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Let $x,y \in \mathbb{R}^k$ ($k\geq 3$), $|x-y|=d>0$ and $r>0$. Prove that if $2r>d$, then there are infinitely many $z\in \mathbb{R}^k$ such that $|z-x|=|z-y|=r$.

Here's what I have proved;

  1. The existence of such $z$, and
  2. $|z-x|=|z-y|=r$ iff $(z-(x+y)/2)\cdot(x-y)=0$ and $|z-(x+y)/2|=\sqrt{r^2 - d^2/4}$

I know exactly what's happening here and that there are infinitely many such $z$, but cannot show this logically. This prob is on 'analysis by rudin' so no topology please..


Edited: Only thing i need to prove here is that 'There exist infinitely many $d\in \mathbb{R}^k$ such that $(x-y)â€Ēd=0$ and $|d|=1$.'

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    @Brian Exactly and that was a typo thanks. – 2012-07-14

2 Answers 2

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I will assume that $d=|x-y|$ the distance between $x$ and $y$. First, take the hyperpan $\mathbf{P}$ defined by $\{z, |z-x| = |z-y|\}$. Now, let's consider $u=\displaystyle\frac{x+y}{2}$, $\mathbf{C} = \{z, |z-u| = r'\}$, with $r' = r-\frac{d}{2}$ ($r'>0$ by hypothesis) the hypersphere of center $u$ and of radius $r$.

All we have to prove now is that $\mathbf{S}=\mathbf{C}\cap\mathbf{P}$ is infinite. We have $dim \mathbf{P}=k-1\geq2$, so $\mathbf{P}$ has a base orthonormal $(v_1,\dots,v_{k-1})$. We now have $w_1= u+r' v_1$ and $w_2= u+r' v_2$ that belong to $\mathbf{S}$. Let's define $\mathbf{K}=\{\cos\theta w_1+\sin\theta w_2, \theta\in[0,2 \Pi]\}$. We can verify that any point $z$ in $\mathbf{K}$ verifies $|z-x|= |z-y| = r$: $z-x=\frac{x+y}{2}+\cos\theta w_1+\sin\theta w_2-x = \cos\theta w_1+\sin\theta w_2-x = (u+r' v_1)\cos\theta+(u+r' w_2)\sin\theta-x = \frac{y-x}{2}+r'(v_1 \cos\theta+v_2\sin\theta)$.

Now $y-x$ is orthogonal to $P$, and then to $v_1$ and $v_2$, so $|z-x| = \frac{|y-x|}{2} + |v_1\cos\theta + v_2\sin\theta| = \frac{d}{2}+r'=r$ and same to prove that $|z-y|=r$.

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    @Katlus yes, I made that mistake. I hope you are happy with the reasoning otherwise. – 2012-07-15
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Introduce $u=x-y$ and assume that $u\ne0$. The set of solutions of the equation $\langle u,d\rangle=0$ is the hyperplane $H_u$ of $\mathbb R^{k}$ which is orthogonal to $u$. The set of solutions of the equation $|d|=1$ is the unit sphere $S^{k-1}$ of $\mathbb R^{k}$. The intersection $H_u\cap S^{k-1}$ is the image of $S^{k-2}$ by any isometry from $\mathbb R^{k-1}$ to $H_u$. Thus, if $k=2$, $H_u\cap S^{k-1}$ is a pair and, for every $k\geqslant3$, $H_u\cap S^{k-1}$ is infinite because $S^{k-2}$ is infinite, or, for example, because $H_u\cap S^{k-1}$ contains a copy of the unit circle $S^{1}$.

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    Let us try to picture the $k=3$ case. Then $H_u$ is a plane (in the usual sense of the term) passing through the origin of $\mathbb R^3$ and $S^2$ is the unit sphere (in the usual sense of the term, say, the surface of the Earth). Thus, $H_u$ slices $S^2$ by the middle. Remember that only the *surface* of the Earth interests you, hence $H_u\cap S^2$ is indeed a circle (more precisely, an Equator), isometric to $S^1$. The same reasoning applies to every $k\geqslant3$. – 2012-07-14