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 Strike a Chord (..Any Chord) (Posted on 2003-10-09)
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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 You never know.. | Comment 9 of 51 |
Bryan's idea is to pick a point on the perimeter, and then consider the distribution in question to be chords originating from that point to any other point. Then, since the analysis will be the same for any given point on the perimeter, and every chord in the circle will be equally represented (exactly twice) in such an analysis of every point, that is sufficient. That makes sense.
However, the analysis also depends on the assumption that any given range of angles gives an equal probability of random selection. That is to say, you need to assume the the chance of getting between 0¡Æ and 10¡Æ is the same as the chance of getting between 80¡Æ and 90¡Æ. How can you prove that this is true? I'm not saying it's right or wrong, but you need to support that part of the analysis.

Now, consider this: the perpendicular bisector of any chord will be a diameter of the circle. So, take a single diameter of the circle, and move along its length. At each 'point,' draw the chord perpendicular to the diameter at this point. The chords thus formed will be unique; there is only one specific diameter and distance that will form any given chord. When this chord is exactly equal in length to the radius, you are at a point on the diameter (√3)r/2 from the center. Farther than this from the center, the chord will be shorter than the radius; closer to the center, you will draw a chord that is longer than the radius.
At this point, np_rt said (in effect) that the probability is the ratio of points closer to the center to the total number of points on the whole diameter (which, if we assume even distribution of points along the diameter, is the ratio of these lengths). That ratio then becomes (2(√3)r/2)/2r, or just (√3)/2. Since it is a safe assumption that the distribution of points along the diameter is uniform, and such an analysis for every diameter of the circle would cover every possible chord exactly once, it could follor that this ratio is also the probability that any chord in the entire circle is longer than the radius.

Here's another way to look at it:
Each chord has a unique midpoint within the area of the circle. Thus, you could actually select any random point, draw the chord that has that point as its midpoint, and let that be your random chord. Then, assuming uniform distribution of points over the area of the circle, the ratio of longer chords (than the radius) to shorter ones will be the ratio of the areas of two circles. The 'top' part of this ratio will be the area of the circle, concentric to the original one, whose radius is small enough that any point within the circle will yield a chord whose length is greater than the radius. The radius of this circle is already determined; it is (√3)r/2. The 'bottom' of the ratio is, of course, the entire original circle. The ratio of the areas of two circles is the square of the ratio of their radii; thus this ratio is ((√3)/2)² = 3/4.
This means that, if you pick a random point in the circle, there is a 75% chance it will be less than (√3)r/2 from the center, and thus is the unique perpendicular bisector of a chord which is longer than the radius of the circle.

These all seem like fairly sound analyses, yet they come up with completely different answers.

Edited on October 9, 2003, 6:00 pm
 Posted by DJ on 2003-10-09 17:58:38

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