
Re: probability of a triangle (SPOILER)
Posted:
Jun 5, 1996 5:39 AM


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 3space 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 3 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 little triangle whose vertices are the midpoints of the edges of T. Therefore Prob( the 3 pieces will form a triangle ) is onefourth.
Dan Asimov

