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Difference between revisions of "1986 AJHSME Problems/Problem 23"

(New page: ==Problem== The large circle has diameter <math>\text{AC}</math>. The two small circles have their centers on <math>\text{AC}</math> and just touch at <math>\text{O}</math>, the center o...)
 
 
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==Solution==
 
==Solution==
  
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{{Solution}}
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The small circle has radius <math>1</math>, thus its area is <math>\pi</math>.
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The large circle has radius <math>2</math>, thus its area is <math>4\pi</math>.
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The area of the semicircle above <math>AC</math> is then <math>2\pi</math>.
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The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is <math>\pi</math>. This means that the area of the shaded part is <math>2\pi-\pi=\pi</math>.
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This is equal to the area of a small circle, hence the correct answer is <math>\boxed{\text{(B)}\ 1}</math>.
  
 
==See Also==
 
==See Also==
  
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[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=22|num-a=24}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 20:32, 3 July 2013

Problem

The large circle has diameter $\text{AC}$. The two small circles have their centers on $\text{AC}$ and just touch at $\text{O}$, the center of the large circle. If each small circle has radius $1$, what is the value of the ratio of the area of the shaded region to the area of one of the small circles?

[asy] pair A=(-2,0), O=origin, C=(2,0); path X=Arc(O,2,0,180), Y=Arc((-1,0),1,180,0), Z=Arc((1,0),1,180,0), M=X..Y..Z..cycle; filldraw(M, black, black); draw(reflect(A,C)*M); draw(A--C, dashed); label("A",A,W); label("C",C,E); label("O",O,SE); dot((-1,0)); dot(O); dot((1,0)); label("$1$",(-.5,0),N); label("$1$",(1.5,0),N); [/asy]

$\text{(A)}\ \text{between }\frac{1}{2}\text{ and 1} \qquad \text{(B)}\ 1 \qquad \text{(C)}\ \text{between 1 and }\frac{3}{2}$

$\text{(D)}\ \text{between }\frac{3}{2}\text{ and 2} \qquad \text{(E)}\ \text{cannot be determined from the information given}$

Solution

The small circle has radius $1$, thus its area is $\pi$.

The large circle has radius $2$, thus its area is $4\pi$.

The area of the semicircle above $AC$ is then $2\pi$.

The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $\pi$. This means that the area of the shaded part is $2\pi-\pi=\pi$. This is equal to the area of a small circle, hence the correct answer is $\boxed{\text{(B)}\ 1}$.

See Also

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AJHSME/AMC 8 Problems and Solutions

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