Random Triangles

What's the probability that three "randomly chosen" points in the 
plane will form an acute-angled triangle?  In attempting to answer
this question, one approach that often occurs to people is to
arbitrarily choose the longest "side" first, and then the third 
point is restricted to a finite area.  Assuming the probability 
density for the third point is uniformly distributed over this 
region, we can compute the probability that the third point will
fall in the subset of this region such that the triangle has only 
acute angles.  However, as plausible as this might sound, this
doesn't really represent three randomly chosen points.  In fact,
choosing thee points "randomly" in the plane is not really well 
defined without giving the probability distribution for the points.
Unfortunately there is no uniform distribution over the plane, so 
the premise of choosing three points randomly on the place is

To understand why the naive solution doesn't actually give an
unique answer, let the three vertices de denoted by A,B,C, and
recall that the naive procedure is to arbitrarily fix the points A
and B (which is permissible because scale and orientation are both 
irrelevant to the shape of the triangle), and then let point C be 
chosen from the finite region that leaves AB as the longest edge.
Now, if the triangles are required to be constructed in this way, 
then the percentage of triangles with three acute angles would 
indeed be 1-(3/(8-(6sqrt(3)/pi))) = 36.0617%.

However, we could just as well choose point C from the finite region
that leaves AB as the SECOND longest leg of the triangle, in which
case only  1-(3/(2+(3sqrt(3)/pi))) = 17.897%  of the triangles would
have three acute angles.  There is also a third possibility, which 
is to choose point C such that AB is the SHORTEST leg of the trangle,
but in that case the region of possible C points is infinite (as is 
the subset that results in three acute angles), so there is no 
well-defined ratio at all.

Furthermore, notice that in each of these three alternatives the 
location of point C is assumed to be constrained by the positions 
of A and B, whereas the problem implies the points are chosen 
independently.  To be consistent with the problem statement we 
would need to fix the points A and B, and then let C be chosen 
from a uniform distribution over the entire plane, which, as 
mentioned above, is not a well-defined proposition.

Another idea that sometimes occurs to people is to calculate the 
ratio of positions for C that give an accute triangle, assuming the
probability density for C uniformly distributed over some finite
region, and then take the limit as the size of this region increases
to infinity.  On this basis, it might be argued that any "reasonable"
choice of the parameterized bounded region will lead us to conclude
that the probability of three acute angles is ZERO, because if we 
put the segment AB on the x axis, there will be an obtuse angle unless
C's x coordinate is between those of A and B, and this will become
an arbitrarily small fraction of the region as its size increases.
However, the concept of a "reasonable choice" of regions is itself
problematical.  Often when dealing with probability problems we can
impose a solution by stating that the problem implicitly assumed 
some "reasonable choice", but unless the boundaries of reasonableness
are defined in advance, this sort of argument is questionable.  The
point of rigorous probability problems is to include ALL possibilities,
no matter how remote.  The "improbability" or "unreasonableness" of 
certain remote cases is precisely what probability theory is intended
to quantify, so it would be misguided to gloss over such ambiguities
in the statements of problems.

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