In a triangle with sides $x \le y \le z$, is it true that $P\left(\frac{x^a + z^a}{y^a} < \frac{x+z}{y}\right) = \frac{1}{a}$?

Question

Let $x \le y \le z$ be the sides of a triangle whose vertices are uniformly distributed on a cirlce. In this question, it was proved that for any $a \ge 1$, the probability

$$ P\left(x^a + y^a > z^a\right) = \dfrac{1}{a^2}. $$

On a similar note, experimental evidences show that there is an elegant closed formula for the probability given below for $a \ge 2$

$$ P\left(\frac{x^a + z^a}{y^a} < \frac{x+z}{y}\right) = \frac{1}{a}. $$

Can this be proved or disproved?

Answer 1

Using argument similar to this answer, we can prove that the probability evaluates to $\frac{1}{a}$ for $a\ge 2$.

Let $$x=2\sin(\alpha+\beta),y=2\sin\alpha,z=2\sin\beta$$ With $$\alpha,\beta>0;\alpha+\beta<\pi$$ And consider $$s=\frac{z}{y}=\frac{\sin\beta}{\sin\alpha},t=\frac{x}{y}=\frac{\sin(\alpha+\beta)}{\sin\alpha}$$ The mapping $(\alpha,\beta)\mapsto(s,t)$ is bijective onto the set of pairs $(s,t)\in\mathbb{R}^2$ satisfying $$s,t>0;s+t>1>|s-t|$$ And the Jacobian determinant of the mapping is $st$, the condition $x\le y\le z$ has Lebesgue measure $\pi^2/12$

