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 under-specified. 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.Return to MathPages Main Menu