Expected area of a random triangle with fixed perimeter

Question

I'm trying to calculate the expected area of a random triangle with a fixed perimeter of 1.

My initial plan was to create an ellipse where one point on the ellipse is moved around and the triangle that is formed with the foci as the two other vertices (which would have a fixed perimeter) would have all the varying areas. But then I realized that I wouldn't account for ALL triangles using that method. For example, an equilateral triangle with side lengths one third would not be included.

Can anyone suggest how to solve this problem? Thanks.

Answer 1

Since the Heron's formula gives: $$ A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)},$$ assuming that the random choice is made accordingly to the broken stick model (notice that not every choice for the side lengths satisfies the triangle inequality, hence for such cases we set the area as zero): $$\begin{eqnarray*}\mathbb{E}[A]&=&\frac{1}{2}\int_{0}^{1}\int_{x}^{1}\sqrt{\max\left[0,(x^2+(y-x)^2+(1-y)^2)^2-2(x^4+(y-x)^4+(1-y)^4)\right]}\,dy\,dx\\&=&0.00747998\ldots\end{eqnarray*}$$ A random model that makes more sense is to choice a random point (with respect to the uniform measure) on the set $T=\{(u,v,w):u,v,w\geq 0, u+v+w=\frac{1}{2}\}$, then pick the side lengths as $a=v+w,b=u+w,c=u+v$. In such a way the triangle inequality is always fullfilled and the Heron's formula gives simply: $$A = \sqrt{\frac{uvw}{2}},$$ so: $$\mathbb{E}[A]=\frac{8}{\sqrt{6}}\int_{T}\sqrt{uvw}\,d\mu=\frac{1}{\sqrt{3}}\int_{0}^{\pi/2}\int_{0}^{\pi/2}\sin^5\theta \cos^2\theta\sin^4\phi\, d\theta \,d\phi=\frac{\pi}{70\sqrt{3}}.$$

Answer 2

My solution:

Let $U,V \sim \mathcal{U}[0,1]$ be 2 sides of your triangle, and $1-u-v$ being the other one (as by definition they add up to 1).

Not every pair $(u,v)$ is acceptable, as they need to form a triangle. The condition for that is that the biggest vertex is smaller than the sum of the two other sides, that is:

$$ \max(u,v,1-u-v) > 1 - \max(u,v,1-u-v) \Leftrightarrow \max(u,v,1-u-v) < \frac{1}{2}$$

The quantity you're after here is the area, so using Heron's formula, you're looking at:

$\int\int_{(u,v)\in[0,1]^2} \sqrt{0.5(0.5-u)(0.5-v)(u+v-0.5)}\chi(\max(u,v,1-u-v) < \frac{1}{2}) dudv \ = \int_{u=0}^1\int_{v=0}^u 2\sqrt{0.5(0.5-u)(0.5-v)(u+v-0.5)}\chi(\max(u,1-u-v) < \frac{1}{2}) dudv \\ = \int_{u=0}^\frac{1}{2}\int_{v=0}^u 2\sqrt{0.5(0.5-u)(0.5-v)(u+v-0.5)}\chi(1-u-v < \frac{1}{2})\chi(v < u)\chi(u < \frac{1}{2}) dudv \\ = \int_{u=\frac{1}{4}}^\frac{1}{2}\int_{v=\frac{1}{2}-u}^u 2\sqrt{0.5(0.5-u)(0.5-v)(u+v-0.5)} dudv \\ =0.00586765$

Answer 3

Using the feedback given in the comments, I've got a proposal for a solution. I must say that I am not very certain of whether it is correct or not.

Let $0<x<y<1$ be the points at which the "stick is broken", and so $x, y-x, 1-y$ are the lengths of the three segments. For a triangle to be formed, the sum of any two sides must be greater than the third side. Therefore we get the following inequalities: $$x+(y-x)>1-y \\ (y-x)+(1-y)>x \\ (1-y)+x>y-x$$ Plotting these on a coordinate system gives a triangular region with vertices $(0, 1/2), (1/2, 1/2), (1/2, 1)$. So any pair $(x, y)$ contained within that region results in a triangle of perimeter 1. I parameterize these pairs: $$\left(\frac{a_1}{2}, \frac{1+a_2}{2}\right),$$ for $0<a_2<a_1<1$. Now these can be plugged in Heron's formula (and simplified): $$A(a_1, a_2)=\frac{1}{4}\sqrt{(1-a_1)(a_1-a_2)(a_2)}$$ Taking the integral for the average value: $$E(A)=\frac{1}{A(R)}\int_0^1\!\!\!\int_0^{a_1}A(a_1, a_2)\,da_2da_1=0.0299199$$ I'm uncertain about whether the distribution is unbiased and uniform. Anyone who can verify this?