If “random” above means uniform, then I believe the above statement is true. However, the uniform distribution over the entire co-ordinate plane is not a proper probability distribution. For any other distribution, let us say we pick points x1,y1,x2,y2,x3,y3 from that distribution. The lengths of the sides are L1^2 = (x1-x2)^2+(y1-y2)^2, and similarly for L2^2 and L3^2. Then the probability of an acute triangle is the probability that (L1^2+L2^2 > L3^2) and (L2^2+L3^2 > L1^2) and (L3^2+L1^2 > L2^2). I find it inconceivable that this probability will be zero. There may be infinitely more points in the region that lends itself to an obtuse triangle, but the integral of the joint probability distribution over that region is finite, and so it is over the region where the acute triangle is formed.

For a N(0,1) Gaussian distribution on the coordinate points, the calculation of the probabilities above seems workable, but is quite tedious. As I had said before, a simulation yields a probability of 0.25 for the acute triangle.

Again taking to be uniformly distributed in , we find that the likelyhood of an acute triangle is .

Clueless suggested we fix two edges of length and then choose the angle between them uniformly between 0 and 180 degrees. Clearly, if we choose the angle larger than 90 degrees we get an obtuse triangle (with the third side as the longest). Similarly, if we choose the angle smaller than we get an obtuse triangle with the edge of length as the longest. Thus, the probability of an acute triangle in this case is , which ranges between 1/2, when a = b, and approaches 0 when a/b approaches 0.

We can also ask that we take uniformly distributed in the interval . (This amounts to defining a random triangle as a triangle such that at one vertex, the ratio of the two sides which meet there is uniformly distributed in [0, 1] and the included angle is uniformly distributed in [0, 180].) In this case, doing the integration gives , smaller than most values we’ve considered.

My assertion is this: Pick three points from a random distribution over the entire coordinate plane. The triangle they form will only be acute if the projection of one point onto the line defined by the other two points falls between those two points. Since there are infinitely more points whose projections lie outside this interval, there should be an essentially zero chance of the three random points forming an acute triangle. It doesn’t matter which point you select to project onto the line formed by the other two points.

JD, the same thought went through my head! I was considering how if I cut off some of the vertical aspect it would make a nice math-based flag! I, too, have been quite surprised at how low the probability of acute triangles is: 0.25 if picking random angles, ~0 if picking random coordinates. It’s especially odd since in general if you ask someone to “draw a random triangle”, I’m sure it will almost definitely be acute!

This one is actually quite easy, although the answer isn’t very pretty. Then we can extend it: for what values of a and b (if any) is the answer to this version maximal? Minimal? (We can also ask this extension of Clueless’s last question.)

Actually, that leads to another offshoot problem: Given lengths a and b, a>b, find the probabililty of an acute triangle if you can choose the included angle at random between 0 and 180 degrees.

The probability of an acute triangle does go to zero if you restrict one side to be of side 1, and the other point to be “uniformly distributed in the infinite space” around it. Once you start drawing the third point from a probability distribution (such as Gaussian), you start getting other answers. Also, the length of the side you start with also follows a certain probability distribution, so you have to integrate over that also. I believe this overall integration leads to a probability of 0.25 (of obtaining an acute triangle) when all the six coordinate points are chosen from a Gaussian distribution.

It is also clear that a significant portion of the space around the line produces obtuse triangles. Once we start restricting that space, for example, using a uniform distribution over the unit square, we get a higher probability of an acute triangle. If you restrict it even more, say over a unit circle, you get an even higher probability. This is the (after the fact) explanation for my simulation results.

And, fwiw, I’m also bothered by how improbable acute triangles seem, no matter which definition of random we seem to use.

All right, so I too found Clueless’s offshoot intriguing, and as I thought about it, it seems to me that it’s a way of investigating Rdt/Rachel’s speculation above about selecting random coordinates over the entire coordinate plane. I’ve never been very good with infinities, and I don’t really know if my logic is correct—I’m going more with what “seems” right since I don’t remember the actual laws and theorems and such (and that already got me in trouble with SSA earlier on this blog). Someone want to take a look at my reasoning?

So I want to pick three random points from anywhere in the plane. I’ll start by picking two of them. I’ll define my coordinate axes so that one point is at (0,0) and the other is at (1,0). (That’s allowed, right? I can rotate and stretch as I like, though it’s not really necessary for this, but it makes it easier to discuss.)

I’m a visual person, so I again sketched it in Paint. It uses the same color scheme as my previous sketch. The two initial points are represented by the two white points along the horizontal line. The third point can be selected from anywhere in the coordinate plane, and I’ve colored the plane based on the type of triangle that would be formed. The black central horizontal line represents the “one-dimensional” triangles. The blue vertical lines and the tangential circle are the right triangles, and they separate the red (obtuse) from the green (acute) triangles. Again, for fun, I threw in the isosceles curves (the central vertical and the two circles, and they intersect in two yellow points representing the equilateral triangles.

So, it seems to me, that (with the obvious restriction of y≠0, if the third vertex is x 1, the triangle will be obtuse. Only for 0 < x < 1 (and not counting that circle ((x-0.5)² + y² < 0.5²) would we have acute triangles. Wouldn’t this make the probability of an acute triangle tend to zero?

Something about this analysis disturbs me, but I’m not sure what. Is it in my assumption (pick two points at random from the coordinate plane)? Can that not be done? I mean, you could pick all three points randomly to begin with, and than select any two and perform this analysis—in fact, you should be able to do it for all three pairs of vertices. And yet it doesn’t seem quite right to me. I wouldn’t have expected the acute triangles to be so low. Can someone explain it to me?

I cheated and used a simulation the different cases. I am sure an analytical calculation is possible, but I am an engineer, and this is close enough.

Your offshoot problem is a nice one. I won’t post the solution so others can try to solve it as well. =)

If they are picked from a uniform distribution over a square (say, bounded by (0,0),(0,1),(1,0), and (1,1)), the probability of an acute triangle is higher (about 0.2749)

If they are picked from a uniform distribution over a circle, the probability of an acute triangle goes up to about 0.2805

I find it interesting that the first case gives the same answer as the “random” angles.

Also, one interesting (at least I found it so) offshoot: Given two points, find the locus of the third point so that the resulting triangle is obtuse.