# Homework Help: Probability to get a specific triangle

1. Feb 21, 2012

### Pzi

Three points are randomly chosen on a circle. Then they are connected together to make a triangle.

What's the probability to see a
1. Right triangle
2. Triangle with all angles less than 90 degrees
3. Triangle with an angle bigger than 90 degrees

I'm pretty sure to solve one of them means to solve them all... In other words I'm trying to come up with a decent method >.<

Thanks

2. Feb 21, 2012

### Staff: Mentor

pick a point on your circle then pick another point as you move the third point around how many right triangles can you make?

3. Feb 21, 2012

### Pzi

Well, once I have two points set, there are only two possible locations for the third point in order to form a right triangle. I don't know how that's enough though :/

4. Feb 21, 2012

### alan2

The answer to a) is zero. I have to think about the others.

5. Feb 21, 2012

### Staff: Mentor

say you had a 1000 free points around the circumference to choose from you now know that for a right triangle you have only two points out of a thousand. Say you increased it to 10000 then 100,000 ... still only two points. So given that the probability for making a right triangle is vanishingly small.

If you move the third point back and forth a bit what is the chance of getting an acute vs obtuse triangle?

6. Feb 21, 2012

### Pzi

When I come to think of that again... If the distance between my first set two points is equal to the diameter of a circle then ANY third point will complete a right triangle (as long as it doesn't coincide with first two points, but this one is indeed infinitely unlikely event).

It casts some uncertainty about the right triangle unlikeliness :/

7. Feb 21, 2012

### Staff: Mentor

If you move the third point back and forth a bit what is the chance of getting an acute vs obtuse triangle?

8. Feb 21, 2012

### Pzi

As first two points are getting further apart almost to form a diameter, the probability of getting an acute (the same for obtuse) angle is indeed getting closer to 1/2. I'm not sure what happens when the first two points are chosen close to each other...

Edit: I think when the first two points are close to each other one is almost certain to complete an obtuse triangle.

Edit2: Indeed. Now I'm pretty much certain that probability to complete an acute triangle is 1/4

Last edited: Feb 21, 2012
9. Feb 21, 2012

### kai_sikorski

In polar coordinates, we have to pick θ1, θ2 and θ3.

just assume that θ1=0, since we can always rotate the result and redefine our coordinates so that this is true. By Thales theorem in order to get a right angle we need one of the other points to land on π. Using the fact the two random variables are independent we get

P(θ2 = π OR θ3 = π) = P(θ2 = π) + P(θ3 = π) - P(θ2 = π)P(θ3 = π)

I'll let you figure out what the above quantity is for part 1. Now in the figure below, lets say the point B corresponds to θ2, we'll assume that 0 ≤ θ2 ≤ π WLOG, because in we can always just redefine the direction of increasing angle to make this true.

I took the image from wikipedia, so I don't have the ability to modify it but pretend there was a point at π + θ2 labeled D. I contend that the region between A and D is where θ3 can land to satisfy the conditions of the second problem. Use the law of total probability

$$P\left(\text{All Triangle Angles} < \frac{\pi}{2}\right) = \int_0^{\pi} P(\theta_3 \in AD| \theta_2 = y)p(\theta_2 = y) dy$$

I'll let you figure out what the two functions P and p in the integral are, and do the integral (just remember that I basically redefined θ2 to be between 0 and π instead of 0 and 2π). If you figure out part 2 then part 3 is given by

P(Triangle with one angle > π/2) = 1 - P(Triangle all angles < π/2)

Last edited: Feb 21, 2012
10. Feb 21, 2012

### SteveL27

Geometrically, three points make a right triangle if and only if two of them are opposite points on a diameter.

If the three points lie in the same half circle, the triangle contains one obtuse angle. And if the three points fail to lie in a half circle, then all three angles are acute.

It seems to me that the the first two points can be arbitrary. The third point can be chosen to be in the same half-circle or a different half-circle.

The right triangle case has far fewer possibilities (measure-wise) than the other two cases, so that probability is zero. (Needs more rigorous proof).

For the other two cases, once you choose any two points and draw both the possible semicircles (extending each point through the center and to the other side of the circle) you see that there is always a larger area where the third point is still within one of the possible semicircles.

So the chance of getting three acute angles is always a little smaller than the chance of getting one obtuse angle.

Perhaps this might be a useful approach.

(edit) I've convinced myself that the probability of three acute angles is 1/8. That seems counterintuitive but I think it's right.

Once you pick two points you've defined two semicircles, whose union defines the obtuse case [by regarding each point as one endpoint of a diameter, and unioning the diameters. In other words there are two ways the third point can be in the same semicircle with the other two: one corresponding to each of the diameters]. So the problem is reduced to asking what is the average acute angle between two random lines in the plane. By symmetry the largest angle we need to care about is pi/2 (because if we go past pi/2 then its the angle with the negative x-axis that matters).

