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Re: probability of a triangle (SPOILER)
Posted:
Jun 5, 1996 5:39 AM
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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
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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 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 one-fourth.
--Dan Asimov
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