Travel-time minimization with speed proportional to squared radius

For speed $v=k{r}^2$ in the punctured plane, the travel-time element is $ds/(k{r}^2)$. Under the inversion in a circle with radial coordinate $u=1/r$, this becomes $k^{-1}\sqrt{d{u}^2+u^{2}d{\theta }^2}$, the arc length in Euclidean space divided by $k$. Thus a shortest travel-time route corresponds to a straight line segment in the inverted plane, whenever that segment avoids the origin and lies in the allowed region. Between $(a,0)$ and $(0,a)$ the inverted segment has equation $u(\cos \theta +\sin \theta )=1/a$. The route is $r=a(\cos \theta +\sin \theta )$, and its minimum travel time is $\sqrt{2}/(ka)$. The triangle inequality for arc length proves global minimality.