Exercise 1.19

Question

\def\reals{\mathbb{R}}
Exercise 19:
  Suppose $\mathbf{a}\in\reals^k$, $\mathbf{b}\in\reals^k$. Find $\mathbf{c}\in\reals^k$ and $r>0$ such that $$|\mathbf{x}-\mathbf{a}|=2|\mathbf{x}-\mathbf{b}|$$ if and only if
$|\mathbf{x}-\mathbf{c}|=r$.

analambanomenos · Accepted Answer

\begin{align*}
	4|\mathbf{x}-\mathbf{b}|^2 &= |\mathbf{x}-\mathbf{a}|^2 \
	4|\mathbf{x}|^2-8\mathbf{x}\cdot\mathbf{b}+4|\mathbf{b}|^2 &= |\mathbf{x}|^2-2\mathbf{x}\cdot\mathbf{a}+|\mathbf{a}|^2 \
	|\mathbf{x}|^2-2\mathbf{x}\cdot\bigl((1/3)(4\mathbf{b}-\mathbf{a})\bigr) &= (1/3)|\mathbf{a}|^2-(4/3)|\mathbf{b}|^2 \
	|\mathbf{x}|^2-2\mathbf{x}\cdot\bigl((1/3)(4\mathbf{b}-\mathbf{a})\bigr) + \big|(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/3)|\mathbf{a}|^2 - (4/3)|\mathbf{b}|^2 + \big|(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 \
	\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/9)(3|\mathbf{a}|^2-12\mathbf{b}^2+|\mathbf{a}|^2-8\mathbf{a}\cdot\mathbf{b}+16|\mathbf{b}|^2) \
	\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/9)(4|\mathbf{a}|^2 - 8\mathbf{a}\cdot\mathbf{b} + 4|\mathbf{b}|^2) \
	\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (4/9)\big|\mathbf{a}-\mathbf{b}\big|^2
\end{align*}
This describes a sphere with center $(1/3)(4\mathbf{b}-\mathbf{a})$ and radius $(2/3)|\mathbf{a}-\mathbf{b}|$.

Amit Awasthi · Answer

\begin{proof}
      First prove $|\mathbf{x} - \mathbf{a}| = 2|\mathbf{x} - \mathbf{b}|$ implies $|\mathbf{x} - \mathbf{c}| = r$. Squaring both sides of the former, we obtain $|\mathbf{x} - \mathbf{a}|^2 = 4|\mathbf{x} - \mathbf{b}|^2$. By Definition 1.36, we get
      \begin{gather*}
        \sum_{i=1}^k (x_i-a_i)^2 = 4\sum_{i=1}^k (x_i-b_i)^2.
\end{gather*}
Solving for $\Sigma x_i$,
\begin{gather}
        \sum_{i=1}^k x_i^2 = \frac{1}{3} \sum_{i=1}^k
        \left[\begin{aligned}
          2x_i&(-a_i + 4b_i)\
          &+(a_i^2 - 4b_i^2)
        \end{aligned}ight].\label{1.19a}
      \end{gather}
      Similarly, for $|\mathbf{x} - \mathbf{c}| = r$,
      \begin{gather*}
        \sum_{i=1}^k (x_i-c_i)^2 = r^2.
\end{gather*}
Solving for $\Sigma x_i$,
\begin{gather}
        \sum_{i=1}^k x_i^2 = r^2 + \sum_{i=1}^k (2c_ix_i - c_i^2).\label{1.19b}
      \end{gather}
      Setting \eqref{1.19a} and \eqref{1.19b} equal to each other, and solving for $r^2$, we get
      \begin{equation}
        r^2 = \frac{1}{3} \sum_{i=1}^k \left[
        \begin{aligned}
          2x_i&(- a_i + 4b_i - 3c_i)\
          &+(a_i^2 - 4b_i^2 + 3c_i^2)
        \end{aligned}ight].\label{1.19c}
      \end{equation}
      Since we want $r$ to be independent of $\mathbf{x}$, we set $(- a_i + 4b_i - 3c_i) = 0$ for all $i \le k$, that is $3\mathbf{c} = 4\mathbf{b}-\mathbf{a}$. Then \eqref{1.19c} becomes
      \begin{equation*}
        r^2 = \sum_{i=1}^k \left( \frac{1}{3}a_i^2 - \frac{4}{3}b_i^2 + c_i^2 ight).
      \end{equation*}
      By the construction of $\mathbf{c}$,
      \begin{align*}
        r^2 &= \sum_{i=1}^k \left[ \frac{1}{3}a_i^2 - \frac{4}{3}b_i^2 + \left( \frac{4}{3} b_i - \frac{1}{3}a_i ight)^2 ight]\
        &= \sum_{i=1}^k \left( \frac{4}{9}a_i^2 - \frac{8}{9} a_ib_i + \frac{4}{9}b_i^2 ight)\
        3r &= 2\Bigg(\sum_{i=1}^k (a_i-b_i)^2\Bigg)^{1/2} = 2|\mathbf{a}-\mathbf{b}|,
      \end{align*}
      by Definition 1.36, and we get $3r = 2|\mathbf{a}-\mathbf{b}|$.
      \par Now prove the converse, i.e. that $|\mathbf{x} - \mathbf{c}| = r$ implies $|\mathbf{x} - \mathbf{a}| = 2|\mathbf{x} - \mathbf{b}|$. By our construction of $\mathbf{c}$ and $r$ above, we get
      \begin{equation*}
        |3\mathbf{x} - 4\mathbf{b} + \mathbf{a}| = 2|\mathbf{a}-\mathbf{b}| = |2\mathbf{a}-2\mathbf{b}|,
      \end{equation*}
      by Theorem $1.33(c)$. Squaring both sides and applying Definition 1.36,
      \begin{align*}
        \sum_{i=1}^k (3x_i-4b_i+a_i)^2 &= 4\sum_{i=1}^k (a_i-b_i)^2\
        \sum_{i=1}^k (3x_i-4b_i+a_i)^2 &= \sum_{i=1}^k (2a_i-2b_i)^2.
      \end{align*}
      Since this is a difference of squares,
      \begin{align*}
        0 &= \sum_{i=1}^k \left[
        \begin{aligned}
          &(3x_i-6b_i+3a_i)\
          &\cdot(3x_i-2b_i-a_i)
        \end{aligned} ight]\
        &= \sum_{i=1}^k \left[
        \begin{aligned}
          &(x_i-2b_i+a_i)\
          &\cdot(3x_i-2b_i-a_i)
        \end{aligned} ight]\
        &= \sum_{i=1}^k \left[
        \begin{aligned}
          &4(x_i^2 - 2b_ix_i - b_i^2)\
          &- (x_i^2 - 2a_ix_i + a_i^2)
        \end{aligned} ight]\
        &= \sum_{i=1}^k \left[ 4(x-b)^2 - (x-a)^2 ight].
      \end{align*}
      And so,
      \begin{align*}
        \sum_{i=1}^k (x-a)^2 &= \sum_{i=1}^k 4(x-b)^2 \
        |\mathbf{x}-\mathbf{a}| &= 2|\mathbf{x}-\mathbf{b}|.
      \end{align*}
      by taking the squareroot for both sides and applying Definition 1.36.
    \end{proof}

Exercise 1.19

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Exercise 1.19

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