Real Numbers and Proportional Relationships: Solving for x, y, and z | POTW #333

[anemone]

I would like to say a big thank you to greg1313, who stood in for me during the last month to take care of the POTW
duty while I was taking a break.

Here is this week's POTW:

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Suppose $a,\,b$ and $c$ are real numbers such that $abc \ne 0$.

Find $x,\,y$ and $z$ in terms of $a,\,b$ and $c$ such that

$a=bz+cy\\b=cx+az\\c=ay+bx$

Prove also that $\dfrac{1 - x^2}{a^2} = \dfrac{1 - y^2}{b^2} = \dfrac{1 - z^2}{c^2}$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Opalg
2. castor28
3. kaliprasad

Solution from castor28:

Spoiler

We notice first that the system is invariant under the permutation $(a,b,c)(x,y,z)$.
Multiply the three equations respectively by $(-a)$, $b$, and $c$, and add the results. This gives:
$$-a^2 + b^2 + c^2 = 2bcx$$
and, using the symmetry of the system and the fact that $abc\ne0$:
\begin{align*}
x &=\frac{-a^2 + b^2 + c^2}{2bc}\\
y &= \frac{a^2 - b^2 + c^2}{2ac}\\
z &= \frac{a^2 + b^2 - c^2}{2ab}
\end{align*}
We compute:
\begin{align*}
\frac{1-x^2}{a^2} &= \frac{4b^2c^2-a^4-b^4-c^4-2b^2c^2+2a^2b^2+2a^2b^2}{4a^2b^2c^2}\\
&=\frac{-a^4-b^4-c^4+2b^2c^2+2a^2b^2+2a^2b^2}{4a^2b^2c^2}
\end{align*}
The RHS is symmetric with respect to $a,b,c$. Therefore, the permutation $(a,b,c)(x,y,z)$ gives:
$$
\frac{1 - x^2}{a^2} = \frac{1 - y^2}{b^2} = \frac{1 - z^2}{c^2}
$$