What is the expected value of the area of a triangle formed from a broken stick?

[kuruman]

OK, thanks.

 

[WWGD]

OK, thanks.

If you give me some time, I will give it a shot. It would be nice to see how it compares to the product of the expected values for, e.g., Heron's formula or general area formula.

 

[SSequence]

@kuruman
Regarding post#25, it seems to me that you didn't include the factor of 1/(1-x). Any specific reason for not including it?
I am guessing the denominator is evaluation of integral in post#11.

I think (in the area formula of a triangle) we can take the two sides to be x and y and the third side to be 1-x-y ... and I am assuming this what you mean by A(x,y) in post#25.

In post#14 I was basically suggesting/asking about the same formula ... except I was suggesting including the factor of 1/(1-x) ---- basically plugging in area formula directly in integral in post#11 (and dividing the result by probability of forming of a triangle).

Ultimately, 0.193 means: The unit cube represents the collection of all possible values the three sides of a triangle bounded by $0 \leq x,y,z \leq 1$ that may be assumed when we cut two pieces as described. 0.193 represents the portion of the volume of $1= 1\times 1 \times 1$ of all these values that form a "viable" (actual) triangle. This value is obtained by adding the "infinitesimal" volumes of viable pairs. So 19.3% of all cuts produce a triangle.

This is a good visualisation, but I think a couple of things should be made explicit:
(1) Within the unit cube, the only points that "could" correspond to formulation of a triangle are the ones satisfying the constraint (which is the main constraint of the problem):
x+y+z=1

Call the resulting set A (set of points in unit cube satisfying the main constraint above). And when we consider the points belonging to A, only a subset of these will correspond to possible formulation of a triangle.

(2) If we consider the points corresponding to set A, we can associate a corresponding probability density function with each point in A. So it seems we have to be a little cautious with doing a literal interpretation in terms of volume ratios.

Furthermore percentage probability of forming a triangle is linked to points of A and some subset of A. Unless I am missing something, the unit cube doesn't come into it directly.

 

[WWGD]

I found this ttps://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-r

 

[kuruman]

Regarding post#25, it seems to me that you didn't include the factor of 1/(1-x). Any specific reason for not including it?

Please ignore post #25 where I was grappling with the 1/(1-x). Also I found that post #9 has an error in it. I had an epiphany last night and I was able to find the same answer using two different ways. I will write it up and post it as an "insight" if it meets with Greg's approval.

 

[kuruman]

I don't think this warrants an insight, so here are the two methods. My epiphany was realizing that once the first piece is chosen, the third apex of all valid triangles must lie on an ellipse.
Method 1
Let a be the first piece with an upper limit of half the perimeter p (p = (a+b+c)/2). The sum of the other two pieces is constant and equal to 2p - a. This means that the third vertex of the triangle must lie on an ellipse, the separation between the two foci of which is the base of the triangle a. The distance r from one of the foci to the ellipse is given in polar coordinates as
$r(\phi)=\frac{C}{1+\epsilon \cos \phi}$
where $\epsilon$ is the eccentricity. We can find constant C by setting $\phi$ = 0 in which case $r(\phi)$ has minimum and maximum values,
$r_{max}=\frac{C}{1-\epsilon}$; $r_{min}=\frac{C}{1+\epsilon}$
Note that $r_{max}-r_{min}=a$ from the definition of these quantities for an ellipse.
But also when $\phi = 0$, $2r_{min}+a=2p-a \rightarrow r_{min}=p-a$ which gives $C=(1+\epsilon)(p-a)$ in which case
$r(\phi)=\frac{(1+\epsilon)(p-a)}{1+\epsilon \cos \phi}$
The eccentricity is the ratio of the base (distance between foci) to the sum of the two other sides,
$\epsilon=\frac{a}{2p-a}$ so that, finally,
$$r(\phi)=\frac{2p(p-a)}{(2p-a)+a \cos \phi}$$
The probability that the first piece be between $a$ and $a+da$ is uniform and given by
$dP_a=\frac{1}{p}da$ subject to the constraint $0<a<p$.
The probability that the second piece be between $r$ and $r+dr$ is also uniform and given by
$dP_r=\frac{1}{r_{max}-r_{min}}dr=\frac{1}{a}dr$ subject to the constraint $r_{min}<r<r_{max}$.
The latter probability in terms of $\phi$ is
$dP_{\phi}=\frac{1}{a} \frac{\partial r}{\partial \phi}d\phi=\frac{2p(p-a)\sin \phi}{[(2p-a)+a \cos \phi]^2}d \phi$

The area of the triangle is
$A(r,\phi)=\frac{1}{2}\left| \vec{a} \times \vec{r}(\phi) \right |=\frac{ap(p-a)\sin \phi}{(2p-a)+a \cos \phi}$

The expectation value for the area is
$$\left< A \right>=\int{A(a,\phi)dP_a~dP_{\phi}}=2p\int_0^p{a(p-a)^2 da} \int_0^\pi{\frac{\sin^2 \phi}{[(2p-a)+a \cos \phi]^3} d \phi}$$Integration yields
$$\left< A \right>=\frac{\pi p^2}{30}.$$
Method 2
Heron's formula
$A=\sqrt{p(p-a)(p-b)(p-c)}~;~p=(a+b+c)/2$
can be transformed to
$A=\frac{1}{2}\sqrt{p(p-a)(a^2-\eta^2)}~;~\eta \equiv b-c$
where $a$ is the first broken piece. As before, the probability that the first piece be between $a$ and $a+da$ is uniform and given by
$dP_a=\frac{1}{p}da$ subject to the constraint $0<a<p$.
The remaining piece can be broken into two more pieces of which the larger piece is labeled $b$ and the smaller piece is labeled $c$. We define the difference between the pieces $\eta \equiv b-c$. The probability that $\eta$ be between $\eta$ and $\eta+d\eta$ is uniform and given by $dP_{\eta}=\frac{1}{a}d\eta$ subject to the constraint $0<\eta<a$
The expectation value for the area is
$$\left< A \right>=\int{A(a,\eta)dP_a~dP_{\eta}}=\frac{1}{2p} \int_0^p{\frac{1}{a}da} \int_0^a{\sqrt{p(p-a)(a^2-\eta^2)}d\eta}$$
Integration yields$$\left< A \right>=\frac{\pi p^2}{30}.$$