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

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== Solution ==
 
== Solution ==
  
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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.
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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 ==

Revision as of 17:42, 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:

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

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 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

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