Difference between revisions of "1965 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
− | Suppose that each side of the hexagon is <math>3</math>. Then the distance from each vertex of the hexagon to the center is also <math>3</math>, so that the circle has radius <math>3</math>. Since the circle has circumference <math>2\pi(3) = 6\pi</math>, the arc intercepted by any side (which measures <math>60^\circ</math>) has length <math>\frac{1}{6}*6\pi = \pi</math>, and we are done. | + | Suppose that each side of the hexagon is <math>3</math>. Then the distance from each vertex of the hexagon to the center is also <math>3</math>, so that the circle has radius <math>3</math>. Since the circle has circumference <math>2\pi(3) = 6\pi</math>, the arc intercepted by any side (which measures <math>60^\circ</math>) has length <math>\frac{1}{6}*6\pi = \pi</math>, so that our answer is <math>\boxed{\text{(D)}}</math> and we are done. |
== See also == | == See also == |
Latest revision as of 18:42, 16 September 2020
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:
Solution
Suppose that each side of the hexagon is . Then the distance from each vertex of the hexagon to the center is also , so that the circle has radius . Since the circle has circumference , the arc intercepted by any side (which measures ) has length , so that our answer is and we are done.
See also
1965 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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All AHSME Problems and Solutions |
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