Probability – P(ab > c) for Triangle Vertices on Unit Circle

Question

The vertices of a triangle are three uniformly random points on a unit circle. The side lengths are, in random order, $a,b,c$.

Show that $P(ab>c)=\frac12$.

The result is strongly suggested by simulations, and by my attempt shown below.

The simplicity of the result suggests that there may be an intuitive explanation. I am hoping for an intuitive explanation, but if that's not possible then any answer is welcome. (Examples of intuitive explanations are here and here.)

My attempt

Assume that the circle is centred at the origin, and the vertices of the triangle are:
$A(\cos(-2Y),\sin(-2Y))$ where $0\le Y\le\pi$
$B(\cos(2X),\sin(2X))$ where $0\le X\le\pi$
$C(1,0)$

Let:
$a=BC=2\sin X$
$b=AC=2\sin Y$
$c=AB=\left|2\sin\left(\frac{2\pi-2X-2Y}{2}\right)\right|=|2\sin(X+Y)|$

$P\left[ab>c\right]=P\left[2(\sin X)(\sin Y)>|\sin(X+Y)|\right]$

This probability is the ratio of the area of the shaded region to the area of the square in the graph below.

Rotate these regions $45^\circ$ clockwise about the origin and then shrink them by a factor of $\frac{1}{\sqrt2}$, by letting $X=x-y$ and $Y=x+y$.

Using symmetry, we only need to consider the left half of the blue "diamond". Note that in the left half, $0<x<\pi/2$, so $|\sin(2x)|=\sin(2x)$.

$P\left[2(\sin X)(\sin Y)>|\sin(X+Y)|\right]$

$=P\left[2(\sin (x-y))(\sin (x+y))>\sin(2x)\right]$

$=P\left[(\sin^2x)(\cos^2y)-(\cos^2 x)(\sin^2 y)>(\cos x)(\sin x)\right]$

$=P\left[(\sin^2x)(\cos^2y)-(\cos^2 x)(1-\cos^2 y)>(\cos x)(\sin x)\right]$

$=P\left[\cos^2 y>\cos^2 x+(\cos x)(\sin x)\right]$

$=P\left[ -f(x)<y<f(x)\right]$

where $\color{red}{f(x)=\arccos\left((\cos x)\sqrt{1+\tan x}\right)}$.

Noting that $f\left(\frac{\pi}{4}\right)=0$, the probability is

$$\dfrac{\int_{\pi/4}^{\pi/2}f(x)\mathrm dx}{\frac12\left(\frac{\pi}{2}\right)^2}$$

Numerical evidence suggests that $\int_{\pi/4}^{\pi/2}f(x)\mathrm dx=\frac{\pi^2}{16}$, but I don't know how to prove this.

If that's true, then the probability is indeed $1/2$.

Context

This question was inspired by a question, "If $(a,b,c)$ are the sides of a triangle, what is the probability that $ac>b^2$?"

Generalization

I found a generalization, which may or may not help in finding an intuitive explanation.

For $k\in\mathbb{R^+}$, we have

$$P(ab<kc)=\frac{2}{\pi}\arctan k$$

or equivalently,

$$P\left(\frac{ab}{c}<k\right)=\frac{\arctan k}{\frac{\pi}{2}}$$

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Proof:

Using the set-up in the above "My attempt", we have

$P[ab\color{red}{>}kc]$

$=P\left[2(\sin X)(\sin Y)>k|\sin(X+Y)|\right]$

$=P\left[2(\sin (x-y))(\sin (x+y))>k\sin(2x)\right]$

$=P\left[(\sin^2x)(\cos^2y)-(\cos^2 x)(\sin^2 y)>k(\cos x)(\sin x)\right]$

$=P\left[(\sin^2x)(\cos^2y)-(\cos^2 x)(1-\cos^2 y)>k(\cos x)(\sin x)\right]$

$=P\left[\cos^2 y>\cos^2 x+k(\cos x)(\sin x)\right]$

$=P\left[ -g(x)<y<g(x)\right]$

where $\color{red}{g(x)=\arccos\left((\cos x)\sqrt{1+k\tan x}\right)}$.

Noting that $g\left(\arctan k\right)=0$, the probability is

$$\dfrac{\int_{\arctan k}^{\pi/2}g(x)\mathrm dx}{\frac12\left(\frac{\pi}{2}\right)^2}$$

To evaluate the integral in the numerator, I copy the method in @Zacky's answer, adjusted for the presence of $k$.

$$I=\int_{\arctan k}^{\pi/2} \arccos\left(\cos x \sqrt{1+k\tan x}\right)dx\overset{\cot x\to x}=\int_0^{1/k} \frac{\arccos \sqrt{\frac{x(k+x)}{1+x^2}}}{1+x^2}dx$$

$$=\int_0^{1/k} \frac{\arctan \sqrt{\frac{1}{x}\frac{1-kx}{k+x}}}{1+x^2} dx\overset{\large \frac{1-kx}{k+x}\to x}=\int_0^{1/k} \frac{\operatorname{arccot} \sqrt{\frac{1}{x}\frac{1-kx}{k+x}}}{1+x^2}dx$$

$$\Rightarrow 2I=\frac{\pi}{2}\int_0^{1/k} \frac{1}{1+x^2}dx\Rightarrow \boxed{I=\frac{\pi}{4}\arctan \frac1k}$$

Above it was utilized that $\, \arccos x =\arctan \left(\frac{\sqrt{1-x^2}}{x}\right)$ and $\arctan x+\operatorname{arccot} x=\frac{\pi}{2}$.

$\therefore P(ab>kc)=\frac{2}{\pi}\arctan\frac1k$

$\therefore P(ab<kc)=1-\frac{2}{\pi}\arctan\frac1k=\frac{2}{\pi}\arctan k$

Answer 1

Here's a way to finish the last integral making use of its symmetry.

$$I=\int_\frac{\pi}{2}^\frac{\pi}{4} \arccos\left(\cos x \sqrt{1+\tan x}\right)dx\overset{\cot x\to x}=\int_0^1 \frac{\arccos \sqrt{\frac{x(1+x)}{1+x^2}}}{1+x^2}dx$$

$$=\int_0^1 \frac{\arctan \sqrt{\frac{1}{x}\frac{1-x}{1+x}}}{1+x^2} dx\overset{\large \frac{1-x}{1+x}\to x}=\int_0^1 \frac{\operatorname{arccot} \sqrt{\frac{1}{x}\frac{1-x}{1+x}}}{1+x^2}dx$$

$$\Rightarrow 2I=\frac{\pi}{2}\int_0^1 \frac{1}{1+x^2}dx\Rightarrow \boxed{I=\frac{\pi^2}{16}}$$

Above it was utilized that $\, \arccos x =\arctan \left(\frac{\sqrt{1-x^2}}{x}\right)$ and $\arctan x+\operatorname{arccot} x=\frac{\pi}{2}$.