Find the number of ordered triplets satisfying $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$

Question

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?

Answer 1

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 .

Answer 2

From the $xy+yz+zx=1$ we get $x = \frac{1-yz}{y+z}$. Substitute that into $5 (x+1/x)-12 (y+1/y)$. Taking the numerator of this and the numerator of $12 (y + 1/y) - 13 (z + 1/z)$, you can eliminate $y$ and get $z (z^4-1) = 0$. Since $z=0$ is not allowed, $z = \pm 1$. From that you can find the $y$ and $x$ values.

Answer 3

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 ? :)