Prove $a^2+b^2+c^2=x^2+y^2+z^2$ given that $a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$ Prove $$a^2+b^2+c^2=x^2+y^2+z^2$$ given that $$a^2+x^2=b^2+y^2=c^2+z^2=\\(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$$ where $a,b,c,x,y,z \in\mathbb R$
A friend forwarded me this problem which recently appeared in a math contest at Costa Rica. I've been working on it but I can't find a definite proof, though I have discovered several facts:

 Let $u$ be equal to the value of the equation, then $ab+xy=\frac{-u}{2}$ and $a^2+b^2+c^2+x^2+y^2+z^2=3u$ which implies that $(a+b+c)^2+(x+y+z)^2 = 0 \implies a+b+c=x+y+z=0$.


 I also noticed that, because of the equalities, working with complex numbers or vectors would be really tempting, as if $r=a+ix$, $s=b+iy$, $t=c+iz$, then we have $|r|=|s|=|t|=|r+s|=|s+t|=|r+t|$, but I can't find a way to make it useful.

Any ideas/hints/solutions?

Your second observation is astute! Simply for interest, I would make one further observation, which is that the question geometrically asks us to prove

Given three $2D$ vectors (equivalently, complex numbers $r,s,t$) with given equality of lengths, prove the two $3D$ vectors $$\mathbf{u}=(a \quad b \quad c) , \qquad \mathbf{v}=(x \quad y \quad z)$$ are also of equal length (see the paper linked below for the condition that three vectors in the plane are images under orthogonal projection of an orthornormal basis in $\mathbb{R}^3$).

As others have pointed out, since $$|r|=|s|=|t|=|r+s|=|s+t|=|t+r|,$$ then, by the parallelogram law of addition, $r,s,t$ form an equilateral triangle, with centre of mass on the origin, in $\mathbb{C}=\mathbb{R}^2$. We know the cube roots of unity $$1,\quad \omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}, \quad \omega^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$$ form an equilateral triangle in $\mathbb{C}$. We know also that there exists a complex number $\alpha$ such that the transformation $$z\mapsto \alpha z$$ maps $\triangle rst$ onto $\triangle 1\omega \omega^2$. Under this transformation, the lengths of $r,s,t$ are all scaled by some real number $R$. Since, as you can check for yourself, the result holds in this special case, and since rotation is an isometry (distance preserving map), we see that $$R|\mathbf{u}|=R|\mathbf{v}|$$ and so the result holds in general. $\square$
Alternatively, for a more purely complex algebraic proof note that, taking $r,s,t$ as points on a circle centred at the origin, $$\tag{1}\triangle rst\space \mathrm{is \space equilateral} \iff \begin{cases} r+s+t=0\\r^2+s^2+t^2=rs+st+tr\end{cases}$$ Equating the real parts of each equation in $(1)$ gives $$\tag{2}a+b+c=0\implies ab+bc+ca=-\frac{1}{2}(a^2+b^2+c^2)$$ and $$a^2+b^2+c^2-(x^2+y^2+z^2)=ab+bc+ca-(xy+yz+zx)\tag{3}.$$ Since $(2)$ also holds for imaginary parts, we conclude $$a^2+b^2+c^2=x^2+y^2+z^2\tag{4}.\quad \square$$ This path to the result is perhaps the cleanest, but, as presented, it offers less clue as to why the symmetry occurs. The proofs of the equations in $(1)$, however, recover some of this why and point to a wider theory of orthogonal projections. Both equations can also in fact be deduced from rotations of the plane, which is a good exercise.
