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Difference between revisions of "1969 AHSME Problems/Problem 13"

(Solution to Problem 13)
(→‎Solution)
 
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<cmath>\frac{a}{b} \div \frac{a-b}{b} = \frac{R^2}{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{a}{a-b} = \frac{R^2}{r^2}</cmath>
−
<cmath>\frac{\sqrt{a}}{\sqrt{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>.
 
The answer is <math>\boxed{\textbf{(B)}}</math>.
  

Latest revision as of 03:28, 7 June 2018

Problem

A circle with radius $r$ is contained within the region bounded by a circle with radius $R$. The area bounded by the larger circle is $\frac{a}{b}$ times the area of the region outside the smaller circle and inside the larger circle. Then $R:r$ equals:

$\text{(A) }\sqrt{a}:\sqrt{b} \quad \text{(B) } \sqrt{a}:\sqrt{a-b}\quad \text{(C) } \sqrt{b}:\sqrt{a-b}\quad \text{(D) } a:\sqrt{a-b}\quad \text{(E) } b:\sqrt{a-b}$

Solution

The area of the larger circle is $\pi R^2$, and the area of the region outside the smaller circle and inside the larger circle $\pi R^2 - \pi r^2$. Thus, \[\pi R^2 = \frac{a}{b} \cdot (\pi R^2 - \pi r^2)\] \[R^2 = \frac{a}{b} \cdot R^2 - \frac{a}{b} \cdot r^2\] \[\frac{a}{b} \cdot r^2 = (\frac{a}{b} -1)R^2\] \[\frac{a}{b} \div \frac{a-b}{b} = \frac{R^2}{r^2}\] \[\frac{a}{a-b} = \frac{R^2}{r^2}\] \[\frac{\sqrt{a}}{\sqrt{a-b}} = \frac{R}{r}\] The answer is $\boxed{\textbf{(B)}}$.

See also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AHSME Problems and Solutions

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