Difference between revisions of "1961 AHSME Problems/Problem 32"
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That means each of these angles between the two lines from the center to the two vertices of a side equals <math>\frac{360}{n}</math> degrees. Thus, the area of the polygon is | That means each of these angles between the two lines from the center to the two vertices of a side equals <math>\frac{360}{n}</math> degrees. Thus, the area of the polygon is | ||
| − | <cmath>n \cdot \frac{1}{2}R^2\sin(\frac{360}{n}^{\circ}) = 3R^2</cmath> | + | <cmath>n \cdot \frac{1}{2}R^2\sin\left(\frac{360}{n}^{\circ}\right) = 3R^2</cmath> |
Dividing both sides by <math>R^2</math> yields | Dividing both sides by <math>R^2</math> yields | ||
| − | <cmath>\frac{n}{2}\sin(\frac{360}{n}^{\circ}) = 3</cmath> | + | <cmath>\frac{n}{2}\sin\left(\frac{360}{n}^{\circ}\right) = 3</cmath> |
Multiply both sides by <math>\frac{2}{n}</math> to get | Multiply both sides by <math>\frac{2}{n}</math> to get | ||
| − | <cmath>\sin(\frac{360}{n}^{\circ}) = \frac{6}{n}</cmath> | + | <cmath>\sin\left(\frac{360}{n}^{\circ}\right) = \frac{6}{n}</cmath> |
At this point, use trial-and-error for each of the answer choices. When checking <math>n = 12</math>, the equation results in <math>\sin(30^{\circ}) = \frac{1}{2}</math>, which is correct. Thus, the answer is <math>\boxed{\textbf{(C)}}</math>. | At this point, use trial-and-error for each of the answer choices. When checking <math>n = 12</math>, the equation results in <math>\sin(30^{\circ}) = \frac{1}{2}</math>, which is correct. Thus, the answer is <math>\boxed{\textbf{(C)}}</math>. | ||
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==See Also== | ==See Also== | ||
Latest revision as of 16:13, 18 April 2019
Problem
A regular polygon of 
sides is inscribed in a circle of radius 
. The area of the polygon is 
. Then equals:
Solution
Note that the distance from the center of the circle to each of the vertices of the inscribed regular polygon equals the radius . Since each side of a regular polygon is the same length, all the angles between the two lines from the center to the two vertices of a side is the same.
That means each of these angles between the two lines from the center to the two vertices of a side equals 
degrees. Thus, the area of the polygon is
![\[n \cdot \frac{1}{2}R^2\sin\left(\frac{360}{n}^{\circ}\right) = 3R^2\]](http://latex.artofproblemsolving.com/0/2/d/02d272300e8b2fc9280c94073780fa4d3ee3ee0c.png)
Dividing both sides by 
yields
![\[\frac{n}{2}\sin\left(\frac{360}{n}^{\circ}\right) = 3\]](http://latex.artofproblemsolving.com/b/7/d/b7d12423738fe85702326bfaf780b8260c618670.png)
Multiply both sides by 
to get
![\[\sin\left(\frac{360}{n}^{\circ}\right) = \frac{6}{n}\]](http://latex.artofproblemsolving.com/b/f/9/bf950051b196aef6fe71d3e52d04dc2f27a4c13c.png)
At this point, use trial-and-error for each of the answer choices. When checking 
, the equation results in 
, which is correct. Thus, the answer is 
.
See Also
| 1961 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 31 |
Followed by Problem 33 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
