AMC 10 · 2003 · #19
Grade 8 geometry-2dThree semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A big semicircle of radius $2$ sits on diameter $\overline{AB}$. On that same diameter three small semicircles of radius $1$ are built, their flat sides also on $\overline{AB}$, with centers spaced evenly so they cut $\overline{AB}$ into four equal pieces. Find the area of the part inside the big semicircle but outside all three small ones (the shaded part).
Givens: The large semicircle has radius $2$; its diameter is $\overline{AB}$, of length $4$; Three small semicircles each have radius $1$ and sit on the same diameter; The three small centers split $\overline{AB}$ into four equal segments of length $1$; Shaded = inside the big semicircle AND outside every small semicircle; Answer choices: (A) $\pi-\sqrt3$, (B) $\pi-\sqrt2$, (C) $\frac{\pi+\sqrt2}{2}$, (D) $\frac{\pi+\sqrt3}{2}$, (E) $\frac{7}{6}\pi-\frac{\sqrt3}{2}$
Unknowns: The area of the shaded region
Understand
Restated: A big semicircle of radius $2$ sits on diameter $\overline{AB}$. On that same diameter three small semicircles of radius $1$ are built, their flat sides also on $\overline{AB}$, with centers spaced evenly so they cut $\overline{AB}$ into four equal pieces. Find the area of the part inside the big semicircle but outside all three small ones (the shaded part).
Givens: The large semicircle has radius $2$; its diameter is $\overline{AB}$, of length $4$; Three small semicircles each have radius $1$ and sit on the same diameter; The three small centers split $\overline{AB}$ into four equal segments of length $1$; Shaded = inside the big semicircle AND outside every small semicircle; Answer choices: (A) $\pi-\sqrt3$, (B) $\pi-\sqrt2$, (C) $\frac{\pi+\sqrt2}{2}$, (D) $\frac{\pi+\sqrt3}{2}$, (E) $\frac{7}{6}\pi-\frac{\sqrt3}{2}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
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 $\overline{AB}$ keeps track of which semicircles actually overlap.
Execute — Answer: E
7.G.B.4 Step 1 Set coordinates and the two basic areas
- Put the origin at the midpoint of $\overline{AB}$, so $A=(-2,0)$ and $B=(2,0)$.
- The large semicircle has radius $2$, so its area is half of $\pi r^2$: $\tfrac12\pi(2)^2=2\pi$.
- The three small centers sit at $(-1,0)$, $(0,0)$, and $(1,0)$, each a radius-$1$ semicircle of area $\tfrac12\pi(1)^2=\tfrac{\pi}{2}$.
💡 A semicircle is just half a circle, so its area is half of $\pi r^2$.
7.G.B.6 Step 2 Rewrite the shaded area as a subtraction
- Every small semicircle fits inside the big one, so the white (unshaded) part is exactly the region the three small semicircles cover together.
- That means the shaded area equals the big semicircle minus the white region: $\text{shaded}=2\pi-\text{white}$.
- Now the only real work is measuring the white region.
💡 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.6 Step 3 See which small semicircles overlap
- Adding the three areas $\tfrac{\pi}{2}+\tfrac{\pi}{2}+\tfrac{\pi}{2}=\tfrac{3\pi}{2}$ double-counts wherever two of them cover the same spot.
- The two outer semicircles have centers $(-1,0)$ and $(1,0)$, a distance $2$ apart, which is the sum of their radii, so they only touch at the origin and share no area.
- The middle semicircle sits $1$ away from each outer one (radius plus radius is $2>1$), so it overlaps each outer neighbor in a lens.
- So $\text{white}=\tfrac{3\pi}{2}-(\text{two equal lenses})$.
💡 When shapes overlap, adding their areas counts the shared part twice, so subtract each overlap once.
8.G.B.7 Step 4 Measure one overlap lens
- Take the middle semicircle (center $O=(0,0)$) and the right one (center $C=(1,0)$); both have radius $1$ and their centers are $1$ apart.
- They cross at $P=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$: the height $\tfrac{\sqrt3}{2}$ comes from the right triangle with legs $\tfrac12$ and $h$ and hypotenuse $1$, since $h=\sqrt{1^2-(\tfrac12)^2}=\tfrac{\sqrt3}{2}$.
- Triangle $OCP$ has all sides $1$, so it is equilateral with $60^\circ$ angles and area $\tfrac{\sqrt3}{4}$.
- The overlap is that triangle plus two circular slivers, one against each center; each sliver is a $60^\circ$ sector minus a copy of the triangle: $\tfrac{60}{360}\pi(1)^2-\tfrac{\sqrt3}{4}=\tfrac{\pi}{6}-\tfrac{\sqrt3}{4}$.
