Baby Rudin Chapter 1 Problem 16

Question

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).

Answer 1

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 $2r<d$.

Answer 2

I also want to try to answer this question, even though it may be late. My style is that I first give the intuition and then the calculations. Thus, try to imagine whatever I say in $\mathbb{R}^3$.

Assume two Identical spheres centered respectively around points x and y. Imagine I can change some parameter for the radius length (r), and both spheres would get bigger or smaller accordingly.

At first, if the radius is small enough, the two spheres would have no intersection. As we make r larger, they will eventually intersect at just one point (the middle point between x and y). Still, if we make the radius even bigger, we would have a circle of intersection points. Focus on this circle! What is its center and radius length?

Because all the points in the circle have to have an equal distance from both x and y, their average (which is the center) is the midpoint between x and y. So, the center is located at $\frac{x + y}{2}$.

To find the radius of the intersection circle (which we would call $\lvert w \rvert$), first, we have to examine its relationship with the connecting line between x and y. Notice that the radius $w$ is perpendicular to the connecting line, so we can use the Pythagorean theorem to find the radius length. Since any point on the intersection circle is, by definition, on both spheres, it has a distance of r from the point x. So, the length of the hypotenuse of the right triangle is r.

What is the distance between x and the center of the intersection circle? it's $\lvert x - \frac{x + y}{2} \rvert = \frac{\lvert x-y\rvert}{2} = \frac{d}{2}$. This is the side of the right triangle. Finally, by the Pythagorean Theorem, we have: $$\lvert w\rvert^2 = r^2 - (\frac{d}{2})^2 \Rightarrow \lvert w\rvert^2 = r^2 - \frac{d^2}{4}$$

In our quest to find the $\lvert w \rvert$, we have used one constraint for the $w$ that we did not write in the mathematical notation. It was the fact that the $w$ is perpendicular to the connecting line between x and y, so their inner product is zero. We shall now write this in mathematical terms: $$\left\langle w, \; x - y\right\rangle = 0 \Rightarrow w.(x-y) = 0$$

These two equations are stated at the beginning of the solution from the manual. So now:

1. If $3 < k$, in $\mathbb{R}^k$ we have:

• if $\lvert w \rvert$ is positive, the two equations have infinitely many solutions, since in 3 dimensions and above there are infinitely many vectors of a given "positive" length that are perpendicular to a specific vector. So: $$ 0 < \lvert w \rvert \Rightarrow 0 < \lvert w \rvert^2 \Rightarrow 0 < r^2 - \frac{d^2}{4} \Rightarrow \frac{d^2}{4} < r^2 \Rightarrow d < 2r$$

• If $\lvert w \rvert = 0$, then only one vector satisfies this equation, the zero vector. so we have one solution if: $$ 0 = \lvert w \rvert \Rightarrow 0 = \lvert w \rvert^2 \Rightarrow 0 = r^2 - \frac{d^2}{4} \Rightarrow \frac{d^2}{4} = r^2 \Rightarrow d = 2r$$

• And with a similar argument, if d > 2r we have no solutions.

2. If $k=2$, there are two vectors of positive length that are perpendicular to a specific vector. Other cases stay the same.

3. If $k=1$, there is no vector of positive length that is perpendicular to a specific vector. Other cases stay the same.

Answer 3

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 $A$