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Difference between revisions of "1967 AHSME Problems/Problem 33"

(Created page with "== 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),blac...")
 
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== Solution ==
 
== Solution ==
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<math>\fbox{D}</math>
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 +
To make the problem much simpler while staying in the constraints of the problem, position point <math>C</math> halfway between <math>A</math> and <math>B</math>. Then, call <math>\overline{AC} = \overline{BC}=r</math> . The area of the shaded region is then <cmath>\frac{ \pi r^2 - \pi (r/2)^2 - \pi (r/2)^2}{2}=\frac{\pi r^2}{4}</cmath>
 +
Because <math>\overline{CD}=r</math> the area of the circle with <math>\overline{CD}</math> as radius is <math>\pi r^2</math>.
 +
Our ratio is then <cmath>\frac{\pi r^2}{4} : \pi r^2 = 1:4</cmath>
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 +
Which corresponds with answer <math>\fbox{D}</math>
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1967|num-b=32|num-a=34}}   
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{{AHSME 40p box|year=1967|num-b=32|num-a=34}}   
  
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:40, 16 August 2023

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 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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