jmc

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 $\pi /2$ ($90^{\circ }$).

Now, let the three randomly chosen points be $A_0$, $A_1$, and $A_2$. Let $\theta$ be the measure of minor arc $A_0{A}_1$. Since $\theta$ is equally likely to be any value from 0 to $\pi$, the probability that it is less than $\pi /2$ is 1/2.

Now suppose that $\theta <\pi /2$. For the problem's condition to hold, it is necessary and sufficient for point $A_2$ to lie within $\pi /2$ of both $A_0$ and $A_1$ 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 $\frac{\pi -\theta }{2\pi }=\frac{1}{2}-\frac{\theta }{2\pi }$, since $A_2$ 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 $\theta <\pi /2$ is $\frac{1}{2}-\frac{\pi /4}{2\pi }=\frac{3}{8}$.

Since the probability that $\theta <\pi /2$ is 1/2, our final probability is $\frac{1}{2}\cdot \frac{3}{8}=\boxed{\frac{3}{16}}$.