Using the change of variable $(\alpha,\beta)\mapsto(s,t)$, we get $$P\left(\frac{x^a+z^a}{y^a}<\frac{x+z}{y}\mid x\le y\le z\right)=\frac{12}{\pi^2}\int_{\Omega(a)}\frac{\mathrm ds\,\mathrm dt}{st}$$ With $$\Omega(a)=\left\{(s,t)\in\mathbb{R}^2:s\ge 1\ge t>0 ;s^a+t^a<s+t\right\}$$ Let $$s^a+t^a<s+t\implies 1\le s<u(t)$$ And $u(t)$ is the unique real root of $s^a+t^a=s+t$. By Fubini's theorem: $$\frac{12}{\pi^2}\int_{\Omega(a)}\frac{\mathrm ds\,\mathrm dt}{st}=\frac{12}{\pi^2}\int_{0}^{1}\int_{1}^{u(t)}\frac{\mathrm ds}{s}\frac{\mathrm dt}{t}=\frac{12}{\pi^2}\int_{0}^{1}\frac{\ln u(t)}{t}\,\mathrm dt$$ Since we have $$u(t)^a+t^a=u(t)+t$$ Using idea similar to the linked post from @Tyma Gaidash (thank you so much!), we let $$t=uv,v\ge 0$$ $$\implies u^a+u^av^a=u+uv$$ $$\implies u^a\left(1+v^a\right)=u(1+v)$$ $$\implies u^{a-1}=\frac{1+v}{1+v^a}$$ $$\implies u=\sqrt[a-1]{\frac{1+v}{1+v^a}}$$ $$\implies t=v\sqrt[a-1]{\frac{1+v}{1+v^a}}$$ When $v$ runs from $0$ to $1$, $t$ also runs from $0$ to $1$, and $u(t)$ stays above $1$ (this is what we want) so we may do the substitution $$t=v\sqrt[a-1]{\frac{1+v}{1+v^a}}$$ $$\implies \mathrm dt=-\frac{\left(v^a-av-a+1\right)\sqrt[a-1]{\frac{1+v}{1+v^a}}}{(a-1)(v+1)\left(v^a+1\right)}\,\mathrm dv$$ And the probability becomes $$P\left(\frac{x^a+z^a}{y^a}<\frac{x+z}{y}\mid x\le y\le z\right)=-\frac{12}{\pi^2}\int_{0}^{1}\frac{\left(v^a-av-a+1\right)\ln\sqrt[a-1]{\frac{1+v}{1+v^a}}}{v(a-1)(v+1)\left(v^a+1\right)}\,\mathrm dv$$ $$=-\frac{12}{\pi^2(a-1)^2}\int_{0}^{1}\frac{\ln(1+v)}{v(v+1)}-a\frac{\ln(1+v)}{v\left(v^a+1\right)}-\frac{\ln\left(1+v^a\right)}{v(v+1)}+a\frac{\ln\left(1+v^a\right)}{v\left(v^a+1\right)}\,\mathrm dv$$ Using the idea from @Edward H (thank you so much!), we substitute $w=v^a\implies \mathrm dw=av^{a-1}\,\mathrm dv\implies \mathrm dw/w=a\,\mathrm dv/v$ into the second and fourth term. $$a\int_{0}^{1}\frac{\ln(1+v)}{v\left(v^a+1\right)}\,\mathrm dv=\int_{0}^{1}\frac{\ln(1+w^{1/a})}{w\left(w+1\right)}\mathrm dw$$ $$a\int_{0}^{1}\frac{\ln(1+v^a)}{v\left(v^a+1\right)}\,\mathrm dv=\int_{0}^{1}\frac{\ln(1+w)}{w\left(w+1\right)}\,\mathrm dw$$ $$\implies P\left(\frac{x^a+z^a}{y^a}<\frac{x+z}{y}\mid x\le y\le z\right)$$ $$=-\frac{12}{\pi^2(a-1)^2}\int_{0}^{1}2\frac{\ln(1+v)}{v(v+1)}-\frac{\ln(1+v^{1/a})}{v(v+1)}-\frac{\ln\left(1+v^a\right)}{v(v+1)}\,\mathrm dv$$ $$=-\frac{12}{\pi^2(a-1)^2}\int_{0}^{1}2\frac{\ln(1+v)}{v}-\frac{\ln(1+v^{1/a})}{v}-\frac{\ln\left(1+v^a\right)}{v}\,\mathrm dv\\+\frac{12}{\pi^2(a-1)^2}\int_{0}^{1}2\frac{\ln(1+v)}{v+1}-\frac{\ln(1+v^{1/a})}{v+1}-\frac{\ln\left(1+v^a\right)}{v+1}\,\mathrm dv$$ As we used $\frac{1}{v(v+1)}=\frac{1}{v}-\frac{1}{v+1}$. Consider the integral $$\int_{0}^{1}\frac{\ln\left(1+v^b\right)}{v}\,\mathrm dv$$ With $b\ge 2$. We substitute $w=v^b\implies \mathrm dw=bv^{b-1}\,\mathrm dv\implies \mathrm dw/w=b\,\mathrm dv/v$ $$\int_{0}^{1}\frac{\ln\left(1+v^b\right)}{v}\,\mathrm dv=\frac{1}{b}\int_{0}^{1}\frac{\ln\left(1+w\right)}{w}\,\mathrm dw=\frac{1}{b}\sum_{k=1}^{\infty}\int_{0}^{1}\frac{(-1)^{k-1}}{k}w^{k-1}\,\mathrm dw$$ $$=\frac{1}{b}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{2}{(2k)^2}=\frac{\pi^2}{12b}$$ Thus $$-\frac{12}{\pi^2(a-1)^2}\int_{0}^{1}2\frac{\ln(1+v)}{v}-\frac{\ln(1+v^{1/a})}{v}-\frac{\ln\left(1+v^a\right)}{v}\,\mathrm dv$$ $$=-\frac{12}{\pi^2(a-1)^2}\left(\frac{\pi^2}{6}-\frac{a\pi^2}{12}-\frac{\pi^2}{12a}\right)=\frac{1}{a}$$ Now we consider: $$I(a)=\int_{0}^{1}2\frac{\ln(1+v)}{v+1}-\frac{\ln(1+v^{1/a})}{v+1}-\frac{\ln\left(1+v^a\right)}{v+1}\,\mathrm dv$$ Noting that $I(1)=0$, we differentiate with respect to $a$: $$\implies I'(a)=\int_{0}^{1}\frac{v^{1/a}\ln v}{a^2(v+1)\left(v^{1/a}+1\right)}-\frac{v^a\ln v}{(v+1)\left(v^a+1\right)}\,\mathrm dv$$ For the first term, we substitute $v=w^a\implies \mathrm dv=aw^{a-1}\,\mathrm dw$ $$\implies \int_{0}^{1}\frac{v^{1/a}\ln v}{a^2(v+1)\left(v^{1/a}+1\right)}\,\mathrm dv=\int_{0}^{1}\frac{w^a\ln w}{(w+1)\left(w^a+1\right)}\,\mathrm dw$$ $$\implies I'(a)=0$$ $$\implies I(a)=I(1)=0$$ Plugging all the pieces together, we obtain $$\boxed{P\left(\frac{x^a+z^a}{y^a}<\frac{x+z}{y}\mid x\le y\le z\right)=\frac{1}{a}}$$

