
#1
Feb2112, 01:20 PM

P: 26

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
Feb2112, 02:00 PM

P: 2,475

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
Feb2112, 02:05 PM

P: 26





#4
Feb2112, 02:26 PM

P: 188

Probability to get a specific triangle
The answer to a) is zero. I have to think about the others.




#5
Feb2112, 02:31 PM

P: 2,475

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
Feb2112, 02:44 PM

P: 26

It casts some uncertainty about the right triangle unlikeliness :/ 



#7
Feb2112, 02:46 PM

P: 2,475

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




#8
Feb2112, 02:56 PM

P: 26

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 :)) 



#9
Feb2112, 06:54 PM

PF Gold
P: 162

In polar coordinates, we have to pick θ_{1}, θ_{2} and θ_{3}.
just assume that θ_{1}=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(θ_{2} = π OR θ_{3} = π) = P(θ_{2} = π) + P(θ_{3} = π)  P(θ_{2} = π)P(θ_{3} = 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 θ_{2}, we'll assume that 0 ≤ θ_{2} ≤ 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 + θ_{2} labeled D. I contend that the region between A and D is where θ_{3} can land to satisfy the conditions of the second problem. Use the law of total probability [tex] 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[/tex] 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 θ_{2} 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) 



#10
Feb2112, 06:54 PM

P: 799

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 halfcircle or a different halfcircle. The right triangle case has far fewer possibilities (measurewise) 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 xaxis 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. 



#11
Feb2112, 08:05 PM

PF Gold
P: 162





#12
Feb2112, 08:17 PM

PF Gold
P: 162





#13
Feb2112, 08:29 PM

P: 188

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 pitheta. Integrate this expression from 0 to pi to find your normalization, in this case 2/pi^2. So your density is (2/pi^2)(pitheta). 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
Feb2112, 08:33 PM

PF Gold
P: 162

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




#16
Feb2112, 09:07 PM

P: 799

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




#17
Feb2112, 09:21 PM

PF Gold
P: 162





#18
Feb2112, 09:54 PM

P: 799

Fun problem. 


Register to reply 
Related Discussions  
Probability of measuring specific energy of particle (in a box)  Advanced Physics Homework  5  
Probability of a Specific 5 Card Poker Hand  Set Theory, Logic, Probability, Statistics  2  
Probability: a specific hand of cards  Precalculus Mathematics Homework  1  
Probability of measuring specific energy of particle  Advanced Physics Homework  18  
Probability that the circumcenter lies inside the triangle  Set Theory, Logic, Probability, Statistics  0 