Posted November 15, 2012 What is the length of a random chord drawn through a unit circle? There are at least four answers. 0 Share this post Link to post Share on other sites

0 Posted November 15, 2012 Should we randomly chose the definition of random? I know our friend laid out three ways to draw a random chord while leaving out, arguably, the most intuitive way. I thought I remember reading that there were more ways than that even. Math is just a hobby for me so I am not immersed in it. 0 Share this post Link to post Share on other sites

0 Posted November 15, 2012 I am thinking the average length of a random chord is pi/4 (0.7854) as a first thought. 0 Share this post Link to post Share on other sites

0 Posted November 15, 2012 (edited) pi / 2 * radius based upon a range of distances from the center of the circle Edited November 15, 2012 by curr3nt 0 Share this post Link to post Share on other sites

0 Posted November 15, 2012 This took me to find Bertrand's paradox. Looks like my answer covers method 2. WIll try to calculate method 1 and 3. What's the fourth? 0 Share this post Link to post Share on other sites

0 Posted November 15, 2012 This took me to find Bertrand's paradox. Looks like my answer covers method 2. WIll try to calculate method 1 and 3. What's the fourth? He left out from his three what I thought was the most intuitive way to make a chord Two points are located randomly and independently in the interior of the circle. An extended straight-line connection of the two random points determines a random chord. 0 Share this post Link to post Share on other sites

0 Posted November 15, 2012 He left out from his three what I thought was the most intuitive way to make a chord Two points are located randomly and independently in the interior of the circle. An extended straight-line connection of the two random points determines a random chord. My problem with the method of randomly choosing two points is that it seems to me to result in an estimate of the length squared. Here's my argument. Choose two points in the interior, B and C. Extend to form the chord intersecting the circle at A and D. Now we have a length: the distance from A to D. Notice, no matter what two points we could have chosen between A and D, we would arrive at the very same chord. So our overall average is going to be weighted by the probability of getting two points on chord AZ. But, if we had chosen two points L and M resulting in a much shorter chord KN, it would have been weighted by the (smaller) probability of getting two points on chord KN. So our overall average would incorporate the length of the chord twice, once for the length of the chord, and once for the probability of picking out that particular chord. I would think that a "randomly chosen chord" would uniformly choose between chords, with each different chord having the same weight. Unfortunately, I can imagine (before looking at Bertrand's paradox) at least two ways to choose a chord randomly, and I still see differential weighting. Both my suggested methods involving designating one point P on the circumference of the circle. After all, we know we want something that touches the circumference. It doesn't matter where, because it's a circle. Let's just start with one point. Now: (method 1) uniformly choose an angle between 0 and 180, measured from a tangent to P. Compute the length of the chord, average over all angles. (method 2) uniformly choose a point Q from the circumference of the circle. Compute the length of the chord, average over all points Q. I assume they give different results. Sorry...:-( 0 Share this post Link to post Share on other sites

0 Posted November 15, 2012 My problem with the method of randomly choosing two points is that it seems to me to result in an estimate of the length squared. Here's my argument. Choose two points in the interior, B and C. Extend to form the chord intersecting the circle at A and D. Now we have a length: the distance from A to D. Notice, no matter what two points we could have chosen between A and D, we would arrive at the very same chord. So our overall average is going to be weighted by the probability of getting two points on chord AZ. But, if we had chosen two points L and M resulting in a much shorter chord KN, it would have been weighted by the (smaller) probability of getting two points on chord KN. So our overall average would incorporate the length of the chord twice, once for the length of the chord, and once for the probability of picking out that particular chord. I would think that a "randomly chosen chord" would uniformly choose between chords, with each different chord having the same weight. Unfortunately, I can imagine (before looking at Bertrand's paradox) at least two ways to choose a chord randomly, and I still see differential weighting. Both my suggested methods involving designating one point P on the circumference of the circle. After all, we know we want something that touches the circumference. It doesn't matter where, because it's a circle. Let's just start with one point. Now: (method 1) uniformly choose an angle between 0 and 180, measured from a tangent to P. Compute the length of the chord, average over all angles. (method 2) uniformly choose a point Q from the circumference of the circle. Compute the length of the chord, average over all points Q. I assume they give different results. Sorry...:-( I was just pointing out that Bertrand left out that method for choosing a chord. However, Pick a random point inside the circle and spin a random angle relative to the radius and pivoting on the point creating a chord. 0 Share this post Link to post Share on other sites

