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Re: Random Triangle Problem
Posted:
Jul 23, 1997 10:26 AM
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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 '
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