= Travel-time minimization with speed proportional to squared radius
For speed $v=kr^2$ in the punctured plane, the travel-time element is $ds/(kr^2)$. Under the <inversion in a circle> with radial coordinate $u=1/r$, this becomes $k^{-1}\sqrt{du^2+u^2d\theta^2}$, 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 $\sqrt2/(ka)$. The <triangle inequality> for <arc length> proves global minimality.
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