1967 AHSME Problems/Problem 33

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 $\fbox{D}$

1967 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1967 AHSME Problems/Problem 33

Problem

Solution

See also