<s>[INST] 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?
 [/INST] You are very close. Both your approaches point to the fact that $r$, $s$, $t$ form an equilateral triangle (three points on a circle, with center of mass at the origin). Then, if you use the exponential notation, the points are at $Re^{i\phi}$, $Re^{i\phi+2\pi/3}$, $Re^{i\phi+4\pi/3}$. Then it's easy to show that $$\sin^2\phi+\sin^2\left(\phi+\frac{2\pi}3\right)+\sin^2\left(\phi+\frac{4\pi}3\right)=\cos^2\phi+\cos^2\left(\phi+\frac{2\pi}3\right)+\cos^2\left(\phi+\frac{4\pi}3\right)=\frac32$$
Multiply with $R$ and you get your desired result.
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