Sum of 3 unit vectors being shorter than 1 What is the probability for the sum of three unit vectors to be shorter than 1?
The vectors' direction angles has uniform distribution on $[0, 2\pi]$.
I've made simulations and I also used Wolfram Alpha to solve the final equation below, so I'm pretty sure, that the result is
$\frac 1 4$. How can I prove that?
If I rotate the vectors with the same angle so that the first one's direction angle is 0, than the length doesn't change. So the result is the same as this value:
$$Prob(\vert e^0 + e^{i\alpha} + e^ {i\beta} \vert < 1)$$
($\alpha$ and $\beta$ is uniformly distributed on $[0, 2\pi]$)
I'll denote the sum by z, so
$$z = e^0 + e^{i\alpha} + e^ {i\beta} = 1 + e^{i\alpha} + e^ {i\beta}$$
$$\vert z\vert^2 = z \overline z = (1 + e^{i\alpha} + e^ {i\beta})(1 + e^{-i\alpha} + e^ {-i\beta})$$
$$\vert z\vert^2 = 3 + \left( e^{i\alpha} + e^{-i\alpha} \right)+ \left(e^ {i\beta}+ e^ {-i\beta}\right) + \left(e^ {-i(\alpha - \beta)} + e^ {-i(\alpha - \beta)}\right)= 3 + 2 \cos \alpha + 2 \cos \beta+ 2 \cos (\alpha - \beta)$$
This is non-negative so $\vert z \vert < 1 \Leftrightarrow \vert z \vert^2 < 1 $, that gives us
$$3 + 2 \cos \alpha + 2 \cos \beta+ 2 \cos (\alpha - \beta) < 1$$
$$2 + 2 \cos \alpha + 2 \cos \beta+ 2 \cos (\alpha - \beta) < 0$$
$$1 + \cos \alpha + \cos \beta+ \cos (\alpha - \beta) < 0$$
I need to solve this equation. I have the solution from Wolfram Alpha:

and this gives me the $\frac 14$ probability. But how can I get this solution?

The first two points $z_+$, $z_-$ have an angle $\alpha'$ among themselves, where $\alpha'$ is uniformly distributed in $[0,\pi]$. We may assume them as $\cos\alpha\pm i \sin\alpha$, where $\alpha$ is uniformly distributed in $\bigl[0,{\pi\over2}\bigr]$. This gives $z_++z_-=2\cos\alpha=:z_*$. The unit circle with center $z_*$ has an arc of length $2\alpha$ within the unit circle of the $z$-plane. We have a success iff the third random point is lying on this arc. The total probability that this happens is
$$p={2\over\pi}\int_0^{\pi/2}{2\alpha\over 2\pi}\>d\alpha={1\over4}\ .$$

