# Proving a statement related to norms in euclidean spaces

Taken from Rudin's Principles of Mathematical Analysis, 3rd edition (Chapter 1, Exercise 16):

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

(a) If $2r > d$, there are infinitely many $z\in\mathbb{R}^k$ such that $|z-x|=|z-y|=r$

(b) If $2r=d$ there is exactly one such $z$.

(c) If $2r < d$, there is no such $z$.

Geometrically (at least in $\mathbb{R}^3$), if $2r>d$ then there exists a circumference of points $z$ such that $|z-x|=|z-y|=r$. I guess that for $\mathbb{R}^k$ the set of such points forms a $(k-2)$-sphere.

More formally, let $m$ be the midpoint between $x$ and $y$: $m=\frac{x+y}{2}$. Then we define the plane $\Pi$ as the set of $w\in\mathbb{R}^k$ that satisfy $(\lambda(x-y))w = (\lambda'(x-y))m$ for some $\lambda, \lambda'\in\mathbb{R}$. Let $S = \{p\in\Pi : |p-x|=r\}$. Because of its geometric interpretation it's obvious that $|p-x|=|p-y|$ for every $p\in S$, but I can't figure out how to prove it formally. $S$ can also be defined as the intersection of $\Pi$ and a $(k-1)$-sphere centered at $m$ with radius $\sqrt{r^2-(d/2)^2}$.

(b) is obvious since $m$ is the only possible point that fulfills what's asked.

(c) follows from the fact that, if we set $S$ as the $(k-1)$-sphere centered at $m$ with radius $\sqrt{r^2-(d/2)^2}$, its radius is non-existent, therefore no such sphere exists.

Is this approach correct? How can I prove (a)? Thanks in advance. (c)

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It would be easier to define Pi as the hyperplane passing through m and perpendicular to the vector from m to x. Then you can use the Pythagorean theorem. –  Qiaochu Yuan Dec 21 '10 at 22:21
@Qiaochu: absolutely right. I defined $\Pi$ as a plane because I was thinking of $\mathbb{R}^3$, but it should definitely be an hyperplane. In fact I use the Pythagorean theorem to get the sphere radius. –  F M Dec 21 '10 at 22:46
Some small comments: in the statement of (a) you may want to fix the math to show $z\in\mathbb{R}^k$. In the paragraph about geometric intuition for (a), the set you get is not a $(k-1)$ sphere, but a $(k-2)$ sphere. –  Willie Wong Dec 21 '10 at 23:04
@Willie: Thanks, fixed! –  F M Dec 21 '10 at 23:12
Concerning part a): you can be explicit using vectors.. Let $v$ be any unit vector perpendicular to $y - x$. Then show that $z = m + tv$ satisfies $|z - x| = |z - y| = r$ for the right choice of $t$, determinable via the Pythagorean theorem (which you basically have already done). –  Zarrax Dec 22 '10 at 2:08