Difference between revisions of "1967 AHSME Problems/Problem 33"
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== Solution == | == Solution == | ||
− | <math>\fbox{D}</math> | + | |
+ | 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> | ||
+ | |||
+ | Which corresponds with answer <math>\fbox{D}</math> | ||
== See also == | == See also == |
Latest revision as of 16:39, 17 March 2019
Problem
In this diagram semi-circles are constructed on diameters , , and , so that they are mutually tangent. If , then the ratio of the shaded area to the area of a circle with as radius is:
Solution
To make the problem much simpler while staying in the constraints of the problem, position point halfway between and . Then, call . The area of the shaded region is then Because the area of the circle with as radius is . Our ratio is then
Which corresponds with answer
See also
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.