# 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 $$n \cdot \frac{1}{2}R^2\sin\left(\frac{360}{n}^{\circ}\right) = 3R^2$$ Dividing both sides by $R^2$ yields $$\frac{n}{2}\sin\left(\frac{360}{n}^{\circ}\right) = 3$$ Multiply both sides by $\frac{2}{n}$ to get $$\sin\left(\frac{360}{n}^{\circ}\right) = \frac{6}{n}$$ 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 $\boxed{\textbf{(C)}}$.

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