Answer 2

Partial answer: Here are proofs of OP's conjecture for $a=2$ and $a=3$.

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Notation

• I change OP's $a$ to $n$.
• I change OP's $x,y,z$ to $a,b,c$.

An equivalent statement of OP's conjecture

We relax OP's requirement that $a\le b\le c$.

If $b=\max(a,b,c)$, then $P\left(\dfrac{a^n+c^n}{b^n}<\dfrac{a+c}{b}\right)=1$.
If $b=\min(a,b,c)$, then $P\left(\dfrac{a^n+c^n}{b^n}<\dfrac{a+c}{b}\right)=0$.

This is because

$$P\left(\dfrac{a^n+c^n}{b^n}<\dfrac{a+c}{b}\right)=P\left(a^n+c^n<b^{n-1}(a+c)\right)=P\left(a\color{red}{\left(a^{n-1}-b^{n-1}\right)}+c\color{red}{\left(c^{n-1}-b^{n-1}\right)}<0\right).$$

Now let $P_n=P\left(\dfrac{a^n+c^n}{b^n}<\dfrac{a+c}{b}\right)$ where the sides are $a,b,c$ in random order (not necessarily $a \le b\le c$).

So OP's conjecture is equivalent to:

$$P_n=\frac13\left(1+\frac1n+0\right)=\frac{n+1}{3n}$$

The case $n=2$

Assume that the circle is centred at the origin, and the vertices of the triangle are:

• $A\space(\cos(-2Y),\sin(-2Y))$ where $Y$ is uniformly random in $0\le Y\le\pi$
• $B\space(1,0)$
• $C\space(\cos(2X),\sin(2X))$ where $X$ is uniformly random in $0\le X\le\pi$

Let:

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

We consider the case $n=2$.

$$P_2=P\left[\frac{a^2+c^2}{b^2}<\frac{a+c}{b}\right]$$

$$=P\left[\frac{\sin^2X+\sin^2Y}{\sin^2(X+Y)}<\frac{\sin X+\sin Y}{|\sin(X+Y)|}\right]$$

$P_2$ 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 $|\sin(2x)|=\sin(2x)$.

$$P_2=P\left[\frac{\sin^2(x-y)+\sin^2(x+y)}{\sin^2 2x}<\frac{\sin(x-y)+\sin(x+y)}{\sin 2x}\right]$$

$$=P\left[\frac{\left(\left(\sin x\right)\left(\cos y\right)-\left(\cos x\right)\left(\sin y\right)\right)^{2}+\left(\left(\sin x\right)\left(\cos y\right)+\left(\cos x\right)\left(\sin y\right)\right)^{2}}{2\left(\sin x\right)\left(\cos x\right)}<2\left(\sin x\right)\left(\cos y\right)\right]$$

$$=P\left[\left(\sin^{2}x-\cos^{2}x\right)\left(\cos^{2}y\right)-2\left(\sin^{2}x\right)\left(\cos x\right)\left(\cos y\right)+\cos^{2}x<0\right]$$

Noting the quadratic expression in $\cos y$, the upper-right boundary of the left half of the shaded region simplifies to

$$y=\color{red}{\arccos\left(-\frac{\cos x}{\cos 2x}\right)}$$

which intersects the $x$-axis at $x=\pi/3$.

Therefore

$$P_2=\frac{\frac12\left(\frac{\pi}{2}\right)^2-J}{\frac12\left(\frac{\pi}{2}\right)^2}$$

where

$$J=\int_{\pi/3}^{\pi/2}\color{red}{\arccos\left(-\frac{\cos x}{\cos 2x}\right)}dx\overset{x\to\frac{\pi}{2}-x}{=}\int_0^{\pi/6}\arccos\left(\frac{\sin x}{\cos 2x}\right)dx.$$

Wolfram suggests that $J=\dfrac{\pi^2}{16}$, but I do not know how to prove this. EDIT: This has now been proven, here.

