Zac who commented here has a blog, squareCirclez, and I was reading it and found a question he found elsewhere. I would call it a puzzle:

What is the probability a random triangle is acute?

He reprinted a story, and linked to the source, the Winter 2004 bulletin of the North Carolina Assoc of AP Math Teachers. The kid does the work based on the teacher’s definition of a random triangle:

1. choosing angle A, $0 < A < 180$
2. choosing angle B, $0 < B < 180 - A$
3. choosing angle C, $C = 180 - (A + B)$

Whoa?!

• Step 1 includes the assumption that half of all triangles are obtuse. Part of the conclusion is buried in a shaky assumption.
• Even in obtuse triangles, two thirds of the angles are acute.
• Why should we randomly choose angle measures?

What is a random triangle? how can we generate them?

I have 3 ideas:

• Lengths. Randomly choose 3 numbers on a given interval. We will get lots of non-triangles this way. On the other hand, dividing a given number into 3 pieces may skew the results – the constraint makes the process non-random.
• Coordinates. Randomly choose 6 numbers on a given interval: (a,b), (c,d), (e,f). Perhaps we can integrate? multiple integrals?
• Angles. Three numbers from (o,2π). And plot them as points on a unit circle. Will the constraint do something funny. It feels like a relative of my mad carpenter or walking stick puzzles.

Ideas? Thoughts? I can’t believe that any is worse than what the article reported.

September 2, 2007 am30 4:56 am 4:56 am

At some level I think the question itself points towards some of the deep questions of probability and statistics about “priors.”

But the least constrained definition I can think of is to pick 3 points at random in the coordinate plane (no constraints on the coordinates, so all the triangles will be huge).

I’m not sure if you’d get a very different answer if you picked 3 points at random within a unit circle…

— Rachel

September 2, 2007 am30 5:18 am 5:18 am

Yeah, that bothered me too! Choosing the first angle uniformly at random from the interval 0 to 180 is definitely strange, since it means that the second angle is not independent of the first.

Of course, there’s clearly no one “correct” definition of what it means to choose a “random” triangle. However, since the original question had to do with angle measures it makes sense to use angles to select a random triangle. Here’s how I would do it: think of a triangular region in the first quadrant, with one corner at the origin, one at (0,180), and the third corner at (180,0). Think of the x coordinate as determining one angle of a triangle, the y coordinate the other. (Then the third angle is determined by the first two.) Then every point in this region represents a valid triangle, and conversely every triangle is represented by at most three points in the region. To pick a random triangle, just pick a random point in this region (for example, by picking the x and y coordinates independently and uniformly over (0,180), and reflecting the chosen point over the line y = 180 – x if it represents an invalid triangle).

According to this definition, exactly 1/4 of all triangles are acute (those which are represented by points in the region bounded by x <= 90, y <= 90, and x + y >= 90).

@Rachel: I don’t think it’s possible to pick points uniformly from the entire (infinite) coordinate plane; you either have to have some sort of bound (like a unit circle), or else use a non-uniform distribution, e.g. where the probability of choosing a point approaches zero the farther you get from the origin.

3. September 2, 2007 am30 5:31 am 5:31 am

Brent, I’m with you. I also have an idea for a little system of inequalities application in algebra II…. Thanks!

However, I don’t get that the original question asked about angles. I think the teacher chose to think of random in that sense.

Rachel, I think picking points (using reals) in a 1×1 square rather than on a unit circle might approximate choosing points in the plane. The extremes in size are just as great, from length = 1 down to as small as you like.

I’m just not sure if that is a better “random.”

September 2, 2007 am30 6:31 am 6:31 am

The original question asked how many triangles are acute; acuteness is a property of triangles that has to do with angles. That’s why I said that the original question had to do with angle measures.

September 3, 2007 am30 4:15 am 4:15 am

I guess this is where the physicist in me kicks in — what the triangles “are” will determine what random means.

For example, one of the not-very-intuitive bits of statistics in astronomy concerns the angles that binary star orbits make with the “plane of the sky.” Most people’s first guess is that the inclination angles would be uniformly distributed between 0 “face on” and 90 “edge on” (or 0 and 180 if you take into account the direction of the orbital motion).

But since the reality is 3 dimensional, not one dimensional, the problem is more complicated, and in fact the cosines of the inclination angles are uniformly distributed between 1 and -1, and there are as many binaries with inclinations between 60 and 90, as between 0 and 60.

So I guess I don’t think there is any unique definition of a “random” triangle. Different ways of looking at triangles will give you different underlying distributions of angles/sizes etc.

— Rachel

6. September 4, 2007 am30 8:58 am 8:58 am

I have to admit, my idea of a random triangle jumps naturally to angles. I suppose it is my non-mathematician perspective; I tend to view triangles in terms of angles, not their sides. (Especially since, to me, similar triangles are not really unique—the angles define the triangle, in my mind.) I guess I view triangles with two independent variables—two of the angles—with a third independent (but less important) variable being the length (say, of the longest side).

