Distance in vector space Suppose $k≧3$, $x,y \in \mathbb{R}^k$, $|x-y|=d>0$, and $r>0$. Then prove
(i)If $2r > d$, there are infinitely many $z\in \mathbb{R}^k$ such that $|z-x| = |z-y| = r$
(ii)If $2r=d$, there is exactly one such $z$.
(iii)If $2r < d$, there is no such $z$
I have proved the existence of such $z$ for (i) and (ii). The problem is i don't know how to show that there are infinitely many and is exactly one such z resectively. Plus i can't derive a contradiction to show that there is no such z for (iii). Please give me some suggestions
 A: (i) If $r>{d\over2}$ then $\rho:=\sqrt{r^2-d^2/4}>0$. Let $m:={x+y\over 2}$ be the midpoint of $[x,y]$. The median hyperplane
$$H:=\{z\in{\mathbb R}^k\ |\ (z-m)\cdot(x-y)=0\}$$
has dimension $k-1\geq2$. Therefore the sphere with center $m$ and radius $\rho$ intersects $H$ in a sphere $S$ of dimension $k-2\geq1$; in particular $S$ has infinitely many points. For any point $z\in S$ one has $$|z-x|^2=|z-y|^2= \rho^2+{d^2\over4}=r^2$$
by Pythagoras' theorem.
(ii) If $|z-x|=|z-y|$ then
$$0=|z-y|^2-|z-x|^2=(z-y)\cdot(z-y)-(z-x)\cdot(z-x)=2(z-m)(x-y)\ ,$$
which implies $z\in H$. By Pythagoras' theorem we therefore have $$r^2:=|z-x|^2=|z-m|^2+|m-x|^2=|z-m|^2 +{d^2\over 4}\ .$$
If $r={d\over2}$ this is only possible if $z=m$.
Note that for (ii) we had to use the geometry of ${\mathbb R}^k$ somehow: Let $x$ and $y$ be antipodal points on the sphere $S^2$ provided with the spherical metric. Then there are infinitely many points $z\in S^2$ with $$d(x,z)=d(y,z)={\pi\over2}={1\over2}d(x,y)\ .$$
A: If there's a $z$ satisfying $|z-x|=r=|z-y|$ then by the triangle inequality, $d=|x-y|\le|z-x|+|z-y|=2r$
So if $d>2r$ there would've been no such $z$!
