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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

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    @christian I knwo exactly what happens here, but i can't not show the uniqueness of $z$ in (ii) and that there are infinitely many z's satisfying (i) logically .. Help2012-07-13

2 Answers 2

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(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 $d(x,z)=d(y,z)={\pi\over2}={1\over2}d(x,y)\ .$

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    Thanks. I showed 'when $2r≧d$, $|z-x|=|z-y|=r$ iff $|z-(x+y)/2|=\sqrt{r^2 - d^2/2}$ and $(z-(x+y)/2)•(x-y)=0$'. This asserts the uniqueness of $z$ in (ii).2012-07-14
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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 $z$!

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    The triangle inequality comes with a clause that specifies the conditions under which equality holds.2012-07-13