1986 AJHSME Problems/Problem 23

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 (ProblemsAnswer KeyResources)
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

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

Invalid username
Login to AoPS