```Date: Jun 5, 1996 5:39 AM
Author: Daniel A. Asimov
Subject: Re: probability of a triangle (SPOILER)

In article <9604298333.AA833373730@ccmail.odedodea.edu> Pat_Ballew@ccmail.odedodea.edu writes:>     a) If a  unit length segment is randomly broken at two points along >     its length, what is the probability that the three pieces created in >     this fashion will form a triangle?>     >     Pat Ballew>     Misawa, Japan>     Pat_Ballew@ccmail.odedodea.edu--------------------------------------------------------------Here's my answer to question a):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 proportionalto 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 the3 conditions:  x + y > z, x + z > y, and y + z > x.  The subset of T where(x,y,z) satisfy these 3 conditions turns out to correspond to the littletriangle whose vertices are the midpoints of the edges of T.  ThereforeProb( the 3 pieces will form a triangle ) is one-fourth.--Dan Asimov
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