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I'm trying to solve the following problem in Baby Rudin :

Suppose $k\ge 3$, $x,y \in \Bbb{R}^k,|x-y|=d>0$, and $r>0$ Prove:

(a) If $2r>d$, there are infinity many $z$ such that: $$ |z-x|=|z-y|=r $$

I solved the problem through constructing a unit vector perpendicular to $(x-y)$ and showing there are infinity many of these under the constraint. I'm trying to solve the problem by another way , I got a hint that it can be solved through constructing a system of equations which is under-determined here's my attempt (the problem I ran into is stated at the end):

$|z-x|^2=(z-x)(z-x)=|z|^2 -2xz+|x|^2$ similarly, $|z-y|=|z|^2-2yz-|y|^2$

Then $ |z-x|=|z-y|=r $ when $z(x-y)=\frac 1 2(|x|^2-|y|^2)$

Also since we want $ |z-x|=r$ then $|z-x|=|rw|$ where $w \in \Bbb{R}^k$ and $|w|=1$, this gives $z=x+rw$ combining this with $|y-z|^2=r^2$ gives $r^2=|x+rw-y|^2=d^2+2rw(x-y)+r^2\implies w (y-x)=\displaystyle\frac {d^2} {2r}=\frac{d}{2r}|x-y| $ then since the left hand side is a dot product we get $$w_1(x_1+y_1)+w_2(x_2+y_2)+...+w_k(x_k+y_k)=\frac{d}{2r}|x-y| $$

and since $w$ is a unit vector we have $\sqrt{w_1^2+w_2^2+...w_k^2}=1$

Hence we have a system of two equations and $k$ unknowns and $k\ge3$ then the system has infinity many solutions since the number of $w$'s we get is the same as the number of $z$'s.

What I'm not sure about is that the second equation is non-linear, does that matter?

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Be careful with your notations. The inner product is usually denoted by $(x,z)$ of $\langle x,z\rangle$. Also, $w$ unit vector means $w_1^2+\ldots+w_k^2=1$ and not what you said. –  1015 Feb 16 '13 at 16:25
    
I forgot the exponents , I'll edit them in –  user10444 Feb 16 '13 at 16:30
    
You made a mistake: $|z-y|^2=|z|^2-2(y,z)+|y|^2$. So actually $2(x-y,z)=|x|^2-|y|^2$. –  1015 Feb 16 '13 at 16:35
    
Regarding intuition, you are looking for the intersection of two spheres (centered at $x$ and $y$, both with radius $r$). If $r$ is too small ($2r<d$), this is empty. If $r$ is large enough ($2r>d$), this is an infinite set. And if $2r=d$, the is a singleton. –  1015 Feb 16 '13 at 16:37
    
I wrote this on a piece of paper so I have some typos, feel free to edit them. The calculations are right though –  user10444 Feb 16 '13 at 16:39

1 Answer 1

up vote 1 down vote accepted

There is no real way around your initial solution. Your second strategy essentially boils down to the same approach.

The problem is invariant by affine transformation. So without loss of generality, we can assume $y=0$, $|x|=d>0$ and $r=1$.

The condition $|z-x|=|z|=r$ is equivalent to $|z-x|^2=|z|^2$ and $|z|=1$.

Hence your are looking for all $z$ such that $$ (z,x)=\frac{d^2}{2}\quad\mbox{and}\quad |z|=1. $$ ie $$ x_1z_1+\ldots+x_kz_k=\frac{d^2}{2}\quad\mbox{and}\quad z_1^2+\ldots+z_k^2=1. $$

Now $x/2$ is a solution of the lhs linear equation, so setting $z=x/2+h$, this amounts to solving $$ x_1h_1+\ldots+x_kh_k=0\quad\mbox{and}\quad |x/2+h|^2=1. $$ Observe that the lhs means that $x$ and $h$ are orthogonal, so $|x/2+h|^2=|x/2|^2+|h|^2=d^2/4$ so this is now equivalent to $$ x_1h_1+\ldots+x_kh_k=0\quad\mbox{and}\quad |h|^2=1-\frac{d^2}{4}. $$ If $d>2$, there are clearly no solutions. If $d=2$, there is exactly one solution, $h=0$.

So assume $d<2$. The solutions of the lhs constitute a hyperplane, ie a $k-1$ dimensional subspace of $\mathbb{R}^k$. So the solution set is a sphere in subspace of dimension $k-1\geq 2$. This is infinite.

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Thank you for supplying another method, but that is not what I'm asking for. I want to know if the method I used is right, or if not how can it be done using the hint. –  user10444 Feb 16 '13 at 17:32
    
@user10444 What I did is I essentially finished your second method. This is not another method. Yes, your second method is right. –  1015 Feb 16 '13 at 17:34
    
does that mean that such a system where one of the equations is linear has infinitely many solutions? Also what is missing beyond finding 2 equations? –  user10444 Feb 16 '13 at 17:39
    
@user10444 Not really. What this means is that the linear equation in the lhs yields a hyperplane, ie a subspace of dimension $k-1$. So together with the second equation, the set of two equations means: a Euclidean sphere in a $k-1$ dimensional space. Since $k-1\geq 2$, this is an infinite set. It is this observation which is missing beyound finding the 2 equations. –  1015 Feb 16 '13 at 17:41
    
Makes a bit more sense now, thank you –  user10444 Feb 16 '13 at 17:44

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