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 $\boxed{\textbf{(C)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS