Exercise 1.16

Question

Exercise 16: Suppose $k\ge 3$, $\mathbf{x,y}\in\mathbb R^k$, $|\mathbf{x}-\mathbf{y}|=d>0$, and $r>0$. Prove:
    \begin{itemize}
        \item[(a)] If $2r>d$, there are infinitely many $\mathbf{z}\in\mathbb R^k$ such that $$|\mathbf{z}-\mathbf{x}|=|\mathbf{z}-\mathbf{y}|=r.$$
        \item[(b)] If $2r=d$, there is exactly one such $\mathbf{z}$.
        \item[(c)] If $2r<d$, there is no such $\mathbf{z}$.
    \end{itemize}
How must these statements be modified if $k$ is 2 or 1?

analambanomenos · Accepted Answer

\def\ints{\mathbb{Z}} (a): To simplify the calculations, we may assume without losing generality that $\mathbf{x}=(0,\ldots,0)$ and $\mathbf{y}=(d,0,\ldots,0)$. (Translate $\mathbb R^k$ by $-\mathbf x$, then rotate it to place $\mathbf y$ on the positive x-axis. Both transformations preserve distances and angles.) For $0\le heta<1$ let $$\mathbf{z}_ heta=(d/2,\cos heta\sqrt{4r^2-d^2}/4,\sin heta\sqrt{4r^2-d^2}/2,0,\ldots,0).$$ Then $$|\mathbf{z}_ heta-\mathbf{x}|^2=|\mathbf{z}_ heta-\mathbf{y}|^2= \frac{d^2}{4}+(\cos^2 heta+\sin^2 heta)\frac{4r^2-d^2}{4}=r^2.$$ For $k=2$, the hyperplane of equidistant points is the line $(d/2,x_2)$, so there are two points with distance $r>d/2$ from $\mathbf x$ and $\mathbf y$, $(d/2,\pm\sqrt{4r^2-d^2}/2)$. For $k=1$, the hyperplane reduces to the single point $z=d/2$, so there are no points with distance $r>d/2$ from $\mathbf x$ and $\mathbf y$. (b): If $d=2r$, then $|\mathbf{z}-\mathbf{x}|+|\mathbf{z}-\mathbf{x}|=|\mathbf{x}-\mathbf{y}|$, that is, we have equality in Theorem 1.37(f). Looking at the proof, this only happens if $(\mathbf{z}-\mathbf{x})\cdot(\mathbf{z}-\mathbf{y})=|\mathbf{z}-\mathbf{x}| |\mathbf{z}-\mathbf{y}|$. By the solution to exercise 15, this only happens if $\mathbf{z}-\mathbf{x}=t(\mathbf{z}-\mathbf{y})$ for some real number $t$. Since $|\mathbf{z}-\mathbf{x}|=|\mathbf{z}-\mathbf{y}|$, $t$ must be $\pm 1$. If $t=1$, then $\mathbf{x}=\mathbf{y}$, so $t=-1$ and $\mathbf z$ is the midpoint $(\mathbf{x}-\mathbf{y})/2$. (c): $|\mathbf{z}-\mathbf{x}|+|\mathbf{z}-\mathbf{y}|=2r<|\mathbf{x}-\mathbf{y}|$ contradicts Theorem 1.37(f).

Amit Awasthi · Answer

\begin{remark} We will assume, without a loss of generality, that $\mathbf{x} = 0$, $\mathbf{y} = (d,0,\ldots,0)$. This takes care of any situation since we can always redefine our coordinate system such that this is true. \end{remark} \begin{lemma} For any $\mathbf{z}$ such that $|\mathbf{z} - \mathbf{x}| = |\mathbf{z} - \mathbf{y}|$, $z_1 = d/2$. \end{lemma} \begin{proof}[Proof of Lemma] From Definition 1.36, the equation in the exercise text implies, \begin{equation*} \sum_{i=1}^k z_i^2 = \sum_{i=1}^k (z_i-y_i)^2. \end{equation*} For $k=1$, this becomes $z_1^2 = (z_1-d)^2$, and so $z_1 = d/2$. For $k\ge2$, subtracting $(z_2^2+\cdots+z_k^2)$, we get $z_1^2 = (z_1-d)^2$, and so $z_1 = d/2$. \end{proof} \begin{proof}[Proof of $(a)$] If $2r > d$, by \eqref{1.16a}, $r^2 = |\mathbf{z}|^2 > d^2/4$. Then, if $k\ge3$, by Definition 1.36 and the Lemma above, \begin{align}\label{1.16b} r^2 = \sum_{i=1}^k z_i^2 = \frac{d^2}{4} + \sum_{i=2}^k z_i^2 &> \frac{d^2}{4} onumber\ r^2 - \frac{d^2}{4} = \sum_{i=2}^k z_i^2 &> 0, \end{align} which has infinitely many solutions for $\mathbf{z}$. If $k=2$, \eqref{1.16b} becomes $r^2 - d^2/4 = z_2^2 > 0$, which only has two solutions for $\mathbf{z}$, \begin{equation*} \mathbf{z} = \left(\frac{d}{2},\pm\sqrt{r^2 - \frac{d^2}{4}} ight). \end{equation*} If $k=1$, there is no such $\mathbf{z}$ since by the Lemma above, $z_1 = d/2$, and so $|\mathbf{z}| = d/2 ot> d/2$. \end{proof} \begin{proof}[Proof of $(b)$] If $2r = d$, by \eqref{1.16a}, $r^2 = |\mathbf{z}|^2 = d^2/4$. Then, \eqref{1.16b} becomes \begin{equation*} r^2 - \frac{d^2}{4} = \sum_{i=2}^k z_i^2 = 0. \end{equation*} Then, there is only one value of $\mathbf{z}$ for any $k$, which is $\mathbf{z} = (d/2,0,0,\ldots)$. \end{proof} \begin{proof}[Proof of $(c)$] If $2r < d$, we get \begin{equation*} |\mathbf{z} - \mathbf{x}| + |\mathbf{z} - \mathbf{y}| < |\mathbf{x} - \mathbf{y}|, \end{equation*} which contradicts Theorem $1.37(f)$. Therefore, there are no solutions for $\mathbf{z}$ for any $k$. \end{proof}

Exercise 1.16

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Exercise 1.16

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