Solve system $x^2+y^2=\frac{2xy}{z}$; $y^2+z^2=\frac{2yz}{x}$; $x^2+z^2=\frac{2xz}{y}$

Question

A problem from the National Mathematical Olympiad in Bulgaria:

Solve the system:

$$\begin{aligned} x^2+y^2 &= \dfrac{2xy}{z} \\ y^2+z^2 &= \dfrac{2yz}{x} \\ x^2+z^2 &= \dfrac{2xz}{y} \end{aligned}$$

As always: $$\begin{array}{|l} x \ne 0 \\ y \ne 0 \\ z \ne 0 \end{array}$$ Now I am trying to see what I can do to simplify and solve the system. I tried to choose two of the equations and eliminate a variable, but I didn't succeed.

This is a problem from a national olympiad so it must have something tricky that I haven't seen. Can you give me a hint?

Answer 1

Given, $$ x^2+y^2 = \dfrac{2xy}{z} \tag 1$$ $$y^2+z^2 = \dfrac{2yz}{x} \tag 2$$ $$ x^2+z^2 = \dfrac{2xz}{y} \tag 3$$

Take (1)-(2) and (3)-(2) to get

$$(x^2-z^2)(xz-2y) = 0$$ $$(x^2-y^2)(xy-2z) = 0$$

So, there are four cases to examine from the factor combinations of the two equations above:

Case 1) $x^2-z^2=0$ and $x^2-y^2=0$. Plug $x^2=y^2$ into (1),

$$2x(xz-y)=0\implies xz=y\implies z=\pm1$$

which leads to the solutions $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$.

Case 2) $xz-2y=0$ and $xy-2z =0$, which leads to $x^2=4$. Plug it and $xy=2z$ into (1) to get

$$4+y^2 = 4\implies y=0$$

Since $y\ne0$, there are no solutions.

Case 3) $x^2-z^2=0$ and $xy-2z=0$. Then, $x^2=z^2=\frac14x^2y^2\implies y^2=4$. Plus it and $xy=2z$ into (1)

$$x^2+4 = 4\implies x=0$$

Since $x\ne 0$, there are no solutions.

Case 4) $x^2-y^2=0$ and $xz-2y=0$. There are no solutions based on similar argument in Case 3).

Thus, the valid solutions only come from Case 1), i.e. $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$.

Answer 2

Assume $x,y,z > 0$ . WLOG assume further that : $x \ge y \ge z$. Then : $\dfrac{1}{z} = \dfrac{x^2+y^2}{2xy} \ge 1 \implies z \le 1$. By the same argument, we have: $x \le 1, y \le 1$. Thus : $x^2 \le x , y^2 \le y, z^2 \le z$ and $\dfrac{2xy}{z}= x^2 + y^2 \le x+y \implies \dfrac{2}{z} \le \dfrac{x+y}{xy} = \dfrac{1}{x} + \dfrac{1}{y} $. But $ x \ge y \ge z \implies \dfrac{1}{z} \ge \dfrac{1}{y} \ge \dfrac{1}{x} \implies \dfrac{2}{z} \ge \dfrac{1}{x} + \dfrac{1}{y} \implies \dfrac{1}{z} = \dfrac{1}{x} = \dfrac{1}{y} \implies x = y= z = 1$.

Note: if say $x > 0, y < 0, z < 0$ we set $y’ = -y, z’ = -z$ then we are back to all positive solutions $(x,y’,z’)$. This covers the case OP mentioned in the comment below.