Three randomly placed points inside a unit circle. Probability that the formed triangle contains the centre of the circle inside?

Question

Three non-collinear points are placed randomly inside a unit circle.

Question: What is the probability that if you were to connect these points, forming a triangle, the triangle will have the center of the circle contained inside (including the periferi)?

Here are two examples, A and B, illustrating what I mean:

where you can see that A works and B doesn't work.

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What Iv'e tried so far is illustarated in the following image:

Two randomly points are obviously always collinear. So we can connect the two. Now let's draw a staright line, starting from each point, through the center and finish on the circumference of the circle. The grey shaded area represents the region where the third point needs to be in order to create a triangle that works. Now, as you can see there can be all kinds of regions depending on where the first two points are located. So I'm simply looking for a method/approach that can answer this question. If you have a another approach than my example feel free to use it. I'm not sure if this is very complicated or if there is a nice solution.

Answer 1

Let's say the points are $p_1, p_2$ and $p_3$. To simplify discussion, let's also choose coordinates so that the center of the circle is at the origin and $p_1$ is on the positive $x$ axis, somewhere between $(0,0)$ and $(0,1)$. With these coordinates, triangle $p_1 p_2 p_3$ encloses the origin if and only if (1) $p_2$ and $p_3$ lie on opposite sides of the $x$ axis, and (2) the line $p_2 p_3$ intersects the negative portion of the $x$ axis.

By symmetry, $p_2$ lies in the upper half plane with probability $1/2$ and in the lower half plane with probability $1/2$. The same is true of $p_3$, so the probability that $p_2$ and $p_3$ lie on opposite sides of the $x$ axis is $1/2$.

Again by symmetry, line $p_2 p_3$ intersects the positive $x$ axis with probability $1/2$ and the negative $x$ axis with probability $1/2$.

Conditions (1) and (2) are independent in probability, so the probability that both hold is $$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

Answer 2

I presume that your points are distributed uniformly in the interior of the circle.

1. Observe than only the angular coordinates matter, but radial don't. Your points could just as well have been uniformly distributed over the circumference.

2. Now we define a function ($\chi$) which equals 1 when the triangle is good and 0 when it is bad, and integrate it over all possible positions of the three points: $0\leqslant\varphi_1,\varphi_2,\varphi_3<2\pi$.

3. When we fix $\varphi_1$ and $\varphi_2$, the integral over $\varphi_3$ is the arc length of your grey sector. It only depends on $\varphi_2-\varphi_1$: when this difference is less than $\pi$, this is just it, and when greater, this is its complement to $2\pi$.

4. Observe that $\varphi_1$ does not matter at all because of the circular symmetry of the problem. The first point could just as well always sit at $\varphi_1=0$.

5. Now integrate over the remaining variables: $$\iint\chi(\varphi_2,\varphi_3)\rho(\varphi_2)\rho(\varphi_3)d\varphi_2d\varphi_3 = \\ = \left({1\over2\pi}\right)^2\left(\int_0^{\pi}\varphi_2d\varphi_2+\int_\pi^{2\pi}(2\pi-\varphi_2)d\varphi_2\right) = \\ = {1\over4\pi^2}\left(\frac12\pi^2+\frac12\pi^2\right)=\frac14$$ So it goes.