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

 Messages: [ Previous | Next ]
 Mike Housky Posts: 79 Registered: 12/6/04
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 60-degree 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 Graphics--so he gets access to lots
of fun toys--and he Monte-Carloed 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.

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