TITLE: area ratios when random line cuts regular polygon
QUESTION [5 upvotes]: A point is randomly selected on one side of a polygon, and another point is randomly selected on one of the other sides. A line is drawn through those points. What is the mean expected ratio of the areas of the larger and smaller pieces created?

REPLY [2 votes]: Circle
Let's consider a simpler problem for the circle first. Let $\phi_1$ and $\phi_2$ be angular coordinates of these two points. Random variables $\phi_1$ and $\phi_2$ are independent identically distributed from uniform distribution on $\left[\, 0, 2\pi\right)$ interval. 
In a wedge with angular aperture $\theta = \left| \phi_2 - \phi_1 \right|$, the ratio of the smaller area to the larger one is 
$$
    r(\theta) = \min \left( \frac{\theta - \sin \theta}{2 \pi - \theta + \sin(\theta)}, \frac{2 \pi - \theta + \sin(\theta)}{\theta - \sin \theta} \right)
$$
It is not hard to verify that $\theta$ has one-sided triangular distribution with density
$$
    f(\theta) = \frac{2 \pi - \theta}{2 \pi^2} \cdot \left[ 0 < \theta < 2 \pi \right]
$$
Since $r(\theta) = r(2\pi - \theta)$, the mean we seek is (no closed form expression for the integral is known to the poster):
$$
   \mu_\infty = \int_0^\pi \frac{\theta - \sin \theta}{2 \pi - \theta + \sin(\theta)} \frac{\mathrm{d} \theta}{\pi} \approx 0.2216302664785579
$$
The simulation (in Mathematica 8) confirms this value:
In[162]:= 
RandomVariate[
  TransformedDistribution[Min[z/(2 Pi - z), (2 Pi - z)/z], 
   z \[Distributed] 
    TransformedDistribution[
     Abs[f1 - f2] - Sin[Abs[f1 - f2]], {f1, f2} \[Distributed] 
      UniformDistribution[{{0, 2 Pi}, {0, 2 Pi}}]]], 10^7] // Mean

Out[162]= 0.221634


Polygon
Let's now turn to the case of a regular $n$-gon circumscribed into a unit circle seated at the origin. Let's parameterize a point on its circumference with a pair $(k, u)$, where $k$ is uniform discrete variable ranging from $0$ to $n-1$, and $u$ is a uniform continuous variable on a unit circle.
A point $(k,u)$ on the circumference has the following Cartesian coordinates:
$$ \begin{eqnarray}
    x(k,u) &=& (1-u) \cdot \cos \left( 2 \pi \frac{k}{n} \right) + u \cdot \cos \left( 2 \pi \frac{k+1}{n} \right) \\
y(k,u) &=& (1-u) \cdot \sin \left( 2 \pi \frac{k}{n} \right) + u \cdot \sin \left( 2 \pi \frac{k+1}{n} \right)
  \end{eqnarray}
$$
The following Mathematica code straightforwardly simulates the ratio, getting (for the sample size of $10^7$, giving accuracy of about $2 \cdot 10^{-4}$):
$$
\begin{array}{c|c||c|c} 
     n & \mu_n & n & \mu_n \\  
     3 & 0.202013 & 4 & 0.218133\\ 
     5 & 0.219839 & {\color \red{6}} & {\color \red{0.221169}} 
     \\ 7 & 0.221218 & 8 & 0.221444
  \end{array}
$$
For $n=3$ the result can be worked out explicitly. Conditioning on the line resting at different sides, and using exchangeability, gives the factor of $\frac{6}{9} = \frac{2}{3}$:
$$
   \mu_3 = \frac{2}{3} \mathbb{E}\left( \min \left( \frac{U_1 U_2}{1 - U_1 U_2}, \frac{1- U_1 U_2}{U_1 U_2} \right) \right) = \frac{1}{18} \left( \pi^2 + 12 (\log^2(2) - 1) \right)
 \approx 0.201947
$$
here $U_1$ and $U_2$ are independent standard uniform random variables.
Similarly, one can work out the $\mu_4$ explicitly. With probability $\frac{1}{4}$ the line end-points belong to the same side, with probability $\frac{1}{2}$ to adjacent sides, and with probability $\frac{1}{4}$ to the opposite sides:
$$ \begin{eqnarray}
   \mu_4 &=&\mathbb{E} \left( \frac{1}{2} \cdot \frac{U_1 \cdot U_2}{2-U_1 \cdot U_2} + \frac{1}{4} \cdot \min \left(\frac{U_1+U_2}{2-U_1-U_2}, \frac{2-U_1-U_2}{U_1+U_2} \right)  \right) \\
  &=& \frac{\pi^2+3}{12} - \frac{\left( 2- \log(2)\right)^2}{2} \approx 0.218535
 \end{eqnarray}
$$
So it seems that the sequence $\mu_n$ for $n \geqslant 3$ is increasing, asymptotically approaching $\mu_\infty$.