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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:

enter image description here

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

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    $\begingroup$ WA factors $1 + \cos \alpha + \cos \beta+ \cos (\alpha - \beta)$ as $4 \cos(\alpha/2) \cos(\beta/2) \cos(\alpha/2 - \beta/2)$ -- this may help. $\endgroup$ Commented Sep 8, 2020 at 10:12
  • $\begingroup$ @AlexeyBurdin It clearly helps. Thanks. $\endgroup$ Commented Sep 8, 2020 at 10:16
  • $\begingroup$ I have the solution. But creating a proper figure (e.g by tikz) would be time consuming. I have a figure for the signs of the factors in a hand drawn figure. pyedu.hu/tmp/3_unit_vector_sum.jpg I guess it is not the best way to include it in an answer like that. $\endgroup$ Commented Sep 8, 2020 at 11:46
  • $\begingroup$ You picture looks nice to me even to include in the answer. Maybe try to use geogebra instead, if you aren't so confident that hand-drawn image is fine. Thanks. $\endgroup$ Commented Sep 8, 2020 at 11:49
  • $\begingroup$ Ok. I'll include it. I can replace it any time when I have a better one. $\endgroup$ Commented Sep 8, 2020 at 11:51

3 Answers 3

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

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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 $$\cos \left(\frac \alpha 2\right)\cos \left(\frac \beta 2\right) \cos \left(\frac \alpha 2 - \frac \beta 2\right) < 0$$

enter image description here

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 $\frac 1 4$.

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Consider the case in which the sum vector $$ {\bf w} = {\bf v}_{\,0} + {\bf v}_{\,1} + {\bf v}_{\,2} $$ is also unitary. You get the situation depicted in the following sketch.

3_unit_v_sum_1

$OAPB$ is an equilateral parallelogram: clearly $\left| {\bf w} \right|$ will be less than one when ${\bf v}_{\,2} $ lies inside the angle $\alpha$ centered in $P$, and greater when outside of that.
That is we shall have $$ \left\{ \matrix{ 0 \le \alpha \le \pi \hfill \cr \pi \le \beta \le \pi + \alpha \hfill \cr} \right. $$ and of course for negative $\alpha$ the situation is symmetric

Then $$ P\left( {\left| {\bf w} \right| < 1\;\;\left| {\,\alpha } \right.} \right) = {\alpha \over {2\pi }}\quad \Rightarrow \quad P\left( {\left| {\bf w} \right| < 1} \right) = 2\int_{\alpha = 0}^\pi {{\alpha \over {2\pi }}} {{d\alpha } \over {2\pi }} = {1 \over 4} $$

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