A typo leads to a discovery: For a random triangle in a circle, we have $P(a^2+b^2

Question

Background: a classic result in probability

If a triangle's vertices are random points on a circle, then the probability that the triangle is acute is $1/4$. There are intuitive explanations (example).

We can express this fact as $P(a^2+b^2<c^2)=1/4$, where $a,b,c$ are the side lengths in random order (we do not define $c$ as the longest side).

A typo leads to a discovery

I made a random triangle simulator, and I wanted to test it using the result above, but I accidentally typed $a^\color{red}{1}+b^2<c^2$.

Interestingly, the simulator suggested that $P(a+b^2<c^2)=1/3$, which is a rather simple probability.

I then proved that this is true if the circle has radius $1$, but my proof (shown below) is not intuitive.

My question

If the radius of the circle is $1$, is there an intuitive explanation for $P(a+b^2<c^2)=1/3$?

Equivalently, if the radius of the circle is $R$, then $P(aR+b^2<c^2)=1/3$.

(More examples of explanations that are intuitive, for me, are here and here. A list of simple probability results that resist intuitive explanation is here.)

My non-intuitive proof

Assume the radius of the circle is $1$.

Let $x$ and $y$ be independent and uniform in $(0,\pi)$, and let:

• $a=2\sin x$
• $b=2|\sin(x+y)|$
• $c=2\sin y$

$P\left(a+b^2<c^2\right)=P\left(f(x,y)<0\right)$ where $f(x,y)=\sin^2x+2\sin^2(x+y)-2\sin^2y$

Define $g(x,y)=\sin\left(y+\frac{1}{2}x-\frac{7\pi}{12}\right)\sin\left(y+\frac{1}{2}x-\frac{11\pi}{12}\right)\sin x$.

When expanded, $f(x,y)$ and $g(x,y)$ become the same thing, namely, $\sin x+4\left(\cos x\right)\left(\sin x\right)\left(\cos y\right)\left(\sin y\right)+2\left(\sin^{2}x\right)\left(\cos^{2}y\right)-2\left(\sin^{2}x\right)\left(\sin^{2}y\right)$.

$\therefore P(a+b^2<c^2)=P\left(\sin\left(y+\frac{1}{2}x-\frac{7\pi}{12}\right)\sin\left(y+\frac{1}{2}x-\frac{11\pi}{12}\right)\sin x<0\right)$

A cartesian graph showing this region has the straight lines $y=-\frac{1}{2}x+\frac{7\pi}{12}+k\pi$ and $y=-\frac{1}{2}x+\frac{11\pi}{12}+k\pi$ and $x=k\pi$ where $k\in\mathbb{Z}$.

In the square bounded by $0<x<\pi$ and $0<y<\pi$, these lines bound a region with $1/3$ the area of the square.

$\therefore P(a+b^2<c^2)=1/3$.

Variations

Simulations suggest the following variations of $P(a^2+b^2<c^2)=1/4$.

• $P\left(ab+b^2<c^2\right)=1/3$
• $P\left(ac+b^2<c^2\right)=1/6$
• $P\left(a^2+b^2<ac\right)=1/12$
• $P(a^2+b^2<(a+b)c)=1/2$

I'm not asking for proofs of these variations; I'm sharing them in case they might shed some insight.