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three points are randomly chosen on a circle. what the probability that

1.triangle formed is right angled triangle.

2.triangle formed is acute angled triangle.

3.triangle formed is obtuse angled triangle.

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What are your thoughts? What have you tried so far? – Rahul Mar 3 '12 at 6:18
My thinking is that the probability of acute and obtuse is the same = 89/180 and that of right angle triangle is 1/90.. I correlated the problem to the hands of a clock and the probability of clock being 90 degrees apart in 60 minutes of duration - and that is 4 times in 60 minutes.. The acute and obtuse have equal probability of 89/180 – Sharat Chandra Mar 3 '12 at 6:42
The right-angled question has an immediate answer: the probability is $0$. – André Nicolas Mar 3 '12 at 6:43
@Sharat Chandra: Your number suggests that you have a discrete model in mind. Perhaps you have mentally marked $360$ points, equally spaced around the circle, and are choosing from these, and only from these. Is that what you have in mind? That is perhaps not the most reasonable geometric model, since $1$ degree has no special geometric significance. Division into $360$ equal parts is "local." That's probably not what a native of Mars would choose. – André Nicolas Mar 3 '12 at 6:51
I'm sure you agree that if the angle of the triangle is $89^\circ$ or $91^\circ$ then it is not a right angle. But you also have to remember that it is also not a right angle if it is $90.1^\circ$, or $90.01^\circ$, or $90.000001^\circ$, and so on. Only when it is $90.000000000\!\ldots^\circ$ is it a right angle. So maybe now you can appreciate why this can only occur with vanishingly small probability. – Rahul Mar 3 '12 at 7:22

The question is about points on a circle, so that's the one I'm answering here. The triangle is right angled precisely if two of the three points lie on a diameter of the circle. This is Thales' theorem. As already noted, the probability that this occurs is $0$. The triangle is obtuse precisely if all three points lie on the same side of some diagonal of the circle. The probability for that is $\frac{3}{4}$ (assuming a uniform distribution on the circle).

One way of seeing this is to introduce for each vertex $v_k$ a random variable $X_k$ as follows:

$$ X_k = \begin{cases} 1 & \textrm{if the other vertices lie on the half circle starting at } v_k \textrm{ in clockwise direction}\\ 0 & \textrm{otherwise} \end{cases} $$

Then at most one of $X_1, X_2, X_3$ can be equal to one. Therefore

$$ \mathbb{P}(\textrm{triangle is obtuse}) = \mathbb{P}(X_1+X_2+X_3 = 1) = \mathbb{E}(X_1+X_2+X_3) = 3 \mathbb{E}(X_1). $$

The last equality follows from the fact that $X_1, X_2, X_3$ all have the same probability distribution. Now $\mathbb{E}(X_1) = \frac{1}{4}$ since both $v_2$ and $v_3$ have a probability of $\frac{1}{2}$ to lie on the half circle starting at $v_1$.

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