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Three points are chosen uniformly at random on a circle. What is the probability that no two of these points form an obtuse triangle with the circle's center?

 Dec 12, 2019
 #1
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The probability is 3/8.

 Dec 12, 2019
 #4
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I did not get 3/8 as my answer.

goreisthebest  Dec 14, 2019
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Well if you know the answer why don't you share it. Along with an outline of your working.

 

I still do not know what uniformaly random means. Sounds contradictory to me.

Melody  Dec 14, 2019
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It is 3/16.  

Let us call the circle's center O . We first note that if A and B are points on the circle, then triangle AOB is isosceles with AO=BO. Therefore, if  is an obtuse triangle, then the obtuse angle must be at O. So AOB is an obtuse triangle if and only if minor arc AB has measure of more than 90 degrees.

Now, let the three randomly chosen points be D, E, and F. Let theta be the measure of minor arc DE . Since theta is equally likely to be any value from 0 to pi (in radians), the probability that it is less than pi/2 is 1/2.

Now suppose that theta is less than pi/2. For the problem's condition to hold, it is necessary and sufficient for point F to lie within pi/2 of both D and E along the circumference. This is the same as saying that  F must lie along a particular arc of measure pi-theta.



The probability of this occurrence is (pi-theta)/2pi, since F is equally likely to go anywhere on the circle. Since the average value of theta between 0 and pi/2 is pi/4 , it follows that the overall probability for  is (1/2) - [(pi/4)/2pi]  or 3/8.

Since the probability that theta is less than pi/2 is 1/2, our final probability is 1/2 * 3/8 or 3/16 .

Hope this helps. 

~~GF

goreisthebest  Dec 15, 2019
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Hi  Gore,

 

You have totally ignored the word uniformly which is probably fair enough since it made absolutely no sense.

 

Your logic is good but I did need to think about this red bit. 

I suppose it does makes sense.  I've just do not remember logic like that used before.

 

The probability of this occurrence is (pi-theta)/2pi, since F is equally likely to go anywhere on the circle.

Since the average value of theta between 0 and pi/2 is pi/4 , it follows that the overall probability for  is (1/2) - [(pi/4)/2pi]  or 3/8.

 

It is ok. I can see that it is logical.

Thanks very much for your very clear explanation.

Melody  Dec 15, 2019
 #8
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Your welcome, uniformly at random means choosing random ... 

goreisthebest  Dec 19, 2019
 #2
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I have no idea what uniformly at random means ?   

Random is self explanatory but uniformly is not.

 Dec 12, 2019
edited by Melody  Dec 12, 2019
edited by Melody  Dec 12, 2019

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