The following is a long description of a computation I'd like to make. You can think of the process described as a spider randomly building a web. I'd like to know how big we can expect the web to be.

Inscribe inside a circle of radius $1$ a regular polygon with $n$-vertices, labeled consecutively. Draw a radius from the center to each vertex. For $i=1,\ldots,n$, place a bead $b_i$ on the radius connected to vertex $i$. We can slide each bead to different positions on its radius, so the state of the beads is described with a vector $(x_1,\ldots,x_n)\in [0,1]^n$, where $x_i$ is the distance of $b_i$ from the center. Let $P(x_1,\ldots,x_n)$ be the polygon formed by connecting each $b_i$ to its immediate neighbors with a straight line segment. Each pair of neighbors along with the center form a triangle, and summing the area of all such triangles gives the area of $P(x_1,\ldots,x_n)$:


The total area of the regular $n$-gon is


The ratio of the two areas defines a function:

If we choose the $x_i$ uniformly at random, what is the probability that $f(x_1,\ldots,x_n)>c$ for a given $c\in[0,1]$?

Since $f$ is continuous, the set $E_c=\{(x_1,\ldots,x_n)\mid f(x_1,\ldots,x_n)>c\}\subset [0,1]^n$ is measurable, so I'm just looking for $m(E_c)$, but I'm not too sure how to find it.


I just look at the case $n=3$; and just $c\ge1/3$. The probability is then $$\int_{(3c-1)/2}^1dx\int_{(3c-x)/(1+x)}^1dy\int_{(3c-xy)/(x+y)}^11dz\\ =\int_{(3c-1)/2}^1dx\int_{(3c-x)/(1+x)}^1dy\left[1-\frac{3c-xy}{x+y}\right]\\ =\int_{(3c-1)/2}^1dx\left[y-3c\ln(x+y)+xy-x^2\ln(x+y)\right]_{(3c-x)/(1+x)}^1\\ =\int_{(3c-1)/2}^1dx\left[1+x-(3c+x^2)\ln(1+x)-(3c-x)+(3c+x^2)\ln\frac{3c+x^2}{1+x}\right]\\ =\frac94(c-1)^2+\left[4c\sqrt{3c}\tan^{-1}\left\{x/\sqrt{3c}\right\}-\frac29x(18c+x^2)+\frac13(9cx+x^3)\ln(3c+x^2)\right.\\ \left.-\frac19(6(9cx+9c+x^3+1)\ln(x+1)-x(54c+2x^2-3x+6))\right]_{(3c-1)/2}^1\\ $$ That is from Wolfram Alpha, which couldn't do the whole integral at once.


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