0
$\begingroup$

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 $(x-y)•d=0$ and $|d|=1$.'

  • 0
    By $|z-x|$, do you mean $||z-x||$, the norm of $z-x$ ?2012-07-14
  • 0
    $d$ should be somehow related to $x$ and $y$. Is $d=|x-y|$, i.e., the distance of $x$ and $y$?2012-07-14
  • 0
    @S4M Aren't those same thing? Since there is no definition of distance in my book so far, i guess it's norm..2012-07-14
  • 0
    @martin oh right sorry for that2012-07-14
  • 0
    I've edited your question, to reproduce Rudin's problem.2012-07-14
  • 0
    You are intersecting two spheres: if the sum of the radii is larger than the distance between the two centers, then the intersection is nonempty. Remark that your problem is rotationally invariant: if $v$ belongs to the intersection, then you can rotate $v$ around the line that joins the centers.2012-07-14
  • 1
    @Siminore you are right, but what i want here is to show that logically, not intuitively2012-07-14
  • 0
    There is nothing intuitive, in rotations. You have two sphere and one intersection point. You draw the hyperplane orthogonal to the line connecting the centers, and passing through your intersection point. Then any point that belongs to both the hyperplane and to one of the spheres belongs also to the second sphere. This is just a property of the euclidean norm.2012-07-14
  • 0
    $\sqrt{r^2-d^2/2}$ is wrong: it should be $\sqrt{r^2-(d/2)^2}=\frac12\sqrt{4r^2-d^2}$. Also, by translating the midpoint to the origin you can reduce it to showing that if $x\ne 0$, there are infinitely many $z$ such that $x\cdot z=0$ and $\|z\|=1$.2012-07-14
  • 1
    @Siminore: The argument via rotations is intuitive rather than rigorous if one doesn’t know how to express it formally; I suspect that this is precisely Katlus’s difficulty.2012-07-14
  • 0
    @siminore Since i don't know that process i called it 'intuitive'. Would you please write that as an answer?2012-07-14
  • 0
    @Brian Exactly and that was a typo thanks.2012-07-14

2 Answers 2

1

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 $|z-y|=r$.

  • 0
    I think $r' = \sqrt{r^2 - d^2/4}$.2012-07-15
  • 0
    @Katlus yes, I made that mistake. I hope you are happy with the reasoning otherwise.2012-07-15
0

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 $S^{1}$.

  • 0
    Thank you, but i started to learn analysis a week ago which means i don't completely understand your proof.. Do you have any easier one?2012-07-14
  • 0
    What is the first step you do not completely understand?2012-07-14
  • 0
    I don't understand why the intersection is the image of $S^{k-2}$.2012-07-14
  • 0
    I have learned perspective geometry before, but the everything taught in the class was intuitive..2012-07-14
  • 0
    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