Therefore $P_2=1/2$, confirming the equivalent statment of OP's conjecture for $n=2$.

(This is another example of a probability question that has answer $1/2$ but resists intuitive explanation.)

The case $n=3$

Using the same method as in $n=2$, we get

$$P_3=P\left[\frac{a^3+c^3}{b^3}<\frac{a+c}{b}\right]$$

$$=P\left[\frac{\sin^3X+\sin^3Y}{\sin^3(X+Y)}<\frac{\sin X+\sin Y}{|\sin(X+Y)|}\right]$$

$$=P\left[\frac{\sin^3(x-y)+\sin^3(x+y)}{\sin^3 2x}<\frac{\sin(x-y)+\sin(x+y)}{\sin 2x}\right]$$

$$=P\left[\left(\cos y\right)^{2}\color{red}{\left(\sin^2x-\frac34\right)}<\left(\cos x\right)^{2}\color{red}{\left(\sin^2x-\frac34\right)}\right]$$

$$P_3=\frac{\left(\frac{\pi}{3}\right)^2}{\left(\frac{\pi}{2}\right)^2}=\frac49$$

This confirms the equivalent statement of OP's conjecture for $n=3$.

Other $n$ values

For $n=4$, I could not obtain an integral expression like I did for $n=2,3$.

For $n=5$, OP's conjecture is true if:

$$\int_{0}^{1}\frac{1}{\sqrt{-x^{2}-2x+15}}\arccos\sqrt{\frac{x^{3}+7x^{2}-25x-15}{8x^{2}-40}}dx=\frac{\pi^{2}}{10}-\left(\arcsin\sqrt{\frac{5}{8}}\right)^{2}$$

Wolfram suggests that this is true. EDIT: It must be true, because @Thinh Dinh's answer proves the general case.

For $n>5$, I suspect it is very difficult to obtain integral expressions like I did for $n=2,3,5$.

Answer 3

There is another way to evaluate @Thinh Dinh’s integral $$\int_0^1\frac{\ln(y(x))}xdx,y^a-y+x^a-x=0$$ using a reversion theorem. Letting $y^{p-1}=w$ and applying it gives:

$$y^a-y+x=0\iff y^{a-1}-1=-xy^{-1}\iff w=1-x w^{-\frac1{a-1}}\\\implies \ln(y)=\frac{\ln(w)}{a-1}=\frac{\ln(1)}{a-1}+\sum_{n=1}^\infty\frac{(-x)^n}{n!}\left.\frac{d^{n-1}}{dt^{n-1}}\left(\left(\frac{\ln(t)}{a-1}\right)’t^{-\frac n{p-1}}\right)\right|_1$$

Evaluating the derivatives and rearranging gives:

$$\ln(y)=\frac1{a-1}\sum_{n=1}^\infty\frac{(-x)^n\Gamma\left(\frac n{1-a}\right)}{n!\Gamma\left(\frac{an}{1-a}+1\right)}\implies \int_0^1\frac{\ln(y(x))}xdx= \frac1{a-1}\sum_{n=1}^\infty\frac{\Gamma\left(\frac n{1-a}\right)}{n!\Gamma\left(\frac{an}{1-a}+1\right)}\int_0^1\frac{(x-x^a)^n}xdx$$

There is essentially a beta function integral, cancelling the gamma functions:

$$\frac1{a-1}\sum_{n=1}^\infty\frac{\Gamma\left(\frac n{1-a}\right)}{n!\Gamma\left(\frac{an}{1-a}+1\right)}\int_0^1\frac{(x-x^a)^n}xdx=\frac1{a-1}\sum_{n=1}^\infty\frac{(-1)^n\Gamma\left(\frac n{1-a}\right)}{n!\Gamma\left(\frac{an}{1-a}+1\right)}\frac{\Gamma\left(\frac{an}{1-a}\right)n!}{(1-a)\Gamma\left(\frac n{1-a}+1\right)}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{an^2}=\boxed{\frac{\pi^2}{12a}}$$