Baby Rudin Chapter 1 Problem 16 I am attempting to self-study through Baby Rudin and I have done every exercise in chapter 1 except for problem 16. Every solution I find on the internet for this problem uses theorems from linear algebra which I have no background in. I don't want to go on to chapter 2 without completing chapter 1. Can anyone provide an explanation of this problem that does not assume prerequisite knowledge of linear algebra?
Suppose $k \geq 3$, $x, y \in R^k$, $|x-y| = d>0$, and $r>0$. Prove:
(a) If $2r>d$ there are infinitely many $z \in R^k$ 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 $z$.
My work so far:
(c) is a trivial consequence of the triangle inequality. In about 2 weeks I have been unable to make any progress on (a) or (b).
 A: I had trouble on this same problem when I was working through Baby Rudin. The easiest solution that doesn't use too complicated linear algebra is this:
Visualize that if $2r>d$ then we're dealing with two spheres that intersect at a circle, and therefore there are infinite many solutions. Now how to prove this?
Let $z = (x+y)/2 + h$ where $h$ is a vector orthogonal to $(x-y)/2$. Note this is a vector, going from the mid point between the spheres up a vector $h$ perpendicularly from the line between between $x$ and $y$ to the circle where they intersect. Using $|x|^2 = x \cdot x$ and expanding $(z-x) \cdot (z-x)=(z-y) \cdot (z-y)=r^2$ we see that this $z$ is a solution iff $h\cdot(y-x)/2 = 0$ and $|h|^2 = r^2 - (|y-x|/2)^2$ that is $|h|^2 = r^2 - d^2/4$. These two equations in dimension three or greater have infinitely many solutions iff $2r>d$, one solution iff $2r=d$, and no solutions iff $2r<d$.
A: It's late, but I'll try to answer it.
WLOG, suppose $k=3$ and $x=0$,then $|y|=d$. Let
$A=${$z\in R^3$$|$ $|z|=|z-y|=r$}, $A'=${$z'=(a_1,a_2,a_3)\in R^3| |z'|=\sqrt{r^2-\frac{d^2}{4}} and <z',y>=0$}
where $<,>$ is the standard inner product in $R^3$. First, we shall prove that
$A=A'+\frac{y}{2}$  (which consists of $z'+\frac{y}{2}$ , $z'\in A'$)
$z\in A \iff |z|=r$ and $|z-y|=r$
$\iff |z|=r$ and $z_1^2+z_2^2+z_3^2-2(z_1y_1+z_2y_2+z_3y_3)+y_1^2+y_2^2+y_3^2=r^2$
$\iff |z|=r$ and $z_1y_1+z_2y_2+z_3y_3=\frac{d^2}{2}$
$\iff |z-\frac{y}{2}|=\sqrt{r^2-\frac{d^2}{4}}$ and $<(z-y),\frac{y}{2}>=<z,y>-<y,\frac{y}{2}>=0$
$*$ ${|z-\frac{y}{2}|}^2=(z_1-\frac{y_1}{2})^2+(z_2-\frac{y_2}{2})^2+(z_3-\frac{y_3}{2})^2=z_1^2+z_2^2+z_3^2-(z_1y_1+z_2y_2+z_3y_3)+\frac{1}{4}(y_1^2+y_2^2+y_3^2)=r^2-\frac{d^2}{2}+\frac{d^2}{4}=r^2-\frac{d^2}{4}$
$\iff z=(z-\frac{y}{2})+\frac{y}{2}\in A'+\frac{y}{2}$
Now (a), if $2r>d$, since $r^2-\frac{d^2}{4}>0$, we can define above $A'$ and suppose $y_3$ is not $0$.Set
$$\alpha=(y_3,0,-y_1)$$$$\beta=(y_1y_2,-(y_1^2+y_3^2),y_2y_3)$$
then $\alpha, \beta$ are not $0$. And define
$$c(z)=\frac{\sqrt{r^2-\frac{d^2}{4}}}{|z|}$$
for $z≠0$, then we can easliy know that $c(\alpha)\alpha, c(\beta)\beta\in A'$ so that $A$ is not empty.For any $0≦t≦1$, define
$$w(t)=\alpha-t(\alpha-\beta)=(1-t)\alpha-t\beta$$
If $w(t)≠0$, $$|c(w(t))w(t)|=\sqrt{r^2-\frac{d^2}{4}}$$$$<c(w(t))w(t),y>=0$$, i.e., $c(w(t))w(t)\in A'$
If we can suppose that if $b_1\alpha+b_2\beta=0$ for some $b_1,b_2 \in R$, then $b_1=b_2=0$, then clealy $w(t)≠0$ and also for any $0≦t_1<t_2≦1$
$$c(w(t_1))w(t_1)≠c(w(t_2))w(t_2)$$
This implies that $A'$ is infinite and so $A$
(b)If $2r=d$, $A'=${$\frac{y}{2}$}, so, $A$ has exactly one element
(c)If $2r<d$, there is no $z'\in R^3$ such that $|z'|=\sqrt{r^2-\frac{d^2}{4}}$.$A$ is the empty set
For $k=1,2$, only (a) has the different result. When k=1, since if $<z',y>=0$ then $z'=0$, $A'$ has no element.and if k=2, $<z',y>=0$ implies that $z'=b(y_2,-y_1)$ for some $b\in R$. $|z'|=|b|d$.If $z'\in A'$, $$|b|=\frac{\sqrt{r^2-\frac{d^2}{4}}}{d}$$.
So $A'$ has only two elements and so is $A$
