Unordered pairs number

Question

How many unordered triplets $(x,y,z)$ , subject to constraints, $(x^4-2x^3)_{cyclic}\leq0$ , satisfy the system of equations: $$\left\{\begin{array}{l}(4x^2-8x+3)\sqrt{2x-x^2}+1=y\\ (8y-4y^2-3)\sqrt{2y-y^2}+1=z\\ (4z^2-8z+3)\sqrt{2z-z^2}+1=x\end{array}\right.$$ $(\cdots)_{cyclic}$ means that the relation holds true for all individual $(x,y,z)$$$$$ Can this be solved using Trigonometric Substitution?$$$$ For example: How many unordered triplets $(x,y,z)$ are in the range $[0,2]$ and satisfy the system of equations: $$\left\{\begin{array}{l}2x^2-4x+2=y\\ 2y^2-4y+2=z\\ 2z^2-4z+2=x\end{array}\right.$$ This could be solved by setting $x'=x-1, y'=y-1, z'=z-1$, then substituting and rearranging the system of equations to get the first equation as $2x'^2-1=y'$ As this is the cosine double angle formula, we can substitute $x'=\cos\theta$ , $y'=\cos2\theta$ and proceed similarly. This would ultimately yield two cases which would yield the required solutions.

Answer 1

Let $(u,v,w)=(x-1,y-1,z-1)$. Then $|u|,|v|,|w|\leq 1$ and the equations become:

$${v=(4u^2-1)\sqrt{1-u^2}\\w=-(4v^2-1)\sqrt{1-v^2}\\u=(4w^2-1)\sqrt{1-w^2}}\tag{1}$$

Square both sides and let $(a,b,c)=(u^2,v^2,w^2)$ then you have:

$${0\leq a,b,c\leq 1\\b=(4a-1)^2(1-a)\\c=(4b-1)^2(1-b)\\a=(4c-1)^2(1-c)}\tag{2}$$

So if $f(x)=(4x-1)^2(1-x)$ you want to find $a\in[0,1]$ so that $f(f(f(a)))=a$. Then let $b=f(a)$ and $c=f(b)$.

Note that if we get a solution $(a,b,c)$, we still have to check eight possible solutions $(u,v,w)=(\pm\sqrt{a},\pm\sqrt{b},\pm\sqrt{c})$ for $(1)$.

$f'(x)=-48x^2+48-9 = 3-48\left(x-\frac{1}{2}\right)^2$, so $f'(x)=0$ when $x=\frac{1}{4}$ and $x=\frac{3}{4}$, and $f(\frac{1}{4})=0$ and $f(\frac{3}{4})=1$, and $f(0)=1$ and $f(1)=0$. So $f([0,1])=[0,1]$.

There are three values $a$ so that $f(a)=a$:

$$a=\frac{1}{2},\frac{2\pm\sqrt{2}}{4}$$

In these cases $(a,a,a)$ is a solution to $(2)$. This leads to three solutions to $(1)$:

$$\begin{align}(u,v,w)=&\left(\frac{\sqrt{2}}2,\frac{\sqrt{2}}2,-\frac{\sqrt{2}}2\right),\\ &\left(\frac{\sqrt{2+\sqrt{2}}}{2},\frac{\sqrt{2+\sqrt{2}}}{2},-\frac{\sqrt{2+\sqrt{2}}}{2}\right),\\ &\left(-\frac{\sqrt{2-\sqrt{2}}}{2},-\frac{\sqrt{2-\sqrt{2}}}{2},\frac{\sqrt{2-\sqrt{2}}}{2}\right) \end{align}$$

There are likely more - points such that $f(a)\neq a$ but $f(f(f(a)))=a$. There are at most $24$ more solutions, and the number of solutions will be a multiple of $3$.

So, the graph of $f$ on $[0,1]$ has $f(0)=1$, $f(1/4)=0$, $f(3/4)=1$, and $f(1)=0$, with monotonic in between - so three monotonic sections that are onto $(0,1)$.

This means that $f(f(x))$ has nine monotonic sections, and $f(f(f(x)))$ has $27$ monotonic sections onto $(0,1)$.

From this, we can conclude that there are $27$ values $a$ with $f(f(f(a)))=a$, one per monotonic interval.

From each solution $(a,f(a),f(f(a)))$, you get one solution $(u,v,w)$, and there is one $(u,v,w)$ per solution of the first equation $(x,y,z)$.

So there are exactly $27$ solutions.

Note, we didn't show how to find the solutions. We don't need to know this to know the number of solutions.

Your approach might work. It is because:

$$\sin3\theta = (4\cos^2\theta -1)\sin\theta\\\cos3\theta=-(4\sin^2\theta-1)\cos\theta$$

The $\sqrt{}$ becomes a problem, though, you get lots of $\pm$ in the equation.

Pick $\theta\in[0,\pi]$ so that $u=\cos\theta$. Then:

$$v=\sin(3\theta)\\ w=\pm\cos(9\theta)\\u=\pm \sin(27\theta)$$

So you need to solve $\cos\theta = \pm\sin(27\theta)$ where then $\pm$ must be the sign of $\sin(9\theta)$.