0 Posted November 15, 2012 Indeed, more than one way. Bonanova, what is the goal of this puzzle? a) show us that the notion of "randomly chosen" depends on how you define it? b) have us show lots of ingenuity in defining truly different ways to define it? c) elicit from us the really-really-best way to define it? d) all of the above e) none of the above f) other 0 Share this post Link to post Share on other sites

0 Posted November 16, 2012 I was just pointing out that Bertrand left out that method for choosing a chord. However, Pick a random point inside the circle and spin a random angle relative to the radius and pivoting on the point creating a chord. I agree with CaptainEd. Jaynes' solution to the Bertrand paradox shows that random radius (method 2) is the only robust theoretical means of choosing random chord. 0 Share this post Link to post Share on other sites

0 Posted November 16, 2012 I agree with CaptainEd. Jaynes' solution to the Bertrand paradox shows that random radius (method 2) is the only robust theoretical means of choosing random chord. So, then it is more random. I looked up Jaynes' proposal and it based on the fact that the size of the circle is unknown, here the OP has given us a size. Is throwing straws at a circle a better method than selecting a number at random between 0 and Ï€r^{2} , letting that number represent an area and drawing a chord to section off that area? 0 Share this post Link to post Share on other sites

0 Posted November 16, 2012 Indeed, more than one way. Bonanova, what is the goal of this puzzle? a) show us that the notion of "randomly chosen" depends on how you define it? b) have us show lots of ingenuity in defining truly different ways to define it? c) elicit from us the really-really-best way to define it? d) all of the above e) none of the above f) other All of the above, probably. Let's make it a competition to show the greatest and least among any reasonable definition of randomness. 0 Share this post Link to post Share on other sites

0 Posted November 20, 2012 Well since I neglected to use the word "average" in the OP, another answer is anything between 0 and 2. That was the "fourth" answer. 0 Share this post Link to post Share on other sites

0 Posted November 22, 2012 (edited) Well the longest chord possible is the diameter, and the shortest possible is a tangent (0 length). On average, you will get something between the two so the average is half the diameter, AKA, the radius! Edited November 22, 2012 by joef1000 0 Share this post Link to post Share on other sites

0 Posted November 22, 2012 (edited) My above solution assumes that you define random in a way that makes all chord lengths equally likely. (I hate the edit system on this site) Edited November 22, 2012 by joef1000 0 Share this post Link to post Share on other sites

0 Posted November 22, 2012 Well the longest chord possible is the diameter, and the shortest possible is a tangent (0 length). On average, you will get something between the two so the average is half the diameter, AKA, the radius! This is an interesting approach. The radius is a middle measure of the extremes of zero and the diameter, but It is the median, not the mean. One must represent (find an expression for) all the uncountably infinite chords and then examine the distribution of their lengths. If the distribution were found to be uniform from 0 to 2r then r would be correct. All the examples given previously in this thread, however, show the lengths not to be uniformly distributed. 0 Share this post Link to post Share on other sites

0 Posted November 22, 2012 My above solution assumes that you define random in a way that makes all chord lengths equally likely. (I hate the edit system on this site) I'm not sure you can define random in any sense other than an unbiased selection from all available instances. The challenge that arises is that several exhaustive, random [in some sense] and uncountable subsets each yield a distinct value of average length. 0 Share this post Link to post Share on other sites

0 Posted November 27, 2012 After doing some research and none of the math, I've foundThe shortest mean chord length was 1.00 and the longest 1.81 depending on the method used to make a random chord. 0 Share this post Link to post Share on other sites

0 Posted November 27, 2012 After doing some research and none of the math, I've foundThe shortest mean chord length was 1.00 and the longest 1.81 depending on the method used to make a random chord. Inscribe a regular polygon of n sides where n > 1. Take the chord to be one side of the polygon; (n = 2 gives a diameter.) Let n be chosen at random. What is that average chord length? 0 Share this post Link to post Share on other sites

0 Posted November 27, 2012 Inscribe a regular polygon of n sides where n > 1. Take the chord to be one side of the polygon; (n = 2 gives a diameter.) Let n be chosen at random. What is that average chord length? Quite small. The larger n is the smaller the chord length. I would argue that this is not an acceptable way to chose "random chords", however. 0 Share this post Link to post Share on other sites

0 Posted November 28, 2012 Quite small. The larger n is the smaller the chord length. I would argue that this is not an acceptable way to chose "random chords", however. Hokey, at best. I agree. 0 Share this post Link to post Share on other sites

0 Posted November 28, 2012 Hokey, at best. I agree. Another hokey approach has come to my attention Inscribe a random right triangle. Take the diameter as the chord. 0 Share this post Link to post Share on other sites

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What is the length of a random chord drawn through a unit circle? There are at least four answers.

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