# area ratios when random line cuts regular polygon

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?

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Are the original two sides marked in advance, or are you just cutting with the line through two uniform points on the boundary? – Louis Jan 3 '12 at 13:35
The second point is on a side randomly chosen from the remaining sides. – Angela Richardson Jan 3 '12 at 15:14

## 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 $\mu_\infty$.

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