Find the number of ordered triplets satisfying $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$ Find the number of ordered triplets $(x,y,z)$ of real numbers satisfying $$5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$$
and
$$xy+yz+zx=1$$
My try:
Letting:
$$\left(x+\frac{1}{x}\right)=\frac{k}{5}\tag{1}$$
$$\left(y+\frac{1}{y}\right)=\frac{k}{12}\tag{2}$$
$$\left(z+\frac{1}{z}\right)=\frac{k}{13}\tag{3}$$
By $AM-GM$
$$\left(z+\frac{1}{z}\right)^2 \ge 4$$
So $$k^2 \ge 676$$
Adding $(1),(2),(3)$ we get:
$$x+y+z+\frac{xy+yz+zx}{xyz}=\frac{281k}{780}$$
$$x+y+z+\frac{1}{xyz}=\frac{281k}{780}$$
Any clue from here?

Let $\left(x;y;z\right)\rightarrow \left(\tan \alpha ;\tan \beta ;\tan \gamma \right)\left(0<\alpha ,\beta ,\gamma <90^o\right)$
Then we have
\begin{align}
    \begin{cases}
        5\left(\tan \alpha +\frac{1}{\tan \alpha }\right)=12\left(\tan \beta +\frac{1}{\tan \beta }\right)=13\left(\tan \gamma +\frac{1}{\tan \gamma }\right) (1) \\
        \tan \alpha \tan \beta +\tan \gamma \tan \beta +\tan \alpha \tan \gamma =1 (2)
    \end{cases}
\end{align}
Note that: $$\left(1\right)\Leftrightarrow 5\left(\frac{\tan ^2\alpha +1}{\tan \alpha }\right)=12\left(\frac{\tan ^2\beta +1}{\tan \beta }\right)=13\left(\frac{\tan ^2\gamma +1}{\tan \gamma }\right)$$
$$\Leftrightarrow \frac{5}{\sin 2\alpha }=\frac{12}{\sin 2\beta }=\frac{13}{\sin 2\gamma }$$
And $\left(2\right)\Rightarrow \cot \gamma =\tan \left(\alpha +\beta \right)$
$$\Rightarrow \tan \left(\frac{\pi }{2}-\gamma \right)=\tan \left(\alpha +\beta \right)\Leftrightarrow \alpha +\beta +\gamma =\frac{\pi }{2}$$
Can you solve it now ? :)
