Difference between revisions of "1969 AHSME Problems/Problem 13"
(→Solution) |
Rockmanex3 (talk | contribs) (→Solution) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
| − | <math>\ | + | The area of the larger circle is <math>\pi R^2</math>, and the area of the region outside the smaller circle and inside the larger circle <math>\pi R^2 - \pi r^2</math>. Thus, |
| + | <cmath>\pi R^2 = \frac{a}{b} \cdot (\pi R^2 - \pi r^2)</cmath> | ||
| + | <cmath>R^2 = \frac{a}{b} \cdot R^2 - \frac{a}{b} \cdot r^2</cmath> | ||
| + | <cmath>\frac{a}{b} \cdot r^2 = (\frac{a}{b} -1)R^2</cmath> | ||
| + | <cmath>\frac{a}{b} \div \frac{a-b}{b} = \frac{R^2}{r^2}</cmath> | ||
| + | <cmath>\frac{a}{a-b} = \frac{R^2}{r^2}</cmath> | ||
| + | <cmath>\frac{\sqrt{a}}{\sqrt{a-b}} = \frac{R}{r}</cmath> | ||
| + | The answer is <math>\boxed{\textbf{(B)}}</math>. | ||
== See also == | == See also == | ||
| Line 16: | Line 23: | ||
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
| + | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 04:28, 7 June 2018
Problem
A circle with radius 
is contained within the region bounded by a circle with radius 
. The area bounded by the larger circle is 
times the area of the region outside the smaller circle and inside the larger circle. Then 
equals:
Solution
The area of the larger circle is 
, and the area of the region outside the smaller circle and inside the larger circle 
. Thus,
![\[\pi R^2 = \frac{a}{b} \cdot (\pi R^2 - \pi r^2)\]](http://latex.artofproblemsolving.com/e/f/a/efaac7260f3a6808f73aca7a1948856dfb83f2fd.png)
![\[R^2 = \frac{a}{b} \cdot R^2 - \frac{a}{b} \cdot r^2\]](http://latex.artofproblemsolving.com/d/f/0/df0441a5e6ccedf6685d8c2e7bc3cdce575b0f8b.png)
![\[\frac{a}{b} \cdot r^2 = (\frac{a}{b} -1)R^2\]](http://latex.artofproblemsolving.com/2/5/3/2539908e0ac0fb933b503410eb57bf0ed489ca15.png)
![\[\frac{a}{b} \div \frac{a-b}{b} = \frac{R^2}{r^2}\]](http://latex.artofproblemsolving.com/a/b/0/ab0b5cfa7c06ee1bb9ef0aef92ee4ba8c25e5abd.png)
![\[\frac{a}{a-b} = \frac{R^2}{r^2}\]](http://latex.artofproblemsolving.com/6/3/f/63f0ebeda5e59189b9e6e0727eccbf0405cd650b.png)
![\[\frac{\sqrt{a}}{\sqrt{a-b}} = \frac{R}{r}\]](http://latex.artofproblemsolving.com/f/9/6/f96cef1d66155c62e30c9b6ac97f8102037fc687.png)
The answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
