
Random Triangle Problem
Posted:
Jul 16, 1997 9:33 PM


Greetings to the newstgroup. I'm an amateur mathematician at best, with a formal training that does not go beyond lower division calculus and linear math. I'm active in the <alt.algebra.help> newsgroup and that apparently got me noticed by a student at Cal State Sacramento, who sent me the following problem via email. In his words:
> Derive a method of finding the chances of 3 random points placed on > a plane that will yield an obtuse triangle.
My first reaction was that it's impossible to pick a random point in the plane with anything like the usual assuption that (without further qualification) any candidate points are equally probable. Picking three such points is harder. :)
However, I thought about it some more and tried to find a manageable definition of an arbitrary triangle that fits the notion of randomness implied in the original problem statement. This might be my first mistake, but I start by assuming that such a triangle exists. By the law of sines, the largest angle is opposite the longest side. Of course, if the triangle is isosceles or equlateral, then one "longest" side may have to be chosen arbitrarily.
Now, I choose to define a Cartesian coordinate system where the longest side lies on the x axis and the measure of length is 1 for that side. I then choose the positive y direction to be the side of the x axis that contains the third vertex of the triangle. I'm free to pick the zero point on the x axis, and I pick the left point on the long side as (0,0) and the right point is (1,0). Call these A and B, respectively, and the third point C. From the definitions, C can't be farther than 1 from either A or B, so C must lie in the intersection of two circles with radius 1, one centered at A and the other at B. Further, since the orientation of the y axis was chosen to put C on the positive y side, then C must lie on the "upper" half of that intersection.
Everything so far, with the possible exception of the assumption that I can pick an arbitrary triangle in the first place, seems fine to me so far. The next step bothers me more. I assume that all of the candidate points for C, within the upper intersection of the two circles, are equally probable. That is, the probability density function for the location of C is a positive constant that integrates to 1 over the intersection region and is zero everywhere else. This is the assumption that bothers me.
Given that assumption, there's a neat "solution": The intersection of the circles is also the union of 60degree sectors of those circles. Each sector has an area of pi/6, so the sum is pi/3...except we have counted the equilateral triangle region in the middle twice. This triangle has a side of 1, so the area is sqrt(3)/4, and the full area of the intersection is pi/3  sqrt(3)/4. Now, all the candidate points for C that create a right triangle lie on the positive y side of the circle whose diameter is AB, or (0,0)(1,0). This is a semicircle that caps an area of pi/8, so the probability would seem to be (pi/8)/(pi/3  sqrt(3)/4)).
All this depends on the assumption that point C is uniformly distributed over the candidate area formed by the union of the two semicircles. Is this valid?
The probability calculates to a value about 0.64 which is not far from what the original questioner's professor speculated as about 65%. If this is correct, a harder problem follows. I described the problem to a good friend of mine wjp is a physicist that currently works for Silicon Graphicsso he gets access to lots of fun toysand he MonteCarloed the problem to get something like 72.5% probability in favor of an obtuse triangle. However, his random points were taken essentially from the unit square, and that adds a constraint that wasn't present or implied in the original problem statement. So, the harder problem is do figurue out what the probability should be for the random triangle whose vertices are taken uniformly from the unit square.
I hope this tickles someone else's imagination and apologize if if it costs anyone some needed sleep.
Cheers, Mike.

