Difference between revisions of "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: | 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: | ||
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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 12:47, 16 July 2024
Problem
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 | |
| 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. 
