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 $\boxed{\text{(B)}\ 1}$ .

Page

Toolbox

Search

1986 AJHSME Problems/Problem 23

Problem

Solution

See Also