How to show the wheel metric is a metic space

Question

How to show the triangle inequality for the wheel metric when not all points lie on the same line?`

For $x, y \in\Bbb R^2$,

$$d(x,y)=\begin{cases} d_{\operatorname{Euc}}(x,y),&\text{if }x\text{ and }y\text{ lie on a common}\\ &\text{line through the origin}\\ d_{\operatorname{Euc}}(x,0)+d_{\operatorname{Euc}}(y,0),&\text{otherwise.} \end{cases}$$

$d_{\operatorname{Euc}}$ is the Euclidean distance.

Answer 1

Suppose that $p,q$, and $s$ are three points in $\Bbb R^2$ such that $p$ and $q$ are not on the same line through the origin. Then in polar coordinates we can write $p=\langle r_p,\theta_p\rangle$, $q=\langle r_q,\theta_q\rangle$, and $s=\langle r_s,\theta_s\rangle$, where $0\le\theta_p,\theta_q,\theta_r<\pi$, and the radial coordinates $r_p,r_q$, and $r_s$ may be negative. (This is so that points on the same line through the origin will have the same angular coordinate.) Moreover, $\theta_p\ne\theta_q$, $d(p,q)=|r_p|+|r_q|$, and we want to show that this is at most $d(p,s)+p(s,q)$. There are three cases, two of which are very similar.

• The first is easy. Suppose that $s$ is not on the same line through the origin as either $p$ or $q$, so that $\theta_p\ne\theta_s\ne\theta_q$. Then what are $d(p,s)$ and $d(s,q)$? What is their sum? How does it compare with $d(p,s)+p(s,q)$?
• Now suppose that $s$ is on the same line through the origin as $p$, so that $\theta_s=\theta_p$. Then $d(p,s)=|r_p-r_s|$, and $d(s,q)=|r_s|+|r_q|$. Can you finish off this case? Note that $|r_p|=|(r_p-r_s)+r_s|\le|r_q-r_s|+|r_s|$.
• The third case is of course when $s$ is on the same line through the origin as $q$, and it’s exactly like the second case.