Hello everybody,

I am stuck with a proof of which I think that I have an idea of how to solve it.

I have given two vectors $\displaystyle x,y \in R^k$ with $\displaystyle k >= 3$. I know that $\displaystyle |x-y| = d > 0$ and $\displaystyle r > 0$.

Now I am supposed to show a couple of things and what I am left to do is to prove that there is only one $\displaystyle z \in R^k$ s.t. $\displaystyle 2r = d$ and $\displaystyle |z-x| = |z-y| = r$.

OK, now here is what I have done. I know from the Schwarz inequality that $\displaystyle |x-z + z - y| = |x-z| + |z-y|$ is possible in general if I chose the right z. I get this same equality if I set the given definition of d into my other constraint. Also should I be able to solve for x since I have one unknown z and three requirements for z. However, whenever I solve for z I get some nonsence. I am not exactly sure about what I am doing wrong or how else I can show that there is only one z for that all these requirements hold. Can one of you help me. Even a hint would be more than enough. Thanks a ton! Cheers.