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What are the solutions of this system of equations where $x,y,z \in \mathbb{R}$?

$\begin{cases} 5(x+\frac{1}{x})=12(y+\frac{1}{y})=13(z+\frac{1}{z})\\ xy+yz+zx=1 \end{cases}$

First I rewrote the equations as

$\begin{cases} 5(\frac{x^2+1}{x})=12(\frac{y^2+1}{y})=13(\frac{x^2+1}{z})\\ xy+yz+zx=1 \end{cases}$

then by multiplying the first equation with $xyz$ I got

$\begin{cases} 5yz(x^2+1)=12xz(y^2+1)=13xy(z^2+1)\\ xy+yz+zx=1 \end{cases}$

Any idea on how to proceed is appreciated.

Birgitt
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  • Your second equation can be formulated as $\frac{1}{z} + \frac{1}{y} + \frac{1}{x} = \frac{1}{xyz}$. Can you use this? – Dominique Dec 04 '23 at 11:51
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    I'm intrigued by those values: $5$, $12$ and $13$. You do know that $5^2+12^2=13^2$? (But don't ask me how to integrate this in your equations :-) ) – Dominique Dec 04 '23 at 11:53
  • Indeed I can rewrite the second equation as $\frac{1}{z}+\frac{1}{x}+\frac{1}{y}=\frac{1}{xyz}$ – Birgitt Dec 04 '23 at 11:55

2 Answers2

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Hint:

If $x<0,$ so are $y,z$ if $(p,q,r)$ is a solution, so is $(-p,-q,-r)$

WLOG $x,y,z>0;$

let $x=\cot A$ etc. to use Proving $\cot(A)\cot(B)+\cot(B)\cot(C)+\cot(C)\cot(A)=1$ with $A+B+C=\pi$

Now $x+\dfrac1x=\dfrac2{\sin2A}$

If $5\cdot\dfrac2{\sin2A}=\cdots=k\implies\sin2A=\dfrac{10}k$

$$\implies\sin^22B=\sin^22C-\sin^22A=\sin2(C+A)\sin2(C-A)$$

$$\iff-\sin2B\sin2(C+A)=-2\sin2B\sin2(C-A)$$

either $\sin2B=0$ but $y+\dfrac1y>0$

or $0=\sin2(C+A)-\sin(C-A)=2\sin2B\cos2C\implies\cos2C=0\implies\sin2C=?$

Can you take it from here?

lab bhattacharjee
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  • I understand until $\sin 2A=\frac{10}{k}$ from there i do the same for $y+\frac{1}{y}$ and $z+\frac{1}{z}$ I don't understand where did you get $sin^2 2B=sin^22C-sin^22A=...$ – Birgitt Dec 04 '23 at 11:43
  • @Birgitt, I have used $$5^2+12^2=13^2$$ – lab bhattacharjee Dec 04 '23 at 11:49
  • Also if $\cos 2C=0$ then $\sin 2C=1-\cos2C=1-0=1$ from there we get that $\sin2C=\frac{13}{k}=1$ how to go from there? – Birgitt Dec 04 '23 at 12:07
  • @Birgitt, $5\left(x+\dfrac1x\right)=\cdots=k,$ right? – lab bhattacharjee Dec 04 '23 at 12:21
  • Ah, I got it thanks for the help after I get $k$ I can write $\begin{cases} 5(x+\frac{1}{x})=26\newline 12(y+\frac{1}{y})=26\newline 13(z+\frac{1}{z})=26 \end{cases}$ and I get the answer after i solve this equations. Another question that I have is from $\sin^22B=\sin^22C−\sin^22A=sin2(C+A)sin2(C−A)$ how did we get $−sin2Bsin2(C+A)=−2sin2Bsin2(C−A)$. Thank you again for the help and sorry for asking so many questions. – Birgitt Dec 04 '23 at 12:51
  • @Birgitt, remember $$\sin2A$$ can be $-1$ as well – lab bhattacharjee Dec 04 '23 at 13:17
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Note that $x,y,z$ are of the same sign and that if $(a,b,c)$ is a solution, so is $(-a,-b,-c)$. Hence WLOG we may assume that $x=\tan \dfrac{A}{2}, y = \tan \dfrac{B}{2}, z = \tan \dfrac{C}{2}$ where $A,B,C$ are angles of a triangle

Given condition translates to $\dfrac{5}{\sin A} = \dfrac{12}{\sin B} = \dfrac{13}{\sin C} $ from which $z=1$ is immediate

Hari Shankar
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