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What is the least area in the plane required to continuously rotate a needle of unit length and positive thickness $a \in ]0,1[$ around completely (i.e. $360°$)?

The answer is known to be $0$ (we take the infimum) if $a=0$. An obvious and very rough upper bound is $\pi (\sqrt{1+a^2}/2)^2 = \pi(1+a^2)/4$. I've found nothing on this version of Kakeya Needle Problem. If you have any reference on this version, it would be welcome.

Alphonse
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    I remember seeing an amazing paper here about this problem and the Riemann hypothesis... – anomaly Sep 07 '16 at 21:02
  • @anomaly: something like this http://mathoverflow.net/questions/137805/riemann-hypothesis-and-kakeya-needle-problem or http://www.one-zero.eu/resources/Kakeya.pdf "Riemann and Mobius dust the Kakeya Problem"? – Alphonse Sep 07 '16 at 21:04
  • @anomaly:Apparently someone just asked on MO: http://webcache.googleusercontent.com/search?q=cache:pQCNRi62smsJ:mathoverflow.net/questions/247988/the-riemann-hypothesis-proof-and-the-kakeya-needle-problem+&cd=1&hl=fr&ct=clnk&gl=ch&client=safari But this is completely unrelated to my question. – Alphonse Sep 07 '16 at 21:04
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    if you take the needle moving as the chord of a circle of radius $r$, the area is $\pi r^2 - \pi (r-f(r)-a)^2 \approx 2 \pi r (f(r)+a)$ where $f(r) = r (1-\cos(\arcsin(1/2r))) \approx r(1-\cos(1/2r)) \approx \frac{1}{8r}$ – reuns Sep 07 '16 at 21:07
  • It is, in fact, totally unrelated. :) That person pops up here (and on mathoverflow, and various other boards) every couple of weeks with the identical post and paper. – anomaly Sep 07 '16 at 21:09
  • So I think that's how it $\to 0$ when $a \to 0$, but for a fixed $a$ the optimum is rotating it without moving the center (weird ?) – reuns Sep 07 '16 at 21:14

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