1967 AHSME Problems/Problem 33

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Problem

[asy] fill(circle((4,0),4),grey); fill((0,0)--(8,0)--(8,-4)--(0,-4)--cycle,white); fill(circle((7,0),1),white); fill(circle((3,0),3),white); draw((0,0)--(8,0),black+linewidth(1)); draw((6,0)--(6,sqrt(12)),black+linewidth(1)); MP("A", (0,0), W);  MP("B", (8,0), E);  MP("C", (6,0), S); MP("D",(6,sqrt(12)), N); [/asy]

In this diagram semi-circles are constructed on diameters $\overline{AB}$, $\overline{AC}$, and $\overline{CB}$, so that they are mutually tangent. If $\overline{CD} \bot \overline{AB}$, then the ratio of the shaded area to the area of a circle with $\overline{CD}$ as radius is:

$\textbf{(A)}\ 1:2\qquad \textbf{(B)}\ 1:3\qquad \textbf{(C)}\ \sqrt{3}:7\qquad \textbf{(D)}\ 1:4\qquad \textbf{(E)}\ \sqrt{2}:6$

Solution

To make the problem much simpler while staying in the constraints of the problem, position point $C$ halfway between $A$ and $B$. Then, call $\overline{AC} = \overline{BC}=r$ . The area of the shaded region is then \[\frac{ \pi r^2 - \pi (r/2)^2 - \pi (r/2)^2}{2}=\frac{\pi r^2}{4}\] Because $\overline{CD}=r$ the area of the circle with $\overline{CD}$ as radius is $\pi r^2$. Our ratio is then \[\frac{\pi r^2}{4} : \pi r^2 = 1:4\]

Which corresponds with answer $\fbox{D}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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