# Random triangles

By Murray Bourne, 29 Aug 2007

I came across this interesting story at North Carolina Association of Advanced Placement Mathematics Teachers. The problem statement:

If a triangle is chosen at random, what is the probability that it is acute?

It turns out that solving this problem involves not only Euclidean geometry and Probability, but also Average value of a function, from calculus.

The great thing about this story (from Stu Schwartz, of Wissahickon High) is the grade 10 boy:

Kurt is a student of mine, the smartest math student I have ever encountered in 33 years of teaching. He is in 10th grade, having completed Calculus AB in 8th grade, BC in 9th. He is now taking multivariable calculus online given by Stanford. This morning he came into my stat class. I casually said to him, "Hey Kurt, what is the probability that a triangle chosen at random is acute?"

He appeared disinterested.

So when he stopped into my room for a study hall 2 hours later, I asked him if he had a solution. He said that he didn't work long on it but he believed the answer was 19.3%.

I was stunned because when I simulated the problem using Fathom (a terrific statistical program put out by Key Curriculum Press), I suspected the answer was 20%. How in the world did this kid do the problem?

First---my definition of a random triangle:

A = a random number between 0 and 180;

B = a random number between 0 and 180-A;

C = 180 - A - B.## Here is Kurt's logic:

First, using the definition above, the chance of choosing angle A to be obtuse is 0.5. So the chance of all three angles being acute must be less than 0.5.

If A=30, then the other two angles must add to 150. That means that one of the other two angles must be between 60 and 90 in order for the triangle to be acute (neglecting right triangles whose probability is zero). So the probability of the triangle being acute is (90-60)/150 = 30/150.

If A=70, the other two angles must add to 110. That means that one of the other two angles must be between 20 and 90 in order for the triangle to be acute. So the probability of the triangle being acute is (90-20)/150 = 70/150.

Suppose that A is chosen and is acute. That will leave (180-A)-90 and 90. 90-[(180-A)-90] = A. So the probability of the triangle with acute angle A being acute is A/(180-A).

Obviously there are infinite possibilities of angle A giving infinite probabilities of the angle being acute, each infinitely small.

So Kurt decides to sum these probabilities up and takes the average probability. So he integrates the expression x/(180-x) from 0 to 90 and divides by 90. This, of course, is the average value of a function. He ends up with 0.386.

He then multiplies by 0.5 because he must multiply the result by the probability that A was acute to begin with. He gets 0.193.

I have simulated this in Fathom 10,000 times and my results are just about the same. A terrific use of probability theory and calculus. This from a 10th grader.

Still has me in shock.

See the 9 Comments below.

1 Sep 2007 at 3:01 am [Comment permalink]

And if you randomly chose the lengths of the sides instead, would your answer be the same? 20% seems low, but I'm not certain how to dive in.

1 Sep 2007 at 3:14 pm [Comment permalink]

If we randomly chose 6 numbers from 0 to 1, and treated them as three ordered pairs, we would have a triangle on a unit square. I think the probabilities are quite different here. Also, unlike my previous proposal, the probability of generating a non-triangle would be very very small.

But let's look back at what the kid did. The first angle is chosen arbitrarily. OK. But all angles (0,180) are treated equally likely, but we know every triangle has 2 or 3 angles on (0,90) and only 0 or 1 on (90,180). It seems that the work has bias towards non-acute triangles.

1 Sep 2007 at 5:22 pm [Comment permalink]

[...] Prepuzzle puzzle: What is a random triangle? September 1, 2007 pm30 9:22 pm Posted by jd2718 in Math Education, Puzzles, mathematics, Math. trackback 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? [...]

4 Sep 2007 at 6:52 am [Comment permalink]

Hi, nice blog! I saw it mentioned over at Jonathon's blog. It's a cool puzzle but I have to say I'm a bit surprised at the teacher. For one, I agree with Jonathon, the method for generating random triangles is somewhat odd, and definitely skews the distribution, in my opinion (though it makes for more interesting/complicated math!) The method another commenter and I describe seems much more logical and elegant.

Also, I don't understand the teacher's (persistent) shock. Yes, the student's math skills are quite impressive. But the article says he completed BC calculus the previous year. Surely the integration skills involved in this problem should be well within his abilities? What I mean is, Kurt is extremely talented. But one of his teachers, one who knows that he's completed basic calculus courses already, shouldn't be shocked every time the student is able to apply his calculus.

But what really disappoints me is the teacher's use of this "Fathom" program. (I'm assuming the teacher in question is a math teacher and possesses knowledge of calculus.) Aren't math teachers supposed to emphasize and embrace logic and calculation over computers and number-crunching? The solution of course is

ln 2 - (1/2). The integration is quite simple and took me only a couple minutes. And I'm a doctor, not a math teacher.4 Sep 2007 at 9:31 am [Comment permalink]

Hi Darmok and thanks for your kind comment.

1. Your green/red triangle is a neat way to explain your solution.

2. Yeh, the integration is no big deal.

3. Math instructors

alwaysget excited when a student is motivated to complete non-assessed work, and especially when they have an intelligent go at it.4. You seem concerned about the math teacher using a tool to help solve the problem. I would hope that we have moved on from mindless calculation. (Logic is not mindless, I hasten to add. And I am certainly not advocating a calculation-free zone in math classes.) The tools are there to help, after all. But rubbish in, rubbish out, of course.

Isn't the desire to stop math teachers from using computers a bit like asking doctors to perform their duties without the benefit of high end tools like MRI and the like? If you've got the tool, and can use it appropriately, surely it should be used.

5. As a doctor, could I invite you to comment on Vlorbik's statement about doctor math on It's Fun to Hate Math. I still hang on to the naive belief that we are teaching mathematics for a purpose beyond algebra for its own sake, and something more than navel gazing.

13 Sep 2007 at 4:59 am [Comment permalink]

I suppose it depends on the tool and the intended use. I don't propose that computers not be used at all. But if a patient came to me with a leg injury, I'd first use my clinical skills to determine the nature of the injury, and if appropriate, obtain an x-ray. MRIs can be used for diagnosing "hidden" fractures that x-rays don't reveal, but they'd hardly be the first choice.

Computers and calculators are great. When I was taking calculus, I'd often use my calculator to approximate the answer to see if my result was reasonable. But the teacher in this case seems to have gone straight to "Fathom"â€”there's no indication he ever worked the problem out for himself.

I suppose you two and I have different views on mathâ€”the conceptualization and analytical thinking required for these problems is part of the appeal to me.

But the strangest thing for me is that this seems like such a simple problem. I don't know how adept he is at this "Fathom" or how long it takes to set it up, but it seems to me that solving such a trivial problem can't have taken much longer than it would take to set up the simulation. I suppose I just grew up with a different breed of math teacher.

18 Oct 2013 at 6:29 pm [Comment permalink]

Timing 0.5 is wrong.

Of course the probability of getting A less than 90 is 0.5

you time the probability of getting an acute triangle when A is less than 90 to the probability of A, but you don't realize that there are more triangle possible when A is less than 90, so the probability of "given that it is a triangle" A is less than 90 is 0.75 not 0.5

the correct answer of random angle is chosen and getting an acute triangle is 0.25

the probability of getting an acute triangle when 3 points is picked at random is 0 by the way.

The solution of this problem is really easy.

You couldn't see it and you use the hard way, and not even get it right.

Think about it.

18 Oct 2013 at 6:31 pm [Comment permalink]

Timing 0.5 is wrong.

Of course the probability of getting A less than 90 is 0.5

you time the probability of getting an acute triangle when A is less than 90 to the probability of A, but you don't realize that there are more triangle possible when A is less than 90, so the probability of 'given that it is a triangle' A is less than 90 is 0.75 not 0.5

the correct answer of random angle is chosen and getting an acute triangle is 0.25

the probability of getting an acute triangle when 3 points is picked at random is 0 by the way.

The solution of this problem is really easy.

You couldn't see it and you use the hard way, and not even get it right.

7 Mar 2016 at 11:13 pm [Comment permalink]

I'm pretty sure "random triangle" is not a well defined concept. Going about it in different ways yields different results.