Probability of triangle to be acute? Suppose that someone randomly picks $3$ points $A, B$ and $C$ on a fixed circle. What is the probability of triangle $ABC$ to be acute?
 A: Let one point be fixed and let circle have radius $1$. Now find posibility for other two points. First integrate all possibilities:
$$I_1=\int_0^{2\pi}\int_0^{2\pi}dbda=4\pi^2$$
Then integrate required case:
$$I_2=\int_0^{\pi}\int_a^{2\pi}dbda+\int_{\pi}^{2\pi}\int_0^{a}dbda=3\pi^2$$
So, your probability is $\frac {I_2} {I_1}=\frac34$.
Explanation: let second point be $A$ and third be $B$. When we choose $A$, we cannot choose point $B$ between first point and $A$. Because of that reason, we integrate from $0$ to $a$.
A: Put the first point at the top of the circle.  Randomly select the second point uniformly from the circle; without loss of generality, assume the point falls on the right half.  (If it's on the left, the arguments that follow can just be mirror-reversed.)
Let the angle between the two points (as measured from the circle's center) be $x$ radians; clearly, $0 \leq x \leq \pi$.  We claim (and will shortly demonstrate) that the triangle will be acute if and only if the third point is chosen such that no semicircle contains all three points.  This happens if the third point is chosen on the left side, between $\pi-x$ and $x$ radians from the first point at the top of the circle.  The probability of this, given $x$, is $x/(2\pi)$.  Since $x$ ranges freely and uniformly from $0$ to $\pi$, $x/(2\pi)$ ranges freely and uniformly from $0$ to $1/2$, and the desired probability is the average of $x/(2\pi)$, or $1/4$.
Argument that a triangle inscribed in a circle is acute if and only if the three points cannot be contained by a semicircle: Let $A, B, C$ be three points on a circle.  Suppose that all three points are contained in a semicircle, in the order $A, B, C$ (without loss of generality).  Then $B$ is opposite a section of the circle subtending at least $\pi$ radians.  Since m$\angle B$ is half of the section of circle it subtends, $m\angle B \geq \pi/2$ and the triangle is obtuse (or right).
Suppose, on the other hand, that no semicircle contains the three points.  Then no point is opposite a section of circle $\geq \pi$ radians, and therefore each of the three angles is less than half that, or $\pi/2$ radians; the triangle is therefore acute.
