Let $a,b,c,x,y,z$ be real numbers such that $$ a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2 $$ Show that $a^2+b^2+c^2=x^2+y^2+z^2$.
Progress: Note that $$0=\sum_{\text{cyc}} (a + b)^2 + (x + y)^2 - \sum_{\text{cyc}} a^2 + x^2 = (a + b + c)^2 + (x + y + z)^2\implies a+b+c=x+y+z=0.$$
So we need to show $$ab+bc+ca=xy+yz+zx.$$ Also, taking $c=-(a+b),z=-(x+y)$, we get $$a^2+x^2=b^2+y^2=(a+b)^2+(x+y)^2.$$
So we get $$y^2+b^2+2ab+2yx=0\implies \sum_{cyc}y^2+b^2+2ab+2yx=0\implies (a+b)^2+(b+c)^2+(c+a)^2+(x+y)^2+(y+x)^2+(x+z)^2=0\implies a=b=c=x=y=z=0. $$
I am not sure of the last part. So can someone check? Can someone also give an alternate solution to this problem?
Writing this using complex numbers makes it a little easier. The equations mean that these following complex numbers have the same distance from the origin of the complex plane.
$$a+ix=re^{i\theta_1}\mbox{ with } a=r\cos(\theta_1),x=r\sin(\theta_1).$$
$$b+iy=re^{i \theta_2}\mbox{ with } b=r\cos(\theta_2),y=r\sin(\theta_2).$$
$$c+iz=re^{i \theta_3}\mbox{ with } c=r\cos(\theta_3),z=r\sin(\theta_3).$$
Now, $$r=(a+b)^2+(x+y)^2=|(a+ix)+(b+iy)|=r|(e^{i\theta_1}+e^{i\theta_2})|$$
$$\implies |e^{i\theta_1}+e^{i\theta_2}|=1$$
For this to happen, $\theta_1=\frac{2\pi}{3}+\theta_2$. Similarly, $\theta_2=\frac{2\pi}{3}+\theta_3$ and $\theta_3=\frac{2\pi}{3}+\theta_1$. The three points lie on the vertices of an equilateral triangle.
$$a^2+b^2+c^2=r^2(\cos^2(\theta_1)+\cos^2(\frac{2\pi}{3}+\theta_1)+\cos^2(\frac{2\pi}{3}-\theta_1))=\frac {3r^2}{2}\tag 1$$
$$x^2+y^2+z^2=r^2(\sin^2(\theta_1)+\sin^2(\frac{2\pi}{3}-\theta_1)+\sin^2(\frac{2\pi}{3}+\theta_1))=\frac{3r^2}{2}\tag 2$$
which completes our proof.
I have used $re^{i\theta}=r(\cos(\theta)+i\sin(\theta))$ . The proof that $\sin^2(\theta_1)+\sin^2(\frac{2\pi}{3}-\theta_1)+\sin^2(\frac{2\pi}{3}+\theta_1)=\cos^2(\theta_1)+\cos^2(\frac{2\pi}{3}+\theta_1)+\cos^2(\frac{2\pi}{3}-\theta_1)=3/2$ can be found online.