Probability of Forming a TriangleDate: 01/24/2001 at 22:35:45 From: Mike Subject: Probability I don't know where to begin. A rope one unit long is cut in two places. What is the probability that the three resulting pieces can be arranged to form a triangle? Thank you. Mike Date: 01/25/2001 at 00:56:51 From: Doctor Pat Subject: Re: Probability Mike, Here is one way, I got this as an answer from the nephew of the late Isaac Asimov, the Science writer, when I posed this problem on the Internet several years ago. Dan is a very good mathematician on his own, as you can see from his approach to the problem... If we let T denote the planar triangle in 3-space whose vertices are (1,0,0), (0,1,0), and (0,0,1), then choosing a point p at random on T (i.e. with the probability of p lying in a subset S of T being proportional to the area of S, or more precisely Prob(p is in S) = area(S) / area(T) will simulate the random cutting of the unit segment desscribed above, with resulting pieces of lengths x, y, and z respectively. The condition that x, y, and z "will form a triangle" is equivalent to the three conditions: x + y > z, x + z > y, and y + z > x. The subset of T where (x,y,z) satisfy these three conditions turns out to correspond to the little triangle whose vertices are the midpoints of the edges of T. Therefore Prob(the 3 pieces will form a triangle) is one-fourth. To which I added... From the three conditions above, you can also deduce that the conditions x + y > z, x + z > y, and y + z > x with the additional knowledge that x + y + z = 1 reduces to Max(x,y,z) < .5, which offers another easy approach... Hope that helps - Doctor Pat, The Math Forum http://mathforum.org/dr.math/ |
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