Choose three uniformly random points on a disk, and let them be the vertices of a triangle. Call the side lengths, in random order, $a,b,c$.
What is $P(a^2<bc)$ ?
A simulation with $10^7$ such random triangles yielded a proportion of $0.4285833\approx1.00003\times\frac{3}{7}$ satisfying $a^2<bc$, leading me to believe that the probability is $\frac37$.
I find this alleged probability interesting because it seems that the number $7$ rarely appears in the answers to natural geometry or probability questions.
Context
Recently I have learned that some probabilities related to circles, of the form $P(x^2<yz)$, have simple rational values (for example, a probability about a random triangle inscribed in a circle, and a probability about a random pentagram inscribed in a circle). So I wondered if there is a probability like this related to a disk. I may have found one.
My attempt
Let the boundary of the disk be $x^2+y^2=1$.
Using disk point picking, let the three points be
- $C\left(\sqrt{r_1}\cos\theta_1,\sqrt{r_1}\sin\theta_1\right)$
- $B\left(\sqrt{r_2}\cos\theta_2,\sqrt{r_2}\sin\theta_2\right)$
- $A\left(\sqrt{r_3},0\right)$
where each $r$ is a uniformly random real number from $0$ to $1$, and each $\theta$ is a uniformly random real number from $0$ to $2\pi$.
Let $a=BC$, $b=AC$, $c=AB$.
This gives:
$$P(a^2<bc)=P\left(r_1+r_2-2\sqrt{r_1r_2}\cos(\theta_1-\theta_2)<\sqrt{\left(r_1-2\sqrt{r_1r_3}\cos\theta_1+r_3\right)\left(r_2-2\sqrt{r_2r_3}\cos\theta_2+r_3\right)}\right)$$
I do not know how to set up an integral, nor any other way to calculate the probability. Apparently, after the dust settles, we should be left with $\frac37$.