Does $P(X^2 + Y^2 > Z^2) = \log 2$ where $X,Y,Z$ are the angles of a triangle?

Question

Let $x,y,z$ be the sides of a triangle whose vertices are uniformly distributed on a circle and let $X,Y,Z$ be their corresponding opposite angles. In this question "Generalization of the triangle inequality to exponents of the sides" it was proved that for $a \ge 1$,

$$ P(x^a + y^a > z^a) = \frac{1}{a^2} $$

Motivated by this I wondered if $f(a) = P(X^a + Y^a > Z^a)$ has an elegant closed form expression. I have not yet observed general closed form formula yet however, experimental data from a simulation of $3 \times 10^{9}$ trail suggests that $f(1) = \frac{3}{4}$ and $f(2) \approx 0.693150$ which is suspiciously close to $\log 2 = 0.693147$.

Question 1: Is $P(X^2 + Y^2 > Z^2) = \log 2$ ?

Question 2: Does $P(X^a + Y^a > Z^a)$ have a closed form formula?

Answer 1

We have $$X+Y+Z=\pi$$ So the probability in question 1 is equivalent to $$P_2=\frac{\int_{0}^{\pi}\int_{0}^{\pi-X}\mathbf{1}_{X^2+Y^2>(\pi-X-Y)^2}\,\mathrm dY\,\mathrm dX}{\int_{0}^{\pi}\int_{0}^{\pi-X}1\,\mathrm dY\,\mathrm dX}=\frac{2}{\pi^2}\int_{0}^{\pi}\int_{0}^{\pi-X}\mathbf{1}_{X^2+Y^2>(\pi-X-Y)^2}\,\mathrm dY\,\mathrm dX$$ Now we analyze the indicator function $$X^2+Y^2>(\pi-X-Y)^2$$ $$X^2+Y^2>\pi^2+X^2+Y^2-2\pi X-2\pi Y+2XY$$ $$Y(2X-2\pi)-2\pi X+\pi^2<0$$ $$Y>\frac{\pi^2-2\pi X}{2\pi-2X}$$ When $X\le\frac{\pi}{2}$, the lower bound is okay, otherwise it must be improved to $0$. Therefore $$P_2=\frac{2}{\pi^2}\left(\int_{0}^{\pi/2}\pi-X-\frac{\pi^2-2\pi X}{2\pi-2X}\,\mathrm dX+\int_{\pi/2}^{\pi}\pi-X\,\mathrm dX\right)$$ The remaining integrals are easy to deal with.

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Similarly, the probability in question 2 is equivalent to $$1-P_a=\frac{2}{\pi^2}\int_{0}^{\pi}\int_{0}^{\pi-X}\mathbf{1}_{X^a+Y^a\le (\pi-X-Y)^a}\,\mathrm dY\,\mathrm dX$$ To deal with the indicator function, we consider $$x=\frac{X}{\pi-X-Y},y=\frac{Y}{\pi-X-Y}$$ The mapping $(X,Y)\mapsto(x,y)$ is bijective onto the set of pairs $(x,y)\in\mathbb{R}^2$ satisfying $x,y>0$. And the inverse is $$X=\frac{\pi x}{x+y+1},Y=\frac{\pi y}{x+y+1}$$ Therefore the Jacobian determinant is $$\frac{\pi(y+1)}{(x+y+1)^2}\cdot\frac{\pi(x+1)}{(x+y+1)^2}-\left(-\frac{\pi x}{(x+y+1)^2}\right)\left(-\frac{\pi y}{(x+y+1)^2}\right)=\frac{\pi^2}{(x+y+1)^3}$$ $$\implies 1-P_a=2\int_{0}^{\infty}\int_{0}^{\infty}\frac{\mathbf{1}_{x^a+y^a\le 1}}{(x+y+1)^3}\,\mathrm dy\,\mathrm dx=2\int_{0}^{1}\int_{0}^{\sqrt[a]{1-x^a}}\frac{1}{(x+y+1)^3}\,\mathrm dy\,\mathrm dx\\=\int_{0}^{1}\frac{1}{(x+1)^2}-\frac{1}{\left(x+\sqrt[a]{1-x^a}+1\right)^2}\,\mathrm dx=\frac{1}{2}-\int_{0}^{1}\frac{1}{\left(x+\sqrt[a]{1-x^a}+1\right)^2}\,\mathrm dx$$ $$\implies P_a=\frac{1}{2}+\int_{0}^{1}\frac{1}{\left(x+\sqrt[a]{1-x^a}+1\right)^2}\,\mathrm dx$$ WolframAlpha does not give closed form when $a=3$, so I guess the final integral doesn't have a closed form.