Expected area of triangle made by vertices of N-gon polygon inscribed in a unit circle

Question

Consider N vertices of a regular N-sided polygon that is inscribed in a unit circle. How can I calculate the expected area of triangle formed by the vertices of N sided polygon ?

Answer 1

I'll assume that the question was intended as clarified in Robo300's comment. I'll write $n$ instead of your $N$ to conform with the convention of writing random variables in uppercase and constants in lowercase.

The triangle is composed of three triangles formed by the centre of the circle and the three pairs of points. By symmetry and linearity of expectation, its expected area is $3$ times the expected area of one of these triangles.

Without loss of generality, fix one point at angle $0$. The next point counterclockwise is at $2\pi K/n$ with $1\le K\le n-2$. The probability distribution of $K$ is linear, as it's proportional to the number of options left for the third point:

$$ P(K=k)=\frac{2(n-1-k)}{(n-1)(n-2)}\;. $$

Then the expected area of the triangle is

\begin{eqnarray*} E(A)&=&3\sum_{k=1}^{n-2}P(K=k)\frac12\sin\frac{2\pi k}n\\ &=&3\sum_{k=1}^{n-2}\frac{2(n-1-k)}{(n-1)(n-2)}\frac12\sin\frac{2\pi k}n\\ &=&\frac3{(n-1)(n-2)}\sum_{k=1}^{n-2}(n-1-k)\sin\frac{2\pi k}n\\ &=&\frac3{(n-1)(n-2)}\frac12n\cot\frac\pi n\\ &=&\frac32\frac n{(n-1)(n-2)}\cot\frac\pi n\;. \end{eqnarray*}

For instance, for $n=3$ this is $\frac34\sqrt3$, which is indeed the area of a regular triangle inscribed in the unit circle, and for $n=4$ the result is $1$, which is indeed the area of each triangle in the square inscribed in the unit circle.

Note that the limit for $n\to\infty$ is $\frac3{2\pi}$, the expected area of a triangle formed by points uniformly distributed on the unit circle, as it should be.

(Note also that I've tacitly used negative areas for $k\gt\frac n2$; this works out since such negative areas cancel the excess positive areas in the other two triangles.)