<s>[INST] 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?
 [/INST] For every triplet of real numbers $(x,y,z)$, which satisfies $$xy+yz+zx=1,$$ then there exists an unique triplet of angles $(\alpha, \beta, \gamma)$, such that $0 \le \alpha, \beta, \gamma < \pi $ and $\alpha + \beta + \gamma= \pi $, i.e. they are angles of a triangle, and $$tg(\frac{\alpha}{2}) = x, \text{ } tg(\frac{\beta}{2}) = y, \text{ }tg(\frac{\gamma}{2}) = z$$
Indeed, you can check that in every triangle with angles $\alpha, \beta, \gamma$ you have $tg(\frac{\alpha}{2})tg(\frac{\beta}{2}) + tg(\frac{\beta}{2})tg(\frac{\gamma}{2}) + tg(\frac{\alpha}{2})tg(\frac{\gamma}{2}) =1  $
We also know $$sin(\alpha)=\frac{2*tg(\frac{\alpha}{2})}{tg(\frac{\alpha}{2})^2+1}= \frac{2x}{x^2+1}$$
The other condition, written such at it helps our intuition, now becomes: 
$$5\left(\frac{x^2+1}{2x}\right)=12\left(\frac{y^2+1}{2y}\right)=13\left(\frac{z^2+1}{2z}\right)$$ which is just
 $$ \frac{sin (\alpha)}{5}= \frac{sin(\beta)}{12} = \frac{sin(\gamma)}{13} =k_0 $$
From Sine law, we know lengths of sides in a triangle are directly proportional to the sines of the opposing angles, therefore if we denote $a,b,c$ the sides, we have
$$ \frac{a}{5}= \frac{b}{12} = \frac{c}{13} =k $$so we obtain 
$$ a=5k, b=12k, c= 13k $$
Using cosines law, we get $$ cos(\alpha)=\frac{b^2+c^2-a^2}{2bc} $$ and the analogous formulae for $\beta $ and $\gamma$ give us 
$$ cos(\alpha) = \frac{12}{13}, cos(\beta) = \frac{5}{13}, cos(\gamma)= 0 $$
which means that $$x= tg(\frac{\alpha}{2}) = \sqrt{\frac{1-cos(\alpha)}{1+cos(\alpha)}} = \frac{1}{5}$$
$$y= tg(\frac{\beta}{2}) = \sqrt{\frac{1-cos(\beta)}{1+cos(\beta)}} = \frac{2}{3}$$
$$z= tg(\frac{\gamma}{2}) = \sqrt{\frac{1-cos(\gamma)}{1+cos(\gamma)}} = 1$$
We have chosen the positive root from $tg^2(\frac{\alpha}{2}) = \frac{1-cos(\alpha)}{1+cos(\alpha)}$, and the analagous, because $0 \le \alpha, \beta, \gamma < \pi $, therefore $0 <\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2} < \frac{\pi}{2} $, which means that $tg(\frac{\alpha}{2}), tg(\frac{\beta}{2}), tg( \frac{\gamma}{2}) $ are positive . 
</s>