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Strike a Chord (..Any Chord) (Posted on 2003-10-09) Difficulty: 4 of 5
What is the probability that a randomly drawn chord will be longer than the radius of the circle?

Prove it.

No Solution Yet Submitted by DJ    
Rating: 4.5263 (19 votes)

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Some Thoughts keep it simple, ... | Comment 13 of 51 |
(In reply to A needed fact by Federico Kereki)

It is indeed significant that we establish the method by which our chord is randomly drawn. I can see how this was abstracted by both of our solutions, and would like to offer another avenue of consideration.

Let us consider a regular n-gon (since probability is so much more concrete in the discrete world), and the set of line segments joining verticies of it. Let me then ask "what is the probability that a randomly chosen line segment is longer than the "radius" of the n-gon(distance from center of polygon to a vertex). This problem is easy to define and percieve. For example: In a regular hexagon there are 15 unique line segments 9 of which are longer then the "radius" giving a probability of 3/5. Furthermore in a regular 60-agon there are 59(60)/2 = 1770 unique line segments 39(60)/2 = 1170 of which are longer than the radius giving a probability of 1170/1770 = 39/59.

If we take the limit as n approches ∞ and extrapolate this method of randomly choosing a line segment in an ∞-gon, we see that our line segment becomes a chord, our n-gon becomes a circle, and our probability is indeed 2/3.

Should we use similar logic with regard to the "random point on a perpendicular bisector" method, we'd have to construct some kind of index of points along the perpendicular bisector, which frankly, is a construction not warranted by the problem.
  Posted by Eric on 2003-10-09 22:52:33

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