# 2003 AMC 10B Problems/Problem 19

The following problem is from both the 2003 AMC 12B #16 and 2003 AMC 10B #19, so both problems redirect to this page.

## Problem

Three semicircles of radius $1$ are constructed on diameter $\overline{AB}$ of a semicircle of radius $2$. The centers of the small semicircles divide $\overline{AB}$ 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? $[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("A",A,W); label("B",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S);[/asy]$ $\textbf{(A) } \pi - \sqrt{3} \qquad\textbf{(B) } \pi - \sqrt{2} \qquad\textbf{(C) } \frac{\pi + \sqrt{2}}{2} \qquad\textbf{(D) } \frac{\pi +\sqrt{3}}{2} \qquad\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}$

## Solution 1 $[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("A",A,W); label("B",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S); draw((1,0)--(0.5,0.866)); draw((0,0)--(0.5,0.866)); draw((-1,0)--(-0.5,0.866)); draw((0,0)--(-0.5,0.866));[/asy]$

By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $\frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$. This gives the area of the white region as $\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}$. The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2\pi-\left(\frac{5}{6}\pi+\frac{\sqrt3}{2}\right)=\frac{7}{6}\pi-\frac{\sqrt3}{2}$.

Thus the answer is $\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}$.

### Note

The reason why it is $\frac{5}{6}$ of a circle and why the triangles are equilateral are because, first, the radii are the same and they make up the equilateral triangles.

Secondly, the reason it is $\frac{5}{6}$ of a circle is because the middle sector has a degree of $180-2 \cdot 60 = 60$ and thus $\frac{60}{360}=\frac{1}{6}$ of a circle.

The other two have areas of $\frac{180-60}{360}=\frac{1}{3}$ of a triangle each.

Therefore, the total fraction of the circle(since they have the same radii) is $\frac{1}{6} + 2 \cdot \frac{1}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}.$

~mathboy282

## Solution 2(Answer Choices)

Answer choices A and B are impossible since the area is obviously not a circle of radius 1 minus a triangle.

Answer choices C and D are also impossible since they are adding but we are subtracting.

That leaves us with only answer choice $\textbf{(E)}.$

~mathboy282

## See Also

 2003 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
 2003 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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