# The expected area of a triangle made up three points picked randomly inside a circle of radius, and contains the center

Consider the triangle formed by randomly distributing three points Inside a circle. What is the expected area of the triangle that contains the center of the circle. And what is the expected area of the intersection of a set of dependent trinagles that contain the center?

• What are your thoughts on the problem? – Stefan Mar 5 '13 at 10:58
• Randomly ditributed how? Is the probability that a given point lies within a given shape proportional to the area of said shape? – MvG Mar 5 '13 at 15:11
• I mean the points are uniformly distributed inside a circle. – Noureddine Mar 5 '13 at 15:21
• Interesting problem. The probability for the triangle contains the center is $\frac14$ and the expected area of the triangle conditional to it contains the center is $\frac{4}{3\pi}$. Though I wonder whether this is a homework problem. – achille hui Mar 5 '13 at 15:57
• This is not a homework problem, it make part of more complex problem that I try to resolve. Could you please give me more explanation about how you drive the probability $1/4$, I think that this probability is for the case where the three points of the triangle are on the circle, but in my problem the three points area INSIDE the circle. – Noureddine Mar 5 '13 at 16:28

This base part of the problem is not that hard because for suitable chosen coordinates, the condition that the center falls inside the triangle is relatively simple.

WOLOG, assume the circle is the unit circle.

Let $(r_i, \alpha_i), i = 1,2,3$ be the positions of the 3 random points in polar coordinates. Because of the symmetry of the problem. It just suffices to look the case where $\alpha_1 = 0$ and $0 \le \alpha_2 \le \pi$.

Let $\theta = \alpha_2$ and $\phi = 2\pi - \alpha_3$. It is easy to check for the triangle to contain the origin, the condition is given by $0 \le \phi \le \pi$ and $\theta + \phi \ge \pi$. (see image at end for an illustration)

From this, we see the probability for the triangle to contain the origin is given by:

$$\int_0^\pi \frac{d\theta}{\pi}\int_{\pi-\theta}^\pi \frac{d\phi}{2\pi} \iiint_{0\le r_i\le1} dr_1^2 dr_2^2 dr_3^2 = \frac{1}{2\pi^2}\int_0^\pi\theta d\theta = \frac14$$

When the triangle contains the origin, its area is given by:

$$\frac12 r_1 r_2 \sin\theta + \frac12 r_1 r_3 \sin\phi + \frac12 r_2 r_3 \sin(2\pi-(\theta+\phi))\tag{*}$$

Using symmetry of the problem again and notice $\int_0^1 r_i dr_i^2 = \frac23$, the contribution to expected area when the triangle contains the origin is given by: $$\int_0^\pi \frac{d\theta}{\pi}\int_{\pi-\theta}^\pi \frac{d\phi}{2\pi} \left[\frac32 \left(\frac23\right)^2 \sin\theta \right] = \frac{1}{3\pi^2}\int_0^\pi \theta\sin\theta d\theta = \frac{1}{3\pi}\tag{**}$$

As a result, the conditional expected area of the triangle when it contains the origin is $\frac{4}{3\pi}$.

UPDATE

There are higher dimension generalization on the probability of picking a triangle containing the origin. Quoting from an article by R.Howard and Paul Sisson, we have:

Let $\mathbb{R}^n$ be endowed with a probability measure $\mu$ which is symmetric with respect to the origin and such that when $n+1$ points are chosen independently with respect to $\mu$, with probability one their convex hull is a simplex. Then the probability that the origin is contained in the simplex generated by $n+1$ such random points is $\frac{1}{2^n}$.

UPDATE2 What does the contribution to expected area means.

In general, when you uniformly pick 3 points from a circle, it need not enclose the center. Let $T$ be a variable running through all possible configuration of the 3 points. Let $\mathscr{A}(T)$ be the area of corresponding triangle and $d\mu(T)$ be the probability density of occurrence of $T$. Let $\mathscr{C}$ be the set of configuration where the triangle contains the origin.

The expected area of the unconstrained triangle is given by

$$\int \mathscr{A}(T) d\mu(T)$$

The expected area of the triangle conditional to it contains the center is given by:

$$\frac{\int_{\mathscr{C}} \mathscr{A}(T) d\mu(T)}{\int_{\mathscr{C}} d\mu(T)}$$

The numerator here is what I mean "contribution to expected area" and the denominator $\int_{\mathscr{C}} d\mu(T) = \frac14$ is the probability for the triangle to contain the origin.

For $T \in \mathscr{C}$, one can setup coordinates so that $\mathscr{A}(T)$ has the simple form in $(*)$. By symmetry, the contribution for those 3 pieces in $(*)$ equal to each other. This explains the factor $\frac32$ appear in the integrand of L.H.S of $(**)$. The remaining part of the integrand in $(**)$ comes from the partial integral over $r_i$: $$\iiint_{0\le r_i \le 1} dr_1^2 dr_2^2 dr_3^2\;r_1 r_2 \sin \theta = \left(\frac23\right)\left(\frac23\right)\left(1\right) \sin\theta = \left(\frac23\right)^2 \sin\theta$$

• Then The probability for the triangle to contain the origin in the general case (when the radius = R) is $(R^3 R^3 R^3)/3 3 3/4$ ?? – Noureddine Mar 5 '13 at 18:33
• No idea what you are talking about. If you uniformly pick the vertices from a circle of radius $R$. The "radial part" of the probability density picks up an extra factor $\left(\frac{1}{R^2}\right)^3$ but the probability for the triangle to contain the origin remains the same. – achille hui Mar 5 '13 at 18:49
• what about the conditional expected area of the triangle when it contains the origin and the radius is R? – Noureddine Mar 5 '13 at 20:21
• The conditional expected area will be scaled by a factor of $R^2$. – achille hui Mar 5 '13 at 21:01
• @MvG, I've double checked my numbers by a $10^7$ simulations before I compose this answer. My simulations give me an estimate 0.249799 for the probability and 0.424439 for the expected area. – achille hui Mar 5 '13 at 21:13

Achill Hui, A nice derivation you have given. How to modify the derivation to get the expected area of the triangle that is formed by three randomly chosen points inside the unit circle? No condition is imposed on this triangle