Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 1, Problem 16

Problem

Suppose $k \geq 3,\mathbf{x},\mathbf{y} \in \mathbb{R}^k, |\mathbf{x} - \mathbf{y}|=d>0$, and $r>0$. Prove:

(a) If $2r > d$, there are infinitely many $\mathbf{z} \in \mathbb{R}^k$ such that

$$ |\mathbf{z} - \mathbf{x}| = |\mathbf{z} - \mathbf{y}| = r.$$

(b) If $2r = d$, there is exactly one such $\mathbf{z}$.

(c) If $2r < d$, there is no such $\mathbf{z}$.

Answer

(a) Consider $\mathbf{w} \in \mathbb{R}^k$ such that $|\mathbf{w}|^2=r^2-d^2/4$ and $\mathbf{w} \cdot (\mathbf{x} - \mathbf{y}) = 0$. Then, the components of $\mathbf{w}=(w_1,\cdots,w_k)$ satisfy the following equations.

$$w_1(x_1 - y_1) + \cdots + w_k(x_k - y_k) = 0,$$

$$w_1^2 + \cdots + w_k^2 = r^2 - d^2/4.$$

For $k \geq 3$ and $2r > d$, there are infinitely many $w_1,\cdots,w_k$ satisfying the above. For every $\mathbf{w}$ satisfying the above, $\mathbf{z} = \frac{\mathbf{x} + \mathbf{y}}{2} + \mathbf{w}$ satisfies

$$ |\mathbf{z} - \mathbf{x}| = |\mathbf{z} - \mathbf{y}| = r.$$

Hence, there there are infinitely many such $\mathbf{z}$.

(b) Since $ |\mathbf{z} - \mathbf{x}| = |\mathbf{z} - \mathbf{y}| = d/2$, the equality holds in the triangle inequality

$$ |\mathbf{z} - \mathbf{x}| + |\mathbf{z} - \mathbf{y}| \geq |\mathbf{x} - \mathbf{y}|, $$

which implies

$$ \mathbf{x} - \mathbf{y} = \alpha (\mathbf{z} - \mathbf{x}) = \beta (\mathbf{z} - \mathbf{y}). $$

In addition, we have $|\alpha|=|\beta|=1/2$. If $\alpha=\beta$, then we have $\mathbf{x}=\mathbf{y}$, which is contradiction; hence, $\alpha=-\beta$. We can conclude that $\mathbf{z}=(\mathbf{x} + \mathbf{y})/2$ is the only solution.

(c) If $2r < d$, the following triangle inequality cannot be satisfied.

$$ |\mathbf{z} - \mathbf{x}| + |\mathbf{z} - \mathbf{y}| \geq |\mathbf{x} - \mathbf{y}| $$

Hence, there is no such $\mathbf{z}$.

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