If I were asked to pick a random triangle, I suppose I’d do it in the following way: define a region of the coordinate plane bounded by the x- and y-axes and the line from (0,180) to (180,0). Randomly select a point in the region. Two of the coordinates give two of the angles, and the third is fixed by these two, obviously. While I make no claims as to the superiority of its randomness, I am pretty sure that this gives a different distribution than the procedure outlined in the puzzle; this would make it relatively unlikely to give an angle near 180. Hmmmm…this is actually quite intriguing!

7. September 4, 2007 am30 9:40 am 9:40 am

I took the concept I was explaining above and sketched it in Paint. The two axes represent the two independent angles. The black boundaries are the not part of the possible triangles, they are the “one-dimensional” triangles (they include at least one 0-degree angle).

Assuming I did this correctly (sort of off the top of my head), the red regions represent obtuse triangles, the green region acute triangles, and the blue boundary between them are the right triangles. Just for fun, I threw in the three gray lines which I think represent the isosceles triangles (and of course, the point at which they intersect, in yellow, is the equilateral triangle). Any triangle not on those lines is scalene. Did I do that right?

If so, it looks to me that (given this method of randomization), the probability of an acute triangle is one-fourth.

8. September 4, 2007 am30 10:12 am 10:12 am

Ah, I somehow missed the bulk of Brent’s first comment, which better describes the method I was trying to explain.

September 4, 2007 pm30 6:48 pm 6:48 pm

That’s a nice visualization, Darmok! =)

September 5, 2007 pm30 4:44 pm 4:44 pm

If we pick the coordinates from a Gaussian distribution, the probability of an acute triangle is 0.25

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.

September 5, 2007 pm30 7:37 pm 7:37 pm

Clueless — how did you compute those probabilities?

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

September 5, 2007 pm30 7:57 pm 7:57 pm

Brent,

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.

13. September 6, 2007 am30 8:47 am 8:47 am

I’d integrate over the region of R^3 defined by x, y, z >= 0, x + y + z = 180. That is a 2-dimensional figure that, under projection, is isomorphic to the shape Darmok describes.

14. September 7, 2007 am30 11:22 am 11:22 am

Yes Alon! Some time after I posted my numerous comments I realized that the sketch I posted would be the projection of a symmetric, equilateral triangle on to the xy plane. That’d make the symmetry even better!

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?

15. September 7, 2007 am30 11:25 am 11:25 am

Forgot that < and > are used in HTML. The third paragraph should say that if x < 0 or x > 1, it’ll be obtuse.

16. September 7, 2007 pm30 3:12 pm 3:12 pm

Clueless, that offshoot problem? Nice. And Darmok, I like that “flag” answer. It’s obvious now, but that’s because your visual solution made it so.

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

September 7, 2007 pm30 4:24 pm 4:24 pm

Hi Darmok,

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.

September 7, 2007 pm30 4:37 pm 4:37 pm

Some meta-thinking on the low probability of acute triangles: Let us consider a much more constrained problem. You are given the lengths of two sides, and are allowed to choose the included angle at random. (This angle is uniform between 0 and 180 degrees). Then it is clear that a loose upper bound for the probability of an acute triangle is 1/2. This is consistent with the low values we get in the general case.

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.

September 8, 2007 am30 12:34 am 12:34 am

Or, another variation: suppose you’re given sides of length a and b, and the third side length is selected uniformly from the interval $\left[|a - b|, a + b\right]$ — what is the probability of an acute triangle?

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.)

20. September 8, 2007 am30 7:57 am 7:57 am

Clueless, I make no such restriction! I was merely superimposing a new coordinate system onto the plane for ease of discussion.

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!

September 9, 2007 pm30 7:35 pm 7:35 pm

Some more values:

Clueless suggested we fix two edges of length $a < b$ 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 $\cos^{-1}(a/b)$ we get an obtuse triangle with the edge of length $b$ as the longest. Thus, the probability of an acute triangle in this case is $\frac{90 - \cos^{-1}(a/b)}{180}$, which ranges between 1/2, when a = b, and approaches 0 when a/b approaches 0.

We can also ask that we take $a$ uniformly distributed in the interval $[0, b]$. (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 $\frac{1}{2} - \frac{1}{\pi} \approx 0.1817$, smaller than most values we’ve considered.

September 9, 2007 pm30 8:22 pm 8:22 pm

For my question, (taking $a < b$ again) we note that the third side must lie between $\sqrt{b^2 - a^2}$ and $\sqrt{b^2 + a^2}$, so the probability is $\frac{\sqrt{b^2 + a^2} - \sqrt{b^2 - a^2}}{a + b - (b - a)} = \frac{\sqrt{b^2 + a^2} - \sqrt{b^2 - a^2}}{2a}$. This is at is maximum of $\frac{\sqrt{2}}{2}$ when a = b and approaches 0 when a/b approaches 0.

Again taking $a$ to be uniformly distributed in $[0, b]$, we find that the likelyhood of an acute triangle is $\frac{\sqrt{2} - \ln(1 + \sqrt{2})}{2} \approx 0.2664$.