The average of a randomly chosen angle between 0 and pi/2 is pi/4. So the probability that all three points are NOT all in the same semicircle (three acute angles) is pi/4 and the probability that the three points ARE in the same semicircle (one obtuse angle) is 7pi/4.

Last edited: Feb 21, 2012
11. Feb 21, 2012

### kai_sikorski

Yes but the point is that, the probability of the first two falling in such an arrangement is 0. Look at my other post.

12. Feb 21, 2012

### kai_sikorski

No thats not true. Say the point angle is in polar coordinates at 0, the second a 2 π/3, by your argument putting the third point at 4 π/3 doesn't work, where this is clearly the equilateral triangle.

13. Feb 21, 2012

### alan2

Look at Kai's post above. The solution is similar. The probability of an acute triangle is 0.75, while the probability of an obtuse triangle is 0.25. I wish I knew how to use the cool editing features but I don't so here goes, you can construct it. Because of the symmetry of the circle, only the relative positions of the points matter. So the problem is equivalent to choosing a point, say A, and choosing an angle, theta, between 0 and pi from a uniform distribution. The point A is the vertex of that angle. Then you have to consider all of the possible orientations of each angle. For example, an angle of pi can only be oriented one way but a very small angle can be oriented many ways. So the angles are weighted by the possible orientations, which in this case is pi-theta. Integrate this expression from 0 to pi to find your normalization, in this case 2/pi^2. So your density is (2/pi^2)(pi-theta). Integrate from 0 to pi/2 and you find the probability of an acute triangle is 0.75. integrate from pi/2 to pi and you find that the probability of an obtuse triangle is 0.25. I wish I had a blackboard.

14. Feb 21, 2012

### kai_sikorski

The answer using my analysis is 1/4. This is confirmed using the Mathematica commands below.

Code (Text):
f[] := {T1 = Random[]*2 \[Pi]; T2 =
Random[]*2 \[Pi]; T3 = Random[]*2 \[Pi];
p1 = {Cos[T1], Sin[T1]};
p2 = {Cos[T2], Sin[T2]};
p3 = {Cos[T3], Sin[T3]};
a1 = VectorAngle[p2 - p1, p3 - p1];
a2 = VectorAngle[p1 - p2, p3 - p2];
a3 = VectorAngle[p1 - p3, p2 - p3];
{a1, a2, a3}}[[1]]

Code (Text):

1 - Mean[
Table[ Total[(If[# > Pi/2, 1, 0] & /@ f[])], {i, 1, 10000}]] //
N

Also nice explenation here

15. Feb 21, 2012

### kai_sikorski

You got it backwards

16. Feb 21, 2012

### SteveL27

My visualization of the acute case is off by a factor of 2 somewhere ... not sure I see why at the moment.

17. Feb 21, 2012

### kai_sikorski

It's the problem I pointed out to you before. In my diagram when B goes past π/2, you're changing the 'good' region from (π,π + B) to (2π-B,2π) but that's not right. Even when B goes past π/2, the 'good' region is still (π,π + B).

18. Feb 21, 2012

### SteveL27

Oh I see it now. When the angle goes past pi/2 I can NOT just flip my orientation and consider the complementary acute angle. The acute region is always the angle in the positive direction ranging from zero to pi between the first and second points. So now I can take the average of a random angle between 0 and pi to get pi/2 as the average acute region, leaving 3pi/2 as the obtuse region.

Fun problem.

19. Feb 22, 2012

### alan2

Thank you Kai. I got it backwards. The probability of an obtuse angle is 0.75, acute is 0.25.

I did find an easier way. There is an obtuse angle iff the arc length between any two points is greater than one half of the circumference. So the problem is equivalent to picking two points from the interval [0,1], thus subdividing the interval into 3 sub-intervals, and finding the probability that any sub-interval has length >1/2. The result is as Kai found.

20. Feb 22, 2012

### Pzi

Hi guys.

Thanks very much for the thoughts!

A variety of ideas is already presented here, so I don't have much to add besides this little experiment when I simulated 1,000,000 events (circle was simplified to 1,296,000 points) and evaluated the largest angle of every triangle:
http://img542.imageshack.us/img542/7431/angles.png [Broken]
The largest angle is a decisive factor for knowing the type of a triangle.

I found it interesting, because this interval of 90-180 degrees (meaning an obtuse triangle) is three times as big as interval of 60-90 degrees (meaning an acute triangle) while the ratio of P(obtuse) and P(acute) is also exactly three. So I wanted to see how the likelihood of those angles is distributed.

Last edited by a moderator: May 5, 2017