Sum of 3 unit vectors being shorter than 1

As Alexey showed me, I can convert that sum to a multiplication using Computer Algebra Systems, like Wolfram Alpha. One can prove that

$$1 + \cos \alpha + \cos \beta+ \cos (\alpha - \beta) = 4 \cos \left(\frac \alpha 2\right)\cos \left(\frac \beta 2\right) \cos \left(\frac \alpha 2 - \frac \beta 2\right)$$

So we need to solve $$\cos \left(\frac \alpha 2\right)\cos \left(\frac \beta 2\right) \cos \left(\frac \alpha 2 - \frac \beta 2\right) < 0$$

The signs of the factors can be seen in the picture above. In the picture

• the blue crosshatched part means that $\cos \left(\frac \alpha 2\right)\cos \left(\frac \beta 2\right)$ is positive, in the other parts of the figure (but the border) it is negative
• the red crosshatched part means that $\cos \left(\frac \alpha 2 - \frac \beta 2\right)$ is positive

We want to find out, what is the probability that randomly chosen point (uniform distibution for both $\beta/2$ and $\alpha / 2$) in the figure has negative value for that multiplication. That means that there is blue crosshatch or red crosshatch but not both. We have 8 such triangles and the whole big square's area is 4 times as big as the sum of the triangles' area.

So the probability in the question is $\frac 1 4$.

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 $$p={2\over\pi}\int_0^{\pi/2}{2\alpha\over 2\pi}\>d\alpha={1\over4}\ .$$