- So $L=\tfrac{\sqrt3}{4}+2\left(\tfrac{\pi}{6}-\tfrac{\sqrt3}{4}\right)=\tfrac{\pi}{3}-\tfrac{\sqrt3}{4}$.
💡 Two radius-$1$ circles whose centers are $1$ apart make an equilateral triangle, which pins every angle to $60^\circ$.
7.G.B.6 Step 5 Add up the white region
- Subtract the two lenses from the tripled semicircle: $\text{white}=\tfrac{3\pi}{2}-2\left(\tfrac{\pi}{3}-\tfrac{\sqrt3}{4}\right)=\tfrac{3\pi}{2}-\tfrac{2\pi}{3}+\tfrac{\sqrt3}{2}$.
- Combine the $\pi$ terms over a common denominator $6$: $\tfrac{9\pi}{6}-\tfrac{4\pi}{6}=\tfrac{5\pi}{6}$.
- So $\text{white}=\tfrac{5\pi}{6}+\tfrac{\sqrt3}{2}$.
💡 Once each overlap is removed exactly once, the leftover is the true area the semicircles cover.
7.G.B.6 Step 6 Subtract to get the shaded area
- Put the white region back into $\text{shaded}=2\pi-\text{white}$: $\text{shaded}=2\pi-\left(\tfrac{5\pi}{6}+\tfrac{\sqrt3}{2}\right)$.
- Write $2\pi=\tfrac{12\pi}{6}$, so $\tfrac{12\pi}{6}-\tfrac{5\pi}{6}=\tfrac{7\pi}{6}$, leaving $\text{shaded}=\tfrac{7\pi}{6}-\tfrac{\sqrt3}{2}$.
- That matches choice (E).
💡 The shaded area is whatever is left of the big semicircle after the white cover is taken away.
7.G.B.4 Put the origin at the midpoint of $\overline{AB}$, so $A=(-2,0)$ and $B=(2,0)$. 7.G.B.6 Every small semicircle fits inside the big one, so the white (unshaded) part is 7.G.B.6 Adding the three areas $\tfrac{\pi}{2}+\tfrac{\pi}{2}+\tfrac{\pi}{2}=\tfrac{3\pi 8.G.B.7 Take the middle semicircle (center $O=(0,0)$) and the right one (center $C=(1,0) 7.G.B.6 Subtract the two lenses from the tripled semicircle: $\text{white}=\tfrac{3\pi}{ 7.G.B.6 Put the white region back into $\text{shaded}=2\pi-\text{white}$: $\text{shaded} Review
Reasonableness: Numerically $\tfrac{7}{6}\pi-\tfrac{\sqrt3}{2}\approx 3.665-0.866=2.80$. That is comfortably less than the whole large semicircle $2\pi\approx 6.28$ and comes out positive, as a real area must. The white cover is about $\tfrac{5\pi}{6}+\tfrac{\sqrt3}{2}\approx 2.618+0.866=3.48$, and $6.28-3.48=2.80$ agrees. Choices (A) and (B) have no $\tfrac{1}{6}\pi$ piece and choices (C) and (D) add $\sqrt{\ }$ instead of subtracting, so only (E) can carry both the $\tfrac{7}{6}\pi$ and the $-\tfrac{\sqrt3}{2}$ that the computation forces.
Alternative: Instead of subtracting overlaps, build the white region up from its own pieces, as the reference solution does. The four radii drawn to the two crossing points split the covered region into a $120^\circ$ sector of a radius-$1$ circle (the middle sector) plus two equilateral triangles, then the outer semicircle caps; carefully assembled these give the same $\tfrac{5\pi}{6}+\tfrac{\sqrt3}{2}$. Either way $\text{shaded}=2\pi-\big(\tfrac{5\pi}{6}+\tfrac{\sqrt3}{2}\big)=\tfrac{7}{6}\pi-\tfrac{\sqrt3}{2}$.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Finding the area of each semicircle as half of $\pi r^2$ and a $60^\circ$ sector as $\tfrac{60}{360}$ of a circle.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Combining and subtracting the composite regions: shaded = big semicircle minus the overlapping white cover.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Getting the crossing height $\tfrac{\sqrt3}{2}$ and the equilateral-triangle area inside each overlap lens.)
⭐ When a region has a messy edge, measure the whole minus the easy leftover, and remember overlapping shapes share area you must subtract once.
⭐ When a region has a messy edge, measure the whole minus the easy leftover, and remember overlapping shapes share area you must subtract once.
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