AMC 10 · 2003 · #19
Grade 8 geometry-2d
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The shaded piece has a jagged boundary made of four arcs, so measuring it head-on is awkward. Tool #16 (Count the Complement) flips the job: the shaded area is just the whole large semicircle minus the white region the small semicircles cover, and both of those are easy round shapes. To get the white area we still have to handle the overlaps, so tool #7 (Identify Subproblems) breaks the covered region into semicircles minus their shared lenses, and each lens into a triangle plus two circular slivers. Tool #1 (Draw a Diagram) with coordinates on AB keeps track of which semicircles actually overlap.
Set coordinates and the two basic areas
Origin at the midpoint of AB: A=(-2,0), B=(2,0). The big semicircle has area 2π; each small one, at (-1,0), (0,0), (1,0), has area π/2.
A semicircle is just half a circle, so its area is half of π r².
7.G.B.4Draw A DiagramRewrite the shaded area as a subtraction
Every small semicircle fits inside the big one, so shaded = 2π - white, where white is the region the three small ones cover.
It is easier to find the shaded strip by removing the round white part from the whole than to chase its wavy edge directly.
7.G.B.6Change Focus Count The ComplementSee which small semicircles overlap
Summing gives 3π/2, but the outer two only touch at the origin while the middle one overlaps each neighbor in a lens: white = 3π/2 - 2L.
When shapes overlap, adding their areas counts the shared part twice, so subtract each overlap once.
When shapes overlap, adding their areas counts the shared part twice, so each overlap is subtracted once.
▸ Why?
A count of overlapping regions charges the shared part once per region it belongs to.
▸ Why?
The covered region is exactly its pieces put together, so the correction restores the true total.
Measure one overlap lens
With radius 1 and centers 1 apart, each overlap is an equilateral triangle plus two 60° slivers: L = √3/4 + 2(π/6 - √3/4) = π/3 - √3/4.
Two radius-1 circles whose centers are 1 apart make an equilateral triangle, which pins every angle to 60°.
8.G.B.7Identify SubproblemsAdd up the white region
From the sum 3π/2 subtract both lenses: 3π/2 - 2(π/3 - √3/4) = 9π/6 - 4π/6 + √3/2 = 5π/6 + √3/2.
Once each overlap is removed exactly once, the leftover is the true area the semicircles cover.
7.G.B.6Identify SubproblemsSubtract to get the shaded area
Then shaded = 12π/6 - (5π/6 + √3/2) = 7π/6 - √3/2, which is choice (E).
The shaded area is whatever is left of the big semicircle after the white cover is taken away.
7.G.B.6Change Focus Count The ComplementWhen a region has a messy edge, measure the whole minus the easy leftover, and remember overlapping shapes share area you must subtract once.
- Set coordinates and the two basic areas
- Rewrite the shaded area as a subtraction
- See which small semicircles overlap
- Measure one overlap lens
- Add up the white region
- Subtract to get the shaded area
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