Difference between revisions of "1965 AHSME Problems/Problem 2"

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 (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 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 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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