Conjecture: Expected total area of a certain set of random triangles in a unit disk is $1/\pi$.

Question

Choose $3n$ independent uniformly random points in a disk with perimeter $x^2+y^2=1$. Label the points $P_1,P_2,P_3,\dots,P_{3n}$ in order of increasing $x$-coordinates. Form triangles $\triangle P_1P_2P_3,\space \triangle P_4P_5P_6,\space \dots,\space \triangle P_{3n-2}P_{3n-1}P_{3n}$.

Here is an example with $3n=21$.

What does the total area of the triangles approach as $n\to\infty$ ?

I did a simulation of $100$ trials with $3n=99999$, and the average total area was $0.318343178$. This number multiplied by $\pi$ is approximately $1.00010$, so I conjecture that the answer is $1/\pi$.

My attempt

I considered a simpler related question: What is the expected area of a triangle with vertices $(1,a_1),(2,a_2),(3,a_3)$ where the $a$ values are independent uniformly random real numbers in $[0,1]$ ?

The probability that $a_2$ is between $a_1$ and $a_3$ is $1/3$.

We have $\mathbb{E}(\max (a_1,a_2,a_3)-\min (a_1,a_2,a_3))=\frac12$ (explanation). The left and right sub-triangles share a vertical side, which in this case has expected length $\frac18$. So the expected area of the triangle is $2\times\frac12\times\frac18\times 1=\frac18$.

The probability that $a_2$ is not between $a_1$ and $a_3$ is $2/3$.

Again we have $\mathbb{E}(\max (a_1,a_2,a_3)-\min (a_1,a_2,a_3))=\frac12$. The left and right sub-triangles share a vertical side, which in this case has expected length $\frac38$. So the expected area of the triangle is $2\times\frac12\times\frac38\times 1=\frac38$.

So the overall expected area of the triangle is $\frac13\times\frac18+\frac23\times\frac38=\frac{7}{24}$. This value is supported by a simulation.

In the original question, the triangles are not limited to a rectangular box of height $1$, but rather are limited to a unit disk, which has an average height of $2\int_0^1\sqrt{1-x^2}\mathrm dx=\frac{\pi}{2}$. The rectangle and the disk have the same width, but in the disk there are horizontal gaps between neighboring triangles, so we multiply by a factor of $\frac23$. So the total area of the triangles in the disk should approach approximately $\frac{7}{24}\times\frac{\pi}{2}\times\frac23=\frac{7\pi}{72}\approx 0.30543$. This is slightly less than $\frac{1}{\pi}\approx 0.31831$. The difference may be due to the fact that the $x$-coordinates of the points in the disk do not have equal spacing.

EDIT:

I made a mistake in my simulation. Using the shoelace theorem, the area of the triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $\frac12|x_1y_2-x_2y_1+x_2y_3-x_3y_2-x_1y_3+x_3y_1|$, but I was accidentally using $\frac12|x_1y_2-x_2y_1+x_2y_3-x_3y_2-x_1y_3+x_\color{red}{1}y_1|$.

After correcting, my simulations suggest the answer is approximately $0.320$, which agrees with @Rei Henigman's answer. So I no longer think my conjecture is true, but I would still like to know if the limiting area of the triangles has a closed form.

I also did a (correct) simulation using a square lamina instead of a disk, and it seems that the proportion covered by the triangles is about $0.102$, which seems to match the proportion in the disk.

Context

This question was inspired by the question "A disc contains $n$ random points. Each point is connected to its nearest neighbor. What does the average cluster size approach as $n\to\infty$?".

Answer 1

I'll sketch an argument that shows that the constant converges to ~$0.3198$. Hopefully it is not too hand-wavy.

Assume that first of all you randomize all $x$ values. Then you sort the numbers, and only then you randomize the $y$ values. A triple of points have values $x_0$, $x_1$, and $x_2$. Look at the strip $\{(x,y)\ |\ x_0 \le x \le x_2\} \subset D^2$. We can approximate this strip as a rectangle, and re-normalize it such that $x_0 = 0$, $x_2 = 1$, and that the $y$ values are between $0$ and $1$.

Then the $x_1, y_0, y_1, y_2$ are random numbers that are randomized with a uniform distribution between $0$ and $1$. The lengths of the sides of the resulted triangle are:

$d_1 = \sqrt{x_1^2 + (y_1 - y_0)^2}$

$d_2 = \sqrt{(1 + (y_2 - y_0)^2}$

$d_3 = \sqrt{(1-x_1)^2 + (y_2 - y_1)^2}$

and the formula for the size of the triangle is:

$Area(\Delta) = \sqrt{s (s-d_1)(s-d_2)(s-d_3)}$

where $s = \frac{d_1+d_2+d_3}{2}$.

From this we can calculate the expectation of the size of this triangle by:

$\int_0^1 \int_0^1 \int_0^1 \int_0^1 Area(\Delta)dx_1 dy_0 dy_1 dy_2$

Which is a terrible integral, and numerically, it equals to ~$0.1527$.

Now, that means that ~$15.27\%$ of the strip above was covered by the triangle, on average. Note that between each strip and strip, we have an "empty strip", whose width is half of the width of the strips with the triangles, on average. Hence, we expect a total of

$ \frac{2}{3} \cdot 15.27 \% = 10.18 \%$

of the disc to be covered by the triangles. Multiplying by the area of the disc, we get $0.1018\pi = 0.3198$ which is not far from your simulations.

If anyone is brave enough to solve the integral above, we will know what is the actual constant, assuming that my argument is not too hand-wavy. (@Kroki solved the integral, see his answer!)

Answer 2

Let's consider the following problem:

Given a square of size $1$. Choose two random points on the vertical borders and one inside the square. What is the expected area of the triangle covered by the three points?

We can call the value $\tau$. Fix the border points $(0, y_0)$ and $(0, y_1)$ and the abscissa of random inside point as $x$. Let $\mathcal A$ be the area not covered by the triangle:

\begin{align} \mathbb E\left[\mathcal A|y_0, y_1, x\right] &= \frac12\left(y_0 + y_1\right)\left(1-(xy_1 + (1-x)y_0\right) + \int_{xy_1 + (1-x)y_0}^{1} \frac12x\left(1-y_0 + 1-y\right)\mathrm d y + \int_{xy_1 + (1-x)y_0}^{1} \frac12(1-x)\left(1-y_1 + 1-y\right)\mathrm d y + \frac12\left(1-y_0 + 1-y_1\right)\left(xy_1 + (1-x)y_0\right) + \int_{0}^{xy_1+(1-x)y_0} \frac12x\left(y_0+y\right)\mathrm d y + \int_{0}^{xy_1 + (1-x)y_0} \frac12(1-x)\left(y_1 + y\right)\mathrm d y\\ &= \frac12y_0 + \frac12y_1 + xy_1 + (1-x)y_0 - \left(y_0 + y_1\right)\left(xy_1 + (1-x)y_0\right) + \frac12\left(1-xy_0 - (1-x)y_1\right)\left(1-xy_1-(1-x)y_0\right) + \frac14\left(1-xy_1-(1-x)y_0\right)^2 + \frac12\left(xy_0 + (1-x)y_1\right)\left(xy_1 + (1-x)y_0\right) + \frac14\left(xy_1 + (1-x)y_0\right)^2 \end{align}

Now,

\begin{align} 1-\tau &= \mathbb E\left[\mathcal A\right]\\ &= \mathbb E\left[y_0 + y_1\right] - \frac12\mathbb E\left[(y_0 + y_1)^2\right] + \frac12 - \frac12\mathbb E\left[y_0 + y_1\right] + \frac23\mathbb E\left[y_0y_1\right] + \frac16\mathbb E\left[y_1^2 + y_0^2\right] + \frac1{6}\mathbb E\left[y_0y_1\right] + \frac16\mathbb E\left[y_1^2 + y_0^2\right]\\ &= 1 - \frac12\times \frac{7}{6} + \frac23\times \frac14 + \frac16 \times \frac23 + \frac16\times \frac14 + \frac16\times \frac23\\ &= 1-\frac{11}{72} \end{align}

using the argument of @Rei's answer the final answer is $\frac23\tau$.