home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!munnari.oz.au!comp.vuw.ac.nz!canterbury.ac.nz!math!wft
- Newsgroups: sci.math
- Subject: Re: random triangles
- Message-ID: <1992Jul28.131046.6090@csc.canterbury.ac.nz>
- From: wft@math.canterbury.ac.nz (Bill Taylor)
- Date: 28 Jul 92 13:10:42 +1200
- References: <BrvLF2.7q@watserv1.waterloo.edu>
- Distribution: world
- Organization: Department of Mathematics, University of Canterbury
- Keywords: asymptotics, acute triangles
- Nntp-Posting-Host: math.canterbury.ac.nz
- Lines: 33
-
- > What is the probability that a random triangle is acute
-
- I think this problem might have appeared here before. It is more elegant to
- formalize the concept of a random triangle in R^2 , as that formed by three
- lines whose angles are independemtly chosen from uniform[0,2pi].
-
- This determines the (distribution of the) shape and orientation of the triangle
- completely, (the size being irrelevant to the question at hand).
-
- Then, WOLOG, we can consider one of the 3 sides to be horizontal,
- and WOLOG the 2nd side to have angle t (anticlockwise from x-axis) to be
- uniform[ 0 , pi/2 ],
-
- and the 3rd side to have angle w (anticlockwise from x-axis) to be
- uniform[ 0 , pi ].
-
- Then P(acute | t) = P( w in [ pi/2 , pi/2 + t ]
- = t/pi
-
- / pi/2
- |
- thus P(acute) = (2/pi)*| (t/pi) dw = 1/4 .
- |
- / 0
-
-
- There is also a quick slick way of getting this answer:- by considering that
- the angles of the 2nd and 3rd lines (given the first is horizontal) must be
- oppositely inclined, (prob 1/2) ; and given this, the angle between them can
- only be half of the range of possibilities, to ensure acuteness.
- Again giving P(acute) = 1/4 .
- ---------------------------------------------
- Bill Taylor wft@math.canterbury.ac.nz
-