Probability distribution for the perimeter and area of triangle with fixed circumscribed radius

Given a circle with radius R = 1, I'm trying to find either the probability distribution function or the density function for the space of triangle, which is randomly selected on this circle. The same task is for perimeter function of this triangle.

The only thing I've understood is the following. If we fix some point R on the circle, then angles ROA, ROB, ROC (counterclockwise) are uniformly distributed on [0; 2 * Pi]. I've tried expressing the space and the perimeter through those angles, but still had no success.

I would appreciate any help, really. I've tried to solve this problem for three weeks, and it seems to me that soon those triangles and circles will begin to come into my night dreams. Thanks.

Solution 1:

Fix any point $R$. Let $\phi_1$, $\phi_2$ and $\phi_3$ be angles, relative to $R$, then $0 < \phi_1 < \phi_2 < \phi_3 < 2 \pi$, i.e the triple $\{\phi_1, \phi_2, \phi_3\}$ is the first, second and the third order statistics of the uniformly distributed angle on the interval $(0, 2\pi)$.

The area of the triangle, is then $$ A = 4 \sin \left( \frac{\phi_2-\phi_1}{2} \right) \sin \left( \frac{\phi_3-\phi_2}{2} \right) \sin \left( \frac{\phi_3-\phi_1}{2} \right) $$ and the perimeter $$ p = 2 \left( \sin \left( \frac{\phi_2-\phi_1}{2} \right) + \sin \left( \frac{\phi_3-\phi_2}{2} \right) + \sin \left( \frac{\phi_3-\phi_1}{2} \right) \right) $$ Given that $\frac{\phi_3 -\phi_1}{2} = \frac{\phi_3 -\phi_2}{2} + \frac{\phi_2 -\phi_1}{2}$, let $\alpha = \frac{\phi_2-\phi_1}{2}$ and $\beta = \frac{\phi_3-\phi_2}{2}$, then $$ A(\alpha, \beta) = 4 \sin \alpha \sin \beta \sin (\alpha+\beta) \qquad p(\alpha, \beta) = 2 \left( \sin \alpha + \sin \beta + \sin (\alpha+\beta) \right) $$ The distribution of $\alpha$ and $\beta$ are not hard to find. Indeed, distribution of $\phi_1$, $\phi_2$ and $\phi_3$ ($[\chi]$ stands for Iverson bracket): $$ \mathrm{d} F(\phi_1, \phi_2, \phi_3) = \frac{3!}{(2 \pi)^3} \Big[ 0 < \phi_1 < \phi_2 < \phi_3 < 2 \pi \Big] \mathrm{d} \phi_1 \mathrm{d} \phi_2 \mathrm{d} \phi_3 $$ Changing variables to $\alpha = \frac{\phi_2-\phi_1}{2}$, $\beta=\frac{\phi_3-\phi_2}{2}$ and $\gamma = \frac{1}{3} \left( \phi_1 + \phi_2 + \phi_3 \right)$ we get, noting that the Jacobian equals $4$, $$ \mathrm{d} F(\alpha, \beta, \gamma) = \frac{3!}{(2 \pi)^3} \Big[ 0 < \alpha < \pi, 0 < \beta < \pi-\alpha, \frac{4 \alpha + 2 \beta}{3} <\gamma< 2\pi - \frac{2 \alpha + 4 \beta}{3} \Big] \cdot 4 \cdot \mathrm{d} \alpha \, \mathrm{d} \beta \, \mathrm{d} \gamma $$ Integrating over $\gamma$ we find the joint pdf for $(\alpha, \beta)$: $$ \mathrm{d} F\left(\alpha,\beta\right) = \frac{3!}{\pi^3} \left( \pi - \alpha - \beta \right) \Big[ 0 < \alpha < \pi, 0< \beta < \pi - \alpha \Big] \, \, \mathrm{d} \alpha \, \mathrm{d} \beta $$ This means that $\left\{ \frac{\alpha}{\pi}, \frac{\beta}{\pi} \right\}$ follows Dirichlet distribution with parameters $\{1,1,2\}$.

Using the interpretation in terms of the Dirichlet distribution, it is not hard to determine the mean and the variance of the area and the perimeter:

Solution 2:

I get the area to be $$2\sin\left(\frac{X_1}{2}\right)\sin\left(\frac{X_2}{2}\right) \sin\left(\frac{|X_1 - X_2|}{2}\right)$$ where $X_1$ and $X_2$ are independent random variables uniformly distributed on $[0, 2\pi)$ but I don't foresee an easy calculation to get the distribution of the area either. As joriki says, the problem does not seem to have a "nice" answer.

Let A be a fixed point on the circle of radius $1$ and let point B be at distance $X$ measured clockwise along the circle boundary from A. Then, the chord AB has length $L = 2\sin(X/2)$ and if $X$ is uniformly distributed on $[0,2\pi)$, then for $z \in [0, 2)$. the CDF and pdf of $L$ are $$\begin{align*} F_L(z) &= \frac{2}{\pi}\arcsin\left(\frac{z}{2}\right),\\ f_L(z) &=\frac{1}{\pi\sqrt{1-(z/2)^2}}. \end{align*}$$ See for example, here for a derivation. The angle between the tangent at A and the chord AB is $X/2$. If $C$ is also chosen similarly, then the triangle has sides AB and AC of lengths $L_1$ and $L_2$ having angle $(\max\{X_1,X_2\} - \min\{X_1.X_2\})/2$ at A. So the area of triangle ABC is $$\begin{align*}\text{area ABC} &=\frac{1}{2}L_1L_2\sin((\max\{X_1,X_2\} - \min\{X_1.X_2\})/2)\\ &= 2\sin\left(\frac{X_1}{2}\right)\sin\left(\frac{X_2}{2}\right) \sin\left(\frac{\max\{X_1,X_2\} - \min\{X_1.X_2\}}{2}\right)\\ &= 2\sin\left(\frac{X_1}{2}\right)\sin\left(\frac{X_2}{2}\right) \sin\left(\frac{|X_1 - X_2|}{2}\right) \end{align*}$$