Sum of Two ArcsDate: 01/30/2003 at 20:46:53 From: Abishek Subject: Probability Hi Dr. Math, This is the question: Three points are taken at random on the circumference of a circle. What is the chance that the sum of any two arcs so determined is greater than the third? Date: 01/31/2003 at 01:56:00 From: Doctor Greenie Subject: Re: Probability Hi, Abishek - I enjoyed the mental exercise I got thinking this one through! If the sum of two of the three arcs is greater than the third, then the largest arc is less than half the circle - i.e., less than 180 degrees. So to solve your problem, let's look at the conditions under which the largest of the three arcs is GREATER than 180 degrees. Suppose we call the points A, B, and C, and suppose points A and B have been chosen, and let x be the degree measure of minor arc AB (i.e., of the smaller of the two arcs determined by points A and B). Now consider diameters AD and BE (draw a picture...). If the third point C is on the same side of diameter AD as point B, then the largest of the three arcs will be greater than 180 degrees. And if the third point C is on the same side of diameter BE as point A, then the largest of the three arcs will be greater than 180 degrees. So the range where point C can be chosen so that the largest of the three arcs is LESS than 180 degrees is the minor arc DE, whose measure is the same as the measure of minor arc AB, which is x degrees. So if the measure of minor arc AB is x degrees, then the probability P (x) that the largest of the three arcs is less than 180 degrees is P(x) = (x/360) If we graph this probability function over its domain from 0 to 180 degrees, we get a straight line with P(0) = 0 and P(180) = 1/2. The overall probability that the largest of the three arcs is less than 180 degrees is the average value of the function P(x) over its domain, which is 1/4. So... Given three points randomly chosen on a circle, the probability that the largest of the three arcs is less than 180 degrees - that is, the probability that the sum of any two arcs is greater than the third - is 1/4. I looked in the Dr. Math archives after I wrote this response, performing a search using the keywords "probability point circumference arc". That search turned up this link which discusses this same problem but never - as far as I could tell - comes to a final answer: A Triangle in a Circle http://mathforum.org/library/drmath/view/51787.html - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ Date: 02/01/2003 at 19:54:24 From: Abishek Subject: Probability I could follow till the part where you said p(x)= x/360, but after that I don't understand why we have to take the average value for the required probability. Date: 02/02/2003 at 00:22:27 From: Doctor Greenie Subject: Re: Probability Hello, Abishek - The statement P(x) = x/360 says that the probability is x/360 that the largest of the three arcs is less than 180 degrees if the minor arc between the first two of the three points measures x degrees. So, for example, if the minor arc between the first two points is 45 degrees, then the probability of choosing the third point so that the largest arc is less than 180 degrees is 45/360 = 1/8. Or, if the minor arc between the first two points is 120 degrees, then the probability of choosing the third point so that the largest arc is less than 180 degrees is 120/360 = 1/3. To find the overall probability that three points randomly chosen will result in the largest of the three arcs measuring less than 180 degrees, we need to consider the probabilities P(x) over the whole range of possible value of x, which is from 0 to 180 degrees. The probability function in this case is a linear function, with value 0 at x=0 and value 180/360 = 1/2 at x=180. The average value of a linear function over its range is the average of its values at the two endpoints of its range -- in this case, the average of 0 and 1/2, which is 1/4. I hope this helps. Please write back if you still have questions about this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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