## Task 1 - 301241

Determine for each triplet $(a, b, c)$ of positive real numbers $a, b, c$ all those triplets $(x, y, z)$ of real numbers that satisfy the following system of equations (1), (2), (3):

$$
\begin{aligned}
& x^{2}-(y-z)^{2}=a \\
& y^{2}-(z-x)^{2}=b \\
& z^{2}-(x-y)^{2}=c
\end{aligned}
$$
From

$(x+y-z)^{2}(x-y+z)^{2}(-x+y+z)^{2}==((x+y-z)(x-y+z))((y+z-x)(y-z+x))((z+x-y)(z-x+y))=a b c$ one obtains first

$$
(x+y-z)(x-y+z)(-x+y+z)= \pm \sqrt{a b c}
$$

and by substituting in it

$$
a=(x+y-z)(x-y+z), b=(y+z-x)(y-z+x), c=(z+x-y)(z-x+y)
$$

the new system of equations

$$
\begin{aligned}
& \text { (4) }-x+y+z= \pm \sqrt{\frac{b c}{a}} \\
& \text { (5) } x-y+z= \pm \sqrt{\frac{a c}{b}} \\
& \text { (6) } x+y-z= \pm \sqrt{\frac{a b}{c}}
\end{aligned}
$$

Under the premise that the same sign is chosen on the right-hand side of (4), (5), and (6), which we now assume, one can immediately recover the original system of equations from this, i.e., it is equivalent to it.

The system of equations (4), (5), (6) is now linear and thus easily solvable, for example, by adding any two of the three equations. This yields the 2 solutions

$$
x= \pm \sqrt{\frac{a}{b c}} \frac{b+c}{2}, y= \pm \sqrt{\frac{b}{a c}} \frac{a+c}{2}, x= \pm \sqrt{\frac{c}{a b}} \frac{a+b}{2}
$$