Difference between revisions of "1986 AHSME Problems/Problem 21"
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==Solution== | ==Solution== | ||
+ | Well, the shaded sector's area is basically <math>\text{(ratio of } \theta \text{ to total angle of circle)} \times \text{(total area)} = \frac{\theta}{2\pi} \cdot (\pi r^2) = \frac{\theta}{2} \cdot (AC)^2</math>. | ||
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+ | In addition, if you let <math>\angle{ACB} = \theta</math>, then <cmath>\tan \theta = \frac{AB}{AC}</cmath><cmath>AB = AC\tan \theta = r\tan \theta</cmath><cmath>[ABC] = \frac{AB \cdot AC}{2} = \frac{r^2\tan \theta}{2}</cmath>Then the area of that shaded thing on the left becomes <cmath>\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2}</cmath>We want this to be equal to the sector area so <cmath>\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2} = \frac{\theta \cdot r^2}{2}</cmath><cmath>\frac{r^2\tan \theta}{2} = \theta \cdot r^2</cmath><cmath>\tan \theta = 2\theta</cmath> | ||
== See also == | == See also == |
Latest revision as of 17:50, 1 April 2018
Problem
In the configuration below, is measured in radians, is the center of the circle, and are line segments and is tangent to the circle at .
A necessary and sufficient condition for the equality of the two shaded areas, given , is
Solution
Well, the shaded sector's area is basically .
In addition, if you let , then Then the area of that shaded thing on the left becomes We want this to be equal to the sector area so
See also
1986 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
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