+97 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?
+118673 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.
+97 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
+118673 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.
+58 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 \(AOB\) 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 \(\dfrac{\pi}{2}\)\((90^{\circ})\).
Now, let the three randomly chosen points be \(A_0\),\(A_1\), and \(A_2\). Let \(\theta\) be the measure of minor arc \(A_0A_1\). Since \(\theta\) is equally likely to be any value from \(0\) to \(\pi\), the probability that it is less than \(\dfrac{\pi}{2}\) is \(\dfrac{1}{2}\).
Now suppose that \(\theta<\dfrac{\pi}{2}\). For the problem's condition to hold, it is necessary and sufficient for point \(A_2\) to lie within \(\dfrac{\pi}{2}\) of both \(A_0\) and \(A_0\) along the circumference. As the diagram below shows, this is the same as saying that \(A_2\) must lie along a particular arc of measure \(\pi-\theta\).
The probability of this occurrence is \(\dfrac{\pi-\theta}{2\pi}=\dfrac{1}{2}-\dfrac{\theta}{2\pi}\), since \(A_2\) is equally likely to go anywhere on the circle. Since the average value of \(\theta\) between \(0\) and \(\dfrac{\pi}{2}\) is \(\dfrac{\pi}{4}\), it follows that the overall probability for \(\theta<\dfrac{\pi}{2}\) is \(\dfrac{1}{2}-\dfrac{\pi/4}{2\pi}=\dfrac{3}{8}\).
Since the probability that \(\theta<\dfrac{\pi}{2}\) is 1/2, our final probability is \(\dfrac{1}{2}\cdot\dfrac{3}{8}=\dfrac{3}{16}\).
