## On The Expected Area of Various Triangles in Squares (Nov. 2020)

### 1. Problem Introduction

I was recently thinking about the expected area of triangles formed by placing points randomly in regular polygons and designed the following puzzle which I have not seen anywhere else.

Given a square with side lengths of $$1$$, what is the expected area of the largest, second largest, third largest, and smallest triangles formed by placing a point randomly in the square and connecting all of the square's corners to this central point?

Fig 1. Example point placement with triangles.

If you want to try this problem for yourself, there are many ways to solve it. Next, I will provide what I believe to be the most elegant way. Stop reading now to avoid solution spoilers.

### 2. How To Solve It

Initially getting an exact solution to this problem seems to involve lots of cases to handle based on all of the size orderings of the respective triangles; however, it is possible to reduce this all to one case WLOG with one key idea.

To solve this problem, we will first calculate the expected area of the largest triangle, then the expected area of the smallest triangle, and finally we will calculate the expected area of the remaining two middle triangles.

What I intend to provide here is not a formal proof of my solutions but instead all of the steps to get to the solution without any leaps that cannot be easily proved.

#### a. The Key to Solving It

We will start by dividing the square into $$4$$ regions of equal size by placing a point in the center

The key insight is that whichever region our random point is placed within will contain the smallest triangle, and the region opposite this will be associated with the largest triangle. For example:

It is trivial to prove this by contradition with the way the heights of the triangles are related and the definition of the regions we created.

Given any point placement, we can rotate our square so that the top edge is part of the largest triangle like with the figure above.

Next we will divide the bottom region into two equal sub-regions with a vertical line segment as follows:

We know that the two remaining triangles are the second and third largest areas, and we observe that if we place our point in the bottom left region, the remaining triangle on the left is smaller and if we place our point in the bottom right region, then the remaining triangle on the right is smaller. For example:

Given any point placement, after we rotate so that the placement is in the bottom region, we will flip the square over its vertical median if needed to place the point in the bottom right region.

Now WLOG, we can assume that the random point placement is in the bottom right region (as in the diagram above), so now we can calculate the expected area of all of the triangles since we know which one is largest, second largest, etc.

#### b. The Expected Area of The Smallest and Largest Triangles

We know that the largest triangle is formed by the top two vertices and the randomly placed point which is located in the bottom right region.

We know that the smallest triangle is formed by the bottom two vertices and the randomly placed point which is located in the bottom right region.

Using our formula for the area of a triangle $$\frac{1}{2}bh$$, we observe that the base of both triangles is always $$1$$ and the height is connected to the expected $$y$$ position of the random point.

Once we find the expected $$y$$ position of the random point (we can use geometry to find this as well), we can calculate the expected areas of the triangles.

The expected $$y$$ position of our point in the subregion is on the horizontal line that divides this triangle into two regions of equal area. This is illustrated by the diagram below:

We can solve for $$y$$, (the expected y position of our point), by the following equation which relates the areas of the triangle and the trapezoid:

$\frac{1}{2}y^2=(\frac{1}{2}-y)\frac{1+y}{2}$

Solving for $$y$$ we get two solutions $$y=-\frac{1}{2}$$ and $$y=\frac{1}{3}$$.

Discarding the negative solution which is not in our range, we get $$y=\frac{1}{3}$$.

Using this we can calculate the areas of the largest and smallest triangles

The largest triangle's expected area is $$\frac{1}{2}bh=\frac{1}{2}\times1\times(\frac{1}{2}+\frac{1}{3})=\frac{5}{12}$$

The smallest triangle's expected area is $$\frac{1}{2}bh=\frac{1}{2}\times1\times(\frac{1}{2}-\frac{1}{3})=\frac{1}{12}$$

#### c. The Expected Area of The Remaining Two Triangles

By either calculating the expected $$x$$ of the point in our subregion, or by observing that $$\mathbb{E}(x)=\mathbb{E}(y)$$, we can find the remaining two triangles' expected areas.

The 2nd largest triangle's expected area is $$\frac{1}{2}bh=\frac{1}{2}\times1\times(\frac{1}{2}+\frac{1}{2}-\frac{1}{3})=\frac{1}{3}$$

The 3rd largest triangle's expected area is $$\frac{1}{2}bh=\frac{1}{2}\times1\times(\frac{1}{3})=\frac{1}{6}$$

Now we have all of the expected areas of our $$4$$ triangles purely through geometry.

### 3. The solution

From largest to smallest, the expected areas of the triangles are: $$(\frac{5}{12},\frac{1}{3},\frac{1}{6},\frac{1}{12})$$

(With common denominators they are): $$(\frac{5}{12},\frac{4}{12},\frac{2}{12},\frac{1}{12})$$

### 4. Future Questions

For those who made it this far, here is an extra challenge (I will eventually post a writeup of the solution).

Calculate the following function $$f(m,n)$$ for the expected value of the $$m$$-th largest triangle in an $$n$$-sided (unit-distance) regular polygon with our aforementioned triangle construction.

(Hint #1 Manually try the triangle, pentagon, and hexagon cases to find a pattern.)

(Hint #2 If you are getting stuck, try treating the odd and even values of $$n$$ differently)

Email: zachdestefano@gmail (dot) com

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