/360° π R²=1/6π(1)²=π/6

5STEP 5

Segment = sector minus triangle

Segment is sector minus triangle, giving π/6 minus √3/4.

segment=sector-triangle=π/6-√3/4

6STEP 6

Assemble the lune

Subtracting and combining the pi terms gives √3/4 minus 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