Conjecture: In a "happy triangle", the expected product of side lengths is $\pi$.

Question

Definition of happy triangle

On a unit circle, choose three uniformly random independent points $A,B,C$.
Inside the circle, choose a uniformly random independent point $D$.
Triangle $ABC$ is called "happy" if it contains $D$.

Conjecture

Is the following conjecture true: In a happy triangle, the expected product of the three side lengths is $\pi$.

I simulated $2\times10^7$ happy triangles, and the average product was $3.14184188\approx1.00007933\pi$.

My thoughts

Letting $X=\text{product of the three side lengths}$, I tried (but failed) to find the probability distribution function $P(X<x)$ so that I could find the probability density function and thus $E(X)$.

The expected product of the three side lengths is not the cube of the expected side length, because the three side lengths are not independent random variables. (A simulation suggests that the expected side length is approximately $1.509$.)

Other elegant properties?

A happy triangle seems to have many elegant properties.

Answer 1

Given a random triangle $ABC$ with $A,B,C\in S^1$, the probability that $D$ lies inside $ABC$ is $\frac{[ABC]}{\pi}=\frac{abc}{4\pi}$, since $abc=4R\Delta$ by Euler's formula and $R=1$. It follows that the conditional expected value

$$\mathbb{E}[abc | D\in ABC] $$

just depends on the average area of the triangles we may build around $D$ (meaning: having $D$ as an interior point), with $D$ being uniformly distributed over the unit ball. Such area can be written as $2R^2\sin A\sin B\sin C$, which equals $2\sin A\sin B\sin C$.

As a continuation of Thomas Andrews' answer,

$$ 64\iint_{[0,\pi]^2}\sin^2(u-v)\sin^2(u)\sin^2(v)\,du\,dv = 6\pi^2 \tag{N}$$

while

$$ \int_{0}^{2\pi}\int_{0}^{2\pi}2\left|\sin\frac{x-y}{2}\sin\frac{x}{2}\sin\frac{y}{2}\right|\,dx\,dy = 8\iint_{[0,\pi]^2}\left|\sin(u-v)\sin(u)\sin(v)\right|\,du\,dv=6\pi\tag{D}$$

since $\sin(u)\sin(v)=\frac{1}{2}\left(\cos(u-v)-\cos(u+v)\right)$ and

$$ \left|\sin x\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1}.$$

This proves that the wanted expected value is indeed $\pi$.

Answer 2

Not a complete answer, just a start.

Always choose one of the vectors as $1$ on the complex plane. You want to choose two more $w,z$ with $w=e^{ix},z=e^{iy}.$ Then the product of the edge lengths is:

$$p(x,y)=8|\sin((x-y)/2)\sin(x/2)\sin(y/2)|.$$

The probability of $D$ being inside the triangle is $\frac A\pi$ where $A=A(x,y)$ is the area of the triangle.

So you need a formula for $A(x,y)$ and then you want the integral:

$$\dfrac{\int_0^{2\pi}\int_0^{2\pi}A(x,y)p(x,y)\,dy\,dx}{\int_0^{2\pi}\int_0^{2\pi}{A(x,y)}\,dy\,dx}$$

You get: $$A(x,y)=\frac12|\sin((x-y)/2)(1-w)(1-z)|=2|\sin((x-y)/2)\sin(x/2)\sin(y/2)|.$$ This would indicate that if the area of a triangle inscribed in a unit circle is $\frac1{4}$ times the product of the lengths of the sides.

$$16\int_0^{2\pi}\int_0^{2\pi}\sin^2((x-y)/2)\sin^2(x/2)\sin^2(y/2)\,dy\,dx$$

Letting $x=2u,y=2v,$ the numerator becomes:

$$64\int_0^\pi\int_0^\pi\sin^2(u-v)\sin^2(u)\sin^2(v)\,dv\,du$$

That shouldn't be hard to integrate.

The denominator is harder, because of the absolute values.