95. Three points $A, B$ and $C$ are taken at random on a circle. What is the probability that triangle $A B C$ will be acute?
95. Since all points on the circle are equal to each other, we can consider that the first of the three points (point $A$) is somehow fixed. The positions of points $B$ and $C$ will then be determined by the lengths of the arcs $A B$ and $A C$, measured in a certain direction (for example, counterclockwise). Now, let's cut the circle at point $A$ and unfold it into a segment $A A^{\prime}$ (Fig. 83; the fact that both ends of this segment represent the same point on the circle is not important for us). The presence of an obtuse angle in triangle $A B C$ clearly means that one of the three arcs into which its vertices divide the segment exceeds a semicircle. Therefore, after transitioning from the circle to the segment $A A^{\prime}$, our problem takes the following form: two points $B$ and $C$ are taken randomly on the segment $A A^{\prime}$; what is the probability that none of the three parts into which these points divide the circle exceeds half the length of the segment? But calculating this last probability is the content of problem 93 (see the solution to the specified problem); therefore, the answer to this problem coincides with the answer to problem 93: the desired probability is $\frac{1}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-325.jpg?height=420&width=740&top_left_y=458&top_left_x=298)

Fig. 83.