1965 AHSME Problems/Problem 2

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A regular hexagon is inscribed in a circle. The ratio of the length of a side of the hexagon to the length of the shorter of the arcs intercepted by the side, is:

$\textbf{(A)}\ 1: 1 \qquad  \textbf{(B) }\ 1: 6 \qquad  \textbf{(C) }\ 1: \pi \qquad  \textbf{(D) }\ 3: \pi \qquad  \textbf{(E) }\ 6:\pi$

Solution

Suppose that each side of the hexagon is $3$. Then the distance from each vertex of the hexagon to the center is also $3$, so that the circle has radius $3$. Since the circle has circumference $2\pi(3) = 6\pi$, the arc intercepted by any side (which measures $60^\circ$) has length $\frac{1}{6}*6\pi = \pi$, so that our answer is $\boxed{\text{(D)}}$ and we are done.

See also

1965 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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