Geometric probability

Daniel123 · 2008-10-27 01:13:54

Please don't give too much away here, I'm just looking for small hints.

Let A, B and C be three randomly and independently chosen points on the circumference of a circle. What is the probability that angle BAC is acute?

My method:

Fix two points anywhere on the circle (but not part of the same diameter), and let the third point vary.

If the third point lands within the same semicircle as the first two (probability of this is 1/2), then there is the possibility that one of the angles in the triangle is acute (probability that it is angle BAC that is acute is 2/3). The probabiltiy of angle BAC being acute, given that all three points lie within the same semicircle, is therefore 1/2 x 2/3 = 1/3.

If the third point does not land within the same semicircle as the first two (probability of this is 1/2), then all three angles must be acute. The probability of angle BAC being acute, given that the three points do not lie within the same semicircle, is therefore 1/2.

1/3 + 1/2 = 5/6.

Basically,

1) Is my method above correct?
2) How do I deal with the possibility of having a right angled triangle (which I haven't considered). I can't see how to calculate the probability of this?

Thanks.

Last edited by Daniel123 (2008-10-27 01:14:30)

mathsyperson · 2008-10-27 08:25:43

You came up with the brilliant idea that a cyclic triangle will have an obtuse angle iff its vertices all lie on a semicircle's arc. I'd explore that idea more.

[Edit: Just realised 'cyclic' is redundant there. All triangles are cyclic. ]

You also made a few false assumptions though:
- If you're fixing two of the points, then they're not all randomly and independently chosen.
I'd say the most specific you can get is to fix one point at the top (which you can do by rotating the general circle) and put the second point somewhere on the left side (which you can do by reflecting if you need to).
So one point can go anywhere on half of the circle, and the other point can go anywhere on all of the circle.

- It's not a 1/2 chance that the three points lie on a semicircle. In the extreme case, if the first two points are on the same point then the third point will be certain to share a semicircle with them. In general, the closer together the two points are, the more likely a shared semicircle will exist.

To answer your question about right angles, I don't think you need to worry.
Assuming the triangle's vertices can be on any point of the circle (with infinite precision) then two of them sharing a diameter will "almost never" happen.
That's a proper mathematical phrase when working with infinetissimal stuff, and means that even though it could technically happen, it has a probability of 0.

Daniel123 · 2008-10-27 08:32:54

Great, thank you mathsyperson.

And yeah, I was thinking along the lines that "the probability of a right angled triangle occuring is 0, but it could still happen..."

I'll get back to you when I've sorted it all out.

TheDude · 2008-10-29 06:01:43

I'm not sure if this is easier or harder than what you're trying, but I immediately thought of integration when I read this problem. What I'm thinking is something like this: Fix point A to be at the top of the circle, which is legal thanks to mathsyperson's logic. Then imagine that we start point B to the top of the circle too. We then move it incrementally around the circumference of the circle. At every increment we calculate the probablility that point C, if randomly placed, will form an acute angle with A and B. Add these all up and you have your answer.

In mathematical terms we want to integrate around the circumference of the circle, so 0 to 360 degrees (or 2pi, whichever you prefer). Using mathsyperson's logic again we can simply integrate from 0 to 180 and multiply it by 2 since the rest is just a reflection of the first half. All that's left is to determine what function we're integrating.

What you want to integrate is a function of the probability that C, if randomly chosen, will form an acute angle BAC. To get an idea of what we're dealing with let's make a mental picture. We have a circle with point A being fixed to the top, which is equivalent to saying that its position is 0 degrees/radians. B is positioned at some point on the right half of the circle, with a position between 0 and 180 degrees. Where on the circle can we place C such that BAC is acute?

It's easier to determine where BAC is obtuse, so let's do that and then subtract that probability from 1 to get the probability that BAC is acute. First, you can clearly see that C must be on the left half of the circle in order for BAC to be obtuse. You can also see that the degree of BAC will increase as C approaches A. As long as B is not at exactly 0 or 180 degrees there is a point on the left half of the circle where BAC will be equal to 90 degrees. If C is closer to A than this point then BAC is obtuse, otherwise it is acute. I will now point out that BAC forms a right angle if B and C lie on the same diameter. This should be easy enough to prove, and from this point on you can solve the problem.

If you want a solution to check your answer against I came up with 75%. Maybe one of the smarter people here can confirm or deny that result.

mathsyperson · 2008-10-29 07:42:59

I agree with your answer. I'm not sure integration is needed though, use of symmetry should be enough.

Daniel123 · 2008-11-12 07:04:50

Okay, I get the answer as 3/4.

lwttu2bk1trxpic.html

Consider the unit circle.

Define the argument of A,B,C to be the angle made with the positive x-axis, where 0 ≤ θ ≤ 2pi.

Fix A at the point shown. We can confine B to the top half of the circle without any loss of generality i.e. 0 < arg(B) < pi

For the triangle to have no obtuse angles, A must not lie on the same semicircle's arc as B and C i.e. arg(A) < arg(B) + pi.

The probability that angle BAC is acute given the position of point B = the probability that arg(A) is smaller than arg(B) + pi = (arg(B) + pi)/(2pi).

Assume that arg(B) is uniformly distributed over [0,pi]. The expected value of arg(B) is therefore pi/2.

Thus the probabiltiy that BAC is acute = (pi/2 + pi)/(2pi) = 3/4

EDIT: Swap A and C.

Last edited by Daniel123 (2008-11-13 03:45:09)

Math Is Fun Forum

#1 2008-10-27 01:13:54

Geometric probability

#2 2008-10-27 08:25:43

Re: Geometric probability

#3 2008-10-27 08:32:54

Re: Geometric probability

#4 2008-10-29 06:01:43

Re: Geometric probability

#5 2008-10-29 07:42:59

Re: Geometric probability

#6 2008-11-12 07:04:50

Re: Geometric probability

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