1961 AHSME Problems/Problem 32

Problem

A regular polygon of $n$ sides is inscribed in a circle of radius $R$ . The area of the polygon is $3R^2$ . Then $n$ equals:

$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 15\qquad \textbf{(E)}\ 18$

Solution

Note that the distance from the center of the circle to each of the vertices of the inscribed regular polygon equals the radius $R$ . 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 $\frac{360}{n}$ degrees. Thus, the area of the polygon is $n \cdot \frac{1}{2}R^2\sin(\frac{360}{n}^{\circ}) = 3R^2$ Dividing both sides by $R^2$ yields $\frac{n}{2}\sin(\frac{360}{n}^{\circ}) = 3$ Multiply both sides by $\frac{2}{n}$ to get $\sin(\frac{360}{n}^{\circ}) = \frac{6}{n}$ At this point, use trial-and-error for each of the answer choices. When checking $n = 12$ , the equation results in $\sin(30^{\circ}) = \frac{1}{2}$ , which is correct. Thus, the answer is $\boxed{\textbf{(C)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 31	Followed by Problem 33
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1961 AHSME Problems/Problem 32

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