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Difference between revisions of "2003 AMC 10B Problems/Problem 19"

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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #16]] and [[2003 AMC 10B Problems|2003 AMC 10B #19]]}}
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==Problem==
 
==Problem==
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==Problem 19==
 
  
 
Three semicircles of radius <math>1</math> are constructed on diameter <math>\overline{AB}</math> of a semicircle of radius <math>2</math>. The centers of the small semicircles divide <math>\overline{AB}</math> 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?
 
Three semicircles of radius <math>1</math> are constructed on diameter <math>\overline{AB}</math> of a semicircle of radius <math>2</math>. The centers of the small semicircles divide <math>\overline{AB}</math> 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?
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<math>\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}</math>
 
<math>\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}</math>
  
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==Solution==
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==Solution 1==
  
 
<asy>
 
<asy>
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By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into <math>\frac{5}{6}</math> of a circle with radius <math>1</math> and two equilateral triangles with side length <math>1</math>.  
 
By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into <math>\frac{5}{6}</math> of a circle with radius <math>1</math> and two equilateral triangles with side length <math>1</math>.  
 
This gives the area of the white region as <math>\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}</math>.  
 
This gives the area of the white region as <math>\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}</math>.  
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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 <math>2\pi-\frac{5}{6}\pi+\frac{\sqrt3}{2}=\frac{7}{6}\pi-\frac{\sqrt3}{2}</math>.
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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 <math>2\pi-\left(\frac{5}{6}\pi+\frac{\sqrt3}{2}\right)=\frac{7}{6}\pi-\frac{\sqrt3}{2}</math>.
  
 
Thus the answer is <math>\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}</math>.
 
Thus the answer is <math>\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}</math>.
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===Note===
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The reason why it is <math>\frac{5}{6}</math> of a circle and why the triangles are equilateral are because, first, the radii are the same and they make up the equilateral triangles.
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Secondly, the reason it is <math>\frac{5}{6}</math> of a circle is because the middle sector has a degree of <math>180-2 \cdot 60 = 60</math> and thus <math>\frac{60}{360}=\frac{1}{6}</math> of a circle.
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The other two have areas of <math>\frac{180-60}{360}=\frac{1}{3}</math> of a triangle each.
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Therefore, the total fraction of the circle(since they have the same radii) is <math>\frac{1}{6} + 2 \cdot \frac{1}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}.</math>
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~mathboy282
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==Solution 2(Answer Choices)==
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Answer choices A and B are impossible since the area is obviously not a circle of radius 1 minus a triangle.
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Answer choices C and D are also impossible since they are adding but we are subtracting.
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That leaves us with only answer choice <math>\textbf{(E)}.</math>
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~mathboy282
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==Video Solution by WhyMath==
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https://youtu.be/XDbP_bc8rxw
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2003|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2003|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2003|ab=B|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 12:19, 9 July 2021

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

Video Solution by WhyMath

https://youtu.be/XDbP_bc8rxw

~savannahsolver

See Also

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 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 by
Problem 18 		Followed by
Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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