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Topic: Random Triangle Problem
Replies: 57   Last Post: Aug 17, 1997 10:51 PM

 Messages: [ Previous | Next ]
 tony richards Posts: 164 Registered: 12/8/04
Re: Random Triangle Problem
Posted: Jul 23, 1997 10:26 AM

mathwft@math.canterbury.ac.nz (Bill Taylor) wrote:
>Mike Housky <mike@webworldinc.com> writes:
>|>
>|> > Derive a method of finding the chances of 3 random points placed on
>|> > a plane that will yield an obtuse triangle.
>
>Any answers to open-ended problems of this type are bound to be somewhat
>subjective, or a matter of taste.
>
>But in this case, the answer seems fairly clear (and has appeared on Usenet
>many times before).
>
>Most folk would consider a triangle formed by "3 randomly chosen points" to
>be the same as one formed by "3 randomly chosen lines", and this second
>formulation, though tougher-looking at first glance is in fact much cleaner.
>
>The three lines, whatever else they have, should have orientations that are
>independent and uniform in (0 , 2pi) ; and this means choosing three numbers
>at random in that range. Once done, we then see that their actual placements
>ARE IRRELEVANT TO THE SOLUTION ! (Merely replacing triangles by similar ones.)
>
>This last observation leads me to suggest that this is a "complete" solution
>to the problem, and disputes of taste and subjectivity essentially decided.
>
>Thus, the problem is fairly easily solved. Let one line be the x-axis; if the
>other two have the same signed slope (prob 1/2), an obtuse is certain. If they
>have opposite slopes there are equal probs of obtuse or acute, (as the angle
>between then can be on either side of 90 deg).

IMHO, this is where you make an invalid assumption.
The third line is not free to make ANY angle between 0 and 180 degrees with the second
line, since it is constrained to have a slope opposite in sign to the second line,
WHEN MEASURED RELATIVE TO THE FIRST LINE.
You therefore have a skewed distribution for the third line, given the first and the second
lines,
so your assumption of equal probabilities of >90 and <90 degrees
between the second and third lines is unwarranted.
the answer therefore cannot be 0.75, if that is what your argument gives
since it has the above flaw.
>
>So putting together, this gives... P(obtuse) = 3/4 .
> ~~~~~~~~~~~~~~~

--
Tony Richards 'I think, therefore I am confused'
Rutherford Appleton Lab '
UK '

Date Subject Author
7/16/97 Mike Housky
7/21/97 Bill Taylor
7/22/97 tony richards
7/24/97 Brian M. Scott
7/23/97 tony richards
7/23/97 T. Sheridan
7/24/97 Bill Taylor
7/24/97 Bill Taylor
7/25/97 Ilias Kastanas
7/23/97 Robert Hill
7/23/97 tony richards
7/27/97 Bill Taylor
7/24/97 Robert Hill
7/28/97 tony richards
7/30/97 Bill Taylor
7/30/97 tony richards
8/1/97 Bill Taylor
7/24/97 Robert Hill
7/24/97 Robert Hill
7/24/97 Robert Hill
7/25/97 Robert Hill
7/30/97 Bill Taylor
8/1/97 Charles H. Giffen
8/1/97 John Rickard
8/1/97 Chris Thompson
8/1/97 John Rickard
8/4/97 Bill Taylor
8/5/97 John Rickard
7/25/97 Charles H. Giffen
7/25/97 Charles H. Giffen
7/28/97 Hauke Reddmann
7/28/97 Robert Hill
7/28/97 Robert Hill
7/28/97 Robert Hill
7/29/97 tony richards
7/30/97 Keith Ramsay
7/30/97 tony richards
8/2/97 Keith Ramsay
7/29/97 tony richards
8/4/97 Bill Taylor
8/5/97 Charles H. Giffen
8/6/97 Terry Moore
8/7/97 Terry Moore
8/16/97 Kevin Brown
8/17/97 Kevin Brown
7/30/97 Robert Hill
7/31/97 tony richards
8/6/97 Terry Moore
7/31/97 John Rickard
7/30/97 Robert Hill
7/31/97 Robert Hill
7/31/97 Robert Hill
8/1/97 R J Morris
8/4/97 Robert Hill
8/4/97 Robert Hill
8/5/97 Charles H. Giffen
8/6/97 Robert Hill