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Suppose that $n \geq 3$, $x$, $y \in \mathbf{R}^n$, $d \colon= |x-y| > 0$, and $r>0$. Then how to prove the following assertions:

(a) If $2r>d$, there are infinitely many $z \in \mathbf{R}^n$ 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 $z$.

How to modify these statements when $n$ is $2$ or $1$?

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draw a picture and interpret absolute value of difference as distance of points. – Maesumi Feb 10 '13 at 19:02

Without loss of generality, you can suppose that $x,y$ lie on the first coordinate axis, evenly spaced from the origin--in particular, suppose $x=\frac{d}2e_1$ and $y=-\frac{d}2e_1$. Then a vector $z\in\Bbb R^n$ is the same distance from $x$ as from $y$--that is $|z-x|=|z-y|$--if and only if its first component is $0$. The set of all such points form a hyperplane of dimension $n-1$ in $\Bbb R^n$.

You can determine that for $z$ in this hyperplane, we have $$|z-x|^2=|z-0|^2+|0-x|^2=|z|^2+\frac{d^2}4,$$ and similarly $$|z-y|^2=|z|^2+\frac{d^2}4.$$ Our choice of $x,y$ should make this simpler, and this result should make proving (b) and (c) fairly trivial.

Now, given any point $z_0$ with $|z_0-x|=|z_0-y|=r$, and any rotation of $\Bbb R^n$ about the first coordinate axis (say $\rho$), we have $\rho(x)=x,\rho(y)=y$, and since rotations perserve distances between points, then $$|\rho(z_0)-x|=|\rho(z_0)-\rho(x)|=|z_0-x|=r,$$ and likewise $$|\rho(z_0)-y|=r$$ so $$|\rho(z_0)-x|=|\rho(z_0)-y|=r.$$ If $z_0$ can be chosen so that $\rho(z_0)\neq z_0$ (sufficient that $z_0\neq 0$ and $\rho$ a non-trivial rotation), then since there are infinitely-many non-trivial rotations of $\Bbb R^n$ about the first coordinate axis, it will follow that there are infinitely-many points $z$ such that $|z-x|=|z-y|=r$. In justifying that these things can be done, you'll prove (a).

In the $n=2$ case, (b) and (c) stay the same, and (a) changes to "two $z\in\Bbb R^n$", rather than "infinitely-many $z\in\Bbb R^n$. This is because we're describing the intersection sets of two circles in a plane, centered at $x,y$.

In the $n=1$ case, (b) and (c) stay the same, and (a) changes to "no $z\in\Bbb R^n$". This is because the one-dimensional analogue to a circle of radius $r$ about a point $x$ is simply the set of two points $\{x-r,x+r\}$.

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