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Let $S$ be a smooth, closed surface in $\mathbb{R}^3$, and $\gamma$ a geodesic segment on $S$, i.e., a finite-length piece of a geodesic. Define $\gamma(w)$ as all the points of $S$ within a distance $w$ of $\gamma$: all the $x \in S$ such that the shortest distance on $S$ from $x$ to $\gamma$ is at most $w$. One can imagine $\gamma(w)$ representing the path of a paintbrush of width $2w$ as it traces $\gamma$:
           (Image based on one at

Q1. For a given $S$, which $\gamma(w)$ have the properties that (a) all of $S$ is covered by $\gamma(w)$, and (b) the area product $\;|\gamma| \cdot w$ is minimized, where $|\gamma|$ is the length of $\gamma$.

In some sense, this is an optimal paint path: a paintbrush tracing $\gamma$ and spreading paint $\pm w$ covers the surface most efficiently. For example, for $S$ a unit-radius sphere, it seems that $\gamma$ a half-great circle, $|\gamma| = \pi$ is optimal with $w=\pi/2$, area product $\frac{1}{2} \pi^2$.

Q2. Is the semi-great circle indeed optimal for the sphere?

Q3. Are there other clear examples of optimal paintbrush geodesics?

Q4. Has this notion been studied before, perhaps in another guise?

Thanks for ideas/pointers!

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How about $w=\pi$ and arbitrarily short $\gamma$? I think in this case $\gamma(w)$ covers the sphere. – Sergei Ivanov May 5 '13 at 16:29
@Sergei: Ah, very nice! – Joseph O'Rourke May 5 '13 at 16:36
In fact this works for every compact manifold, no? – Will Sawin May 5 '13 at 16:46
@Will: Yes, I think so! Which is why this concept has not been studied---It's essentially empty. :-) – Joseph O'Rourke May 5 '13 at 16:50
A natural way to avoid the trivial answer is to assume that the normal injectivity radius of $\gamma$ is at least $w$. This is consistent with the actual paintbrushes. – Misha May 5 '13 